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Unbroken vs smooth
A graph you can draw without lifting the pen. That is what continuous will mean.
Think about drawing a graph in your notebook. For most graphs you know, the pencil moves in one sweep. You never lift it off the page. Put simply: a graph you can draw without lifting the pen from the paper.
The courier’s card: 40 rupees up to 500 g, 60 rupees just over it. The pen must jump at 500 g.
Now think about a courier’s rate card. A parcel up to 500 grams costs 40 rupees. A parcel just over 500 grams costs 60 rupees. Draw the charge against the weight. At 500 grams the graph jumps up, and you have to lift the pen.
The V of |x| has no jump, but its slope is −1 on the left and 1 on the right. At 0 it turns.
There is a third kind of graph. In Class 11 you drew $y = \left|x\right|$, the V shape. You can draw it without lifting the pen. But at 0 it turns a sharp corner. Coming down the left arm, the slope is $-1$. Going up the right arm, it is $1$. At the corner itself there is no single slope.
One graph, two failures: a jump of 1.25 at x equals 2, and a corner at x equals 4.5 with slopes minus 0.9 and 0.9.
Naming continuous and differentiable
So a graph can fail at a point in two ways. It can jump, or it can turn a corner. A function with no jump at a point is CONTINUOUS there. A function with one slope at a point is DIFFERENTIABLE there. The V has no jump at 0, but its corner leaves it with two slopes.
x² at 1
the card at 500 g
|x| at 0
Three graphs, three points. Which can be drawn through without lifting the pen, and which have one slope there?
Class 11 gave you the derivative as a limit. It also gave you the slopes of $x^n$, $\sin x$ and $\cos x$. Here is the question to carry to the last page. Which graphs can you draw without lifting the pen? Which of those have one slope at every point? And how do you find the slope of the new functions of Class 12?
Four new kinds of function: a composite, an inverse, eˣ with ln x, and a curve given by one equation.
Those new functions come in many kinds. Some are a function inside a function. Some are inverse trig functions, or $e^x$ and $\ln x$. Some curves are given by an equation in $x$ and $y$, or by a third variable. By the end you will have one method that reaches all of them.
A SCHOLAR INDIA REMEMBERS
Aryabhata wrote the Aryabhatiya at Kusumapura, near modern Patna, in 499 CE. He was $23$. It set out numbers in place value, and one general method for a whole class of equations, checked against every case, not just one. Bring that same care to a graph. Come close to a point from the left, then from the right. Do both sides close in on one height, the height at the point? Then the pen need not lift there.
Aryabhata · the hoopoe
Fold the curve at x = 2, and see what breaks: only the slope, not the function.
Check yourself
What is the central lesson this chapter draws out through the continuity/differentiability distinction?
continuity and differentiability never occur together at the same point
a function can be continuous without being differentiable, though differentiable implies continuous
a function can be differentiable at a point without being continuous there
continuity and differentiability are two names for the same property, so establishing one establishes the other
Check your answer
continuity and differentiability never occur together at the same point — The two properties are not opposites: every function is continuous at every point where it is differentiable. What varies is only whether continuity can hold without differentiability.
✓ a function can be continuous without being differentiable, though differentiable implies continuous — (B) Continuity (unbroken) and differentiability (smooth) are not the same, continuity can hold at a point without differentiability holding there too, even though the reverse direction, differentiable implies continuous, always holds.
a function can be differentiable at a point without being continuous there — This reverses the one implication that always holds. Differentiable forces continuous, never the reverse, so a differentiable-but-not-continuous function cannot occur.
continuity and differentiability are two names for the same property, so establishing one establishes the other — It is tempting to treat “smooth” and “unbroken” as the same idea, since a differentiable graph is also unbroken, but continuity and differentiability are two separate properties that split apart, with continuity the weaker, more common condition.
Of the two directions between continuity and differentiability at a point, which one always holds?
differentiable at a point always implies continuous there
continuous at a point always implies differentiable there
neither direction ever holds without additional conditions
both directions hold simultaneously, so the two properties always occur together
Check your answer
✓ differentiable at a point always implies continuous there — (A) Differentiable at a point always implies continuous there, this is the one direction that always holds, with no extra conditions attached.
continuous at a point always implies differentiable there — This reverses the fixed direction. Continuity is the weaker, more common condition, a function can be continuous with a sharp corner, so continuity cannot force smoothness.
neither direction ever holds without additional conditions — Exactly one direction is always true, with no extra conditions attached, differentiable implies continuous. Denying it removes the one settled fact between these two properties.
both directions hold simultaneously, so the two properties always occur together — If the two properties always occurred together, there would be no distinction left to draw, but a continuous function can fail to be differentiable, so they do not always pair up.
How do these later tools (the chain rule, implicit and logarithmic differentiation, parametric and second-order derivatives, Rolle’s theorem, and the Mean Value Theorem) relate to the continuity/differentiability distinction set up earlier?
it replaces the continuity/differentiability distinction, making the earlier definitions unnecessary once the chain rule is introduced
it is required in order to first establish that differentiable implies continuous, so the distinction depends on the machinery
it is built on top of the settled continuity/differentiability foundation, using it without re-deriving it
it applies only to functions that are continuous but not differentiable, since differentiable functions need no further machinery
Check your answer
it replaces the continuity/differentiability distinction, making the earlier definitions unnecessary once the chain rule is introduced — The chain rule and the rest do not retire the continuity/differentiability distinction, they assume it. Nothing in the later chapter reopens whether differentiable implies continuous.
it is required in order to first establish that differentiable implies continuous, so the distinction depends on the machinery — This has the dependency backwards. The continuity/differentiability relationship is settled first, as a foundation, and the chain rule, implicit and logarithmic differentiation, and the rest are built afterward, on top of it.
✓ it is built on top of the settled continuity/differentiability foundation, using it without re-deriving it — (C) It is built on top of the settled continuity/differentiability foundation, using that relationship without re-deriving it each time.
it applies only to functions that are continuous but not differentiable, since differentiable functions need no further machinery — The later machinery is used constantly on functions that are differentiable, the chain rule, for instance, differentiates a composite of differentiable functions. It is not reserved for continuous-but-not-differentiable cases.
You have already made a check like this, long before you met the word continuity. Try each one below.
Left-hand and right-hand limit (Class 11, Mathematics (NCERT Class 11), page 221). Here: This same left and right split decides continuity here, checked against the value of the function at that point. Check: let $f(x)=x+1$ for $x<2$ and $f(x)=2x+1$ for $x \geq 2$. At $x=2$, the left-hand limit is $2+1=3$ and the right-hand limit is $2(2)+1=5$.
Derivative by first principles (Class 11, Mathematics (NCERT Class 11), page 242). Here: This same first-principle limit, split left and right, is the LHD and RHD test used here. Check: for $f(x)=x^2$ at $x=3$ with step $h=1$, the quotient $(f(3+h)-f(3))/h$ equals $(16-9)/1=7$, matching the formula $6+h=7$.
The modulus function (Class 11, Mathematics (NCERT Class 11), page 35). Here: That corner at zero is exactly where the one-sided derivative test here first finds a function that fails it. Check: $\left|-5\right| = 5$ and $\left|5\right| = 5$, the same output from two different inputs.
Algebra of real functions (Class 11, Mathematics (NCERT Class 11), page 36). Here: Adding, multiplying and dividing functions already follow this rule. Here the question becomes whether continuity itself survives each combination. Check: for $f(x)=x^2$ and $g(x)=x+1$, $(f+g)(2) = 4+3 = 7$.
The sine-over-x limit (Class 11, Mathematics (NCERT Class 11), page 235). Here: That same limit is the hidden reason the chain-rule example here can differentiate a sine of an inner function. Check: at $x=0.01$, $\sin(0.01)/0.01 \approx 0.99998$, within $0.001$ of $1$.
If any of these felt new, read the page named before going on.
A courier counter. Up to 500 g costs 40 rupees, anything over costs 60.
40 rupees
60 rupees
499
499.9
500.1
501
499 g and 499.9 g cost 40 rupees; 500.1 g and 501 g cost 60. However close you come, the two sides stay 20 apart.
Go back to the courier’s rate card. Weigh a parcel of 499 grams, then one of 499.9 grams. The charge stays at 40 rupees. Now come from above, with 501 grams and then 500.1 grams. The charge stays at 60 rupees. From the left the charge is close to 40, and from the right it is close to 60. The two sides never meet, so the graph jumps at 500.
From the left the price settles on 40, from the right on 60. Those are the left-hand and right-hand limits.
You met this idea in Class 11. The number a function gets close to from the left is its LEFT-HAND LIMIT, or LHL. The number it gets close to from the right is its RIGHT-HAND LIMIT, or RHL. At $x = a$ they are written $\lim_{x \to a^-} f(x)$ and $\lim_{x \to a^+} f(x)$.
Arrows from the left and the right approach x equals 2 along the curve. LHL, RHL and f of 2 all equal 1.8.
A function $f$ is CONTINUOUS at $x = a$ when three things hold.
$f(a)$ exists, so $f$ has a value at $a$.
The LHL and the RHL both exist and are equal.
That common limit equals $f(a)$.
In short, $\text{LHL} = \text{RHL} = f(a)$. On the rate card, the LHL at 500 is 40 and the RHL is 60. They differ, so the charge is not continuous at 500.
Check y = 2x + 1 at 3: from both sides the values close on 7, and f(3) = 7. Continuous at 3.
Let us check a function that passes. Take $f(x) = 2x + 1$ at $x = 3$. From either side, $2x + 1$ gets close to 7. And $f(3) = 7$. All three numbers agree, so $f$ is continuous at 3.
Three breaks at x equals a: the sides disagree, one side escapes with no number to settle on, and the value misses the shared limit.
A function can fail at a point in three ways. The two sides can get close to different numbers. One side can fail to settle on any number at all. Or the two sides can agree with each other but not with $f(a)$. Each one is a break in the graph.
Continuous at every point of its domain: test any point of y = x² − 2x and the three numbers agree.
A function that is continuous at every point where it is defined is a CONTINUOUS FUNCTION. You test it one point at a time, with the same three numbers.
Your turn: take $f(x) = x + 1$ for $x \leq 2$ and $f(x) = x + 3$ for $x > 2$. Find the LHL and the RHL at 2. Is $f$ continuous at 2?
Check yourself
What three quantities must all agree for $f$ to be continuous at $x=a$?
$\lim_{x \to a^-} f(x)$, $\lim_{x \to a^+} f(x)$, and $f(a)$
$f(a)$, the first derivative of $f$ at $a$, and the second derivative of $f$ at $a$
$\lim_{x \to a^-} f(x)$ and $\lim_{x \to a^+} f(x)$ only, regardless of $f(a)$
the left-hand derivative, the right-hand derivative, and $f(a)$
Check your answer
✓ $\lim_{x \to a^-} f(x)$, $\lim_{x \to a^+} f(x)$, and $f(a)$ — (A) $\lim_{x \to a^-} f(x)$, $\lim_{x \to a^+} f(x)$, and $f(a)$ must all agree, the left-hand limit, the right-hand limit, and the function’s actual value at the point.
$f(a)$, the first derivative of $f$ at $a$, and the second derivative of $f$ at $a$ — the first and second derivatives of $f$ at $a$ belong to differentiability, a separate and stronger property. Continuity at a point never requires any derivative to exist.
$\lim_{x \to a^-} f(x)$ and $\lim_{x \to a^+} f(x)$ only, regardless of $f(a)$ — Two limits agreeing with each other is not enough on its own, a function can approach the same value from both sides while $f(a)$ itself is a different number, or undefined. The comparison with $f(a)$ is the third, necessary piece.
the left-hand derivative, the right-hand derivative, and $f(a)$ — The left-hand and right-hand derivatives (LHD/RHD) measure slope and belong to differentiability, a separate property from continuity. Continuity is tested by the left-hand and right-hand limits (LHL/RHL), which measure function value, not slope.
The left-hand limit (LHL) and right-hand limit (RHL) used to test continuity at $x=a$ are measuring what, exactly?
the rate at which $f(a)$ itself changes
whether $f$ is defined at points other than $a$
how the value behaves as $x$ approaches $a$ from each side
the slope of the tangent line as $x$ approaches $a$ from each side
Check your answer
the rate at which $f(a)$ itself changes — $f(a)$ is a single fixed number, not something that changes over an interval, so there is no rate to measure. LHL and RHL instead track what $f(x)$ approaches as $x$ moves toward $a$.
whether $f$ is defined at points other than $a$ — LHL and RHL are about behavior as $x$ approaches $a$, not about which points besides $a$ are in the domain, a function can be defined everywhere nearby and still have LHL and RHL disagree.
✓ how the value behaves as $x$ approaches $a$ from each side — (C) They measure how the function value $f(x)$ behaves as $x$ approaches $a$ from each side, LHL and RHL are the same two-sided limit split by direction of approach.
the slope of the tangent line as $x$ approaches $a$ from each side — Slope belongs to the left-hand and right-hand derivatives, LHD/RHD. LHL and RHL are limits of the function’s value, not of any rate of change.
Which of the following describes a genuine way $f$ can fail to be continuous at $x=a$?
$f$ is discontinuous only when $f(a)$ is undefined, never when it is defined but mismatched
LHL and RHL both exist and agree with each other, but disagree with $f(a)$
LHL and RHL both exist, agree with each other, and agree with $f(a)$
$f$ is discontinuous whenever $f(a)$ is defined, regardless of the two one-sided limits
Check your answer
$f$ is discontinuous only when $f(a)$ is undefined, never when it is defined but mismatched — Where $f(a)$ is undefined, $a$ is not in the domain, and continuity is not asked there. At a point of the domain, continuity can still fail when the two limits agree with each other but not with $f(a)$.
✓ LHL and RHL both exist and agree with each other, but disagree with $f(a)$ — (B) LHL and RHL can both exist and agree with each other, yet disagree with $f(a)$, the two limits settle on one value while the function’s actual value at $a$ is something else, so continuity fails on the third condition even though the first two hold.
LHL and RHL both exist, agree with each other, and agree with $f(a)$ — This describes the three-way agreement that makes $f$ continuous at $a$, the opposite of what the question asks for, it is not a way discontinuity can occur.
$f$ is discontinuous whenever $f(a)$ is defined, regardless of the two one-sided limits — Whether $f(a)$ is defined says nothing on its own about continuity, the one-sided limits still have to be checked and compared against it.
If $\lim_{x \to a^-} f(x) = \lim_{x \to a^+} f(x) = L$, what can be concluded from this alone, before checking $f(a)$?
$f$ is differentiable at $x=a$
$\lim_{x \to a} f(x)$ exists and equals $L$
$f(a) = L$
$f$ is continuous at $x=a$, since the limit from both sides exists
Check your answer
$f$ is differentiable at $x=a$ — Differentiability is a separate, stronger property tested by a different limit, the difference quotient, not by whether LHL and RHL agree. Nothing here speaks to slope at all.
✓ $\lim_{x \to a} f(x)$ exists and equals $L$ — (B) $\lim_{x \to a} f(x)$ exists and equals $L$, LHL and RHL are the same two-sided limit split by direction, so their agreement establishes that the two-sided limit exists and what it equals.
$f(a) = L$ — The limit’s existence says what $f(x)$ approaches as $x$ nears $a$; it says nothing about what $f(a)$ actually is. The two can disagree even when the limit exists.
$f$ is continuous at $x=a$, since the limit from both sides exists — LHL and RHL agreeing establishes that the two-sided limit exists, but continuity needs a third piece checked too, whether $f(a)$ also equals that same value. Nothing here confirms that yet.
Join two continuous functions by +, −, × or ÷ and the result is continuous, except where you divide by 0.
If $f$ and $g$ are both continuous at $x = a$, then so are $f + g$, $f - g$ and $f g$. The quotient $f/g$ is continuous at $a$ too, but only when $g(a) \neq 0$. Where $g(a) = 0$, the quotient has no value at all.
See why with two straight lines. In the figure, $f$ is $x + 1$ and $g$ is $x - 2$. At every $x$, the height of $f + g$ is the height of $f$ plus the height of $g$. At $x = 1$ that is $2 + (-1) = 1$. Neither line has a break, so adding their heights cannot make one. The product works the same way: at $x = 1$, $f g$ is $2 \times (-1) = -2$, and neither factor breaks there either.
at 0.99 too: 1.99 + (−1.01) = 0.98
f + g adds the heights: at 1, 2 + (−1) = 1. No break in f or g, so none in their sum.
These rules come from the limit rules you used in Class 11. The limit of a sum is the sum of the limits. So as $x$ gets close to $a$, $f(x) + g(x)$ gets close to $f(a) + g(a)$. That is the value of $f + g$ at $a$. Differences, products and quotients work the same way.
3
→
x²
→
√
→
3
3.01
→
x²
→
√
→
3.01
Two functions in a row, each continuous. A small change going in, 3 to 3.01, gives a small change coming out.
Two functions in a row keep continuity too. If $g$ is continuous at $a$, and $f$ is continuous at $g(a)$, then $f(g(x))$ is continuous at $a$. Notice the two points. You check $g$ at $a$, but you check $f$ at $g(a)$, the number $g$ puts out.
Every polynomial is continuous everywhere. So are sin x and cos x.
Now see what this gives you. The function $x$ is continuous everywhere, and so is any constant. Every polynomial is built from these by adding and multiplying. So every polynomial is continuous at every real number. $\sin x$ and $\cos x$ are continuous everywhere too.
y = 1/(x − 2) is continuous at every point where it has a value. At x = 2 it has none.
domain of 1/(x − 2)
Its domain is every real number except 2, and it is continuous on all of it.
Try the quotient rule on $1/(x - 2)$. The top, 1, is continuous everywhere. The bottom, $x - 2$, is 0 only at 2. So the quotient is continuous at every real number except 2. At 2 it has no value.
These rules save you work. For most functions you meet, you can say where they are continuous without working out a single limit.
Your turn: where is $(x^2 + 1)/(x^2 - 9)$ continuous?
Check yourself
If $f$ and $g$ are both continuous at $x=a$, what condition must additionally hold for $f+g$, $f-g$, and $f \cdot g$ to be continuous at $a$?
no extra condition is needed
$f(a) = g(a)$
$f$ and $g$ must both be polynomials
$g(a) \neq 0$
Check your answer
✓ no extra condition is needed — (A) No extra condition is needed, continuity of $f$ and $g$ at $a$ is enough on its own for $f+g$, $f-g$, and $f \cdot g$ to be continuous at $a$ too.
$f(a) = g(a)$ — Nothing requires $f(a)$ and $g(a)$ to match, $f+g$, $f-g$, and $f \cdot g$ stay continuous at $a$ for any values $f(a)$ and $g(a)$ happen to take.
$f$ and $g$ must both be polynomials — The rule holds for any pair of functions continuous at $a$, polynomial or not, it is the continuity of $f$ and $g$ that matters, not what kind of formula defines them.
$g(a) \neq 0$ — $g(a) \neq 0$ is the condition attached to the quotient $f/g$, where a zero denominator is a real hazard. Sums, differences, and products have no denominator, so that hazard never arises.
A student claims that if $f$ and $g$ are continuous at $x=a$, then $f/g$ must also be continuous at $a$, with no extra condition needed. What is wrong with this claim?
the claim omits the condition $f(a) \neq 0$ instead of a condition on $g$
nothing is wrong: the claim is correct as stated
the claim omits $g(a) \neq 0$; a quotient is undefined where $g(a)=0$
the claim is wrong because $f/g$ is never continuous, even when $g(a) \neq 0$
Check your answer
the claim omits the condition $f(a) \neq 0$ instead of a condition on $g$ — It is the denominator’s value, $g(a)$, that creates the hazard of division by zero, the numerator’s value $f(a)$ places no such restriction on the quotient’s continuity.
nothing is wrong: the claim is correct as stated — The claim drops exactly the condition that makes it false in general, continuity of $f$ and $g$ alone is not enough for $f/g$ once $g(a)$ can be zero.
✓ the claim omits $g(a) \neq 0$; a quotient is undefined where $g(a)=0$ — (C) The claim omits the condition $g(a) \neq 0$. Continuity of a quotient is never automatic at a zero of the denominator, $f/g$ is continuous at $a$ only once that condition is added.
the claim is wrong because $f/g$ is never continuous, even when $g(a) \neq 0$ — When $g(a) \neq 0$, $f/g$ is continuous at $a$, the rule is conditional, not a blanket denial. The claim’s error is omitting the condition, not the conclusion being impossible.
Which condition guarantees that the composite $f(g(x))$ is continuous at $x=a$?
$f$ continuous at $a$ only; continuity of $g$ is not required
$g$ continuous at $g(a)$ and $f$ continuous at $a$
$g$ continuous at $a$ and $f$ continuous at $g(a)$
$g$ continuous at $a$ and $f$ continuous at $a$
Check your answer
$f$ continuous at $a$ only; continuity of $g$ is not required — Without $g$’s continuity there is no guarantee: if $g$ jumps at $a$, $f(g(x))$ can jump too, even when $f$ is continuous everywhere. The theorem asks for both.
$g$ continuous at $g(a)$ and $f$ continuous at $a$ — The roles are swapped, it is $g$ that must be continuous at the input point $a$, and $f$ that must be continuous at the output point $g(a)$, not the other way round.
✓ $g$ continuous at $a$ and $f$ continuous at $g(a)$ — (C) If $g$ is continuous at $a$ and $f$ is continuous at $g(a)$, the composite $f(g(x))$ is continuous at $a$. $g$ is checked at the input point, and $f$ at the point $g$ actually outputs.
$g$ continuous at $a$ and $f$ continuous at $a$ — $f$ has to be continuous at the point $g$ actually outputs, $g(a)$, not at $a$ itself, which is where $g$’s input lives, not where $f$’s input lands.
Using only the sum and product rules, and the fact that constant and identity functions are continuous everywhere, why is every polynomial continuous at every real number?
a polynomial is built from finite sums and products of continuous functions, and both preserve continuity
a polynomial is continuous everywhere because it is always differentiable, and continuity is a separate, unrelated fact
a polynomial is continuous only where it does not equal zero, matching the quotient rule’s condition
a polynomial is continuous only at the points where all of its terms individually equal zero
Check your answer
✓ a polynomial is built from finite sums and products of continuous functions, and both preserve continuity — (A) A polynomial is built from finite sums and products of continuous constant and identity functions, and both operations preserve continuity with no extra condition, so the whole polynomial inherits continuity everywhere, term by term.
a polynomial is continuous everywhere because it is always differentiable, and continuity is a separate, unrelated fact — Differentiability is a true fact about polynomials, but it is not the reason cited here. The sum and product rules for continuity are what stack a polynomial’s continuity up term by term, independent of any differentiability argument.
a polynomial is continuous only where it does not equal zero, matching the quotient rule’s condition — A polynomial has no denominator, since it is built purely from sums and products, so the quotient rule’s $g(a) \neq 0$ condition has nothing to attach to here.
a polynomial is continuous only at the points where all of its terms individually equal zero — No term of a polynomial has to equal zero anywhere for the whole to be continuous, the sum and product rules apply regardless of what value each term takes.
The pieces x squared and 2x minus 1 meet at (1, 1). LHL, RHL and f of 1 are all 1.
Close up at 1: 0.99 gives 0.9801 and 1.01 gives 1.02. Both sides close on f(1) = 1.
Worked example
Test continuity of a piecewise function at $x = 1$
$f(x) = x^2$ for $x \leq 1$, and $f(x) = 2x - 1$ for $x > 1$ Two rules meet at 1. Each piece is a polynomial, so 1 is the only point to test.
At $x=1$: $\text{LHL} = \lim_{x \to 1^-} x^2 = 1$ To the left of 1 the rule is $x^2$. From below, $x^2$ gets close to 1.
At $x=1$: $\text{RHL} = \lim_{x \to 1^+} (2x-1) = 2(1)-1 = 1$ To the right of 1 the rule is $2x - 1$. From above, it gets close to 1 as well.
$f(1) = 1^2 = 1$ Now the value itself. The point 1 belongs to the first piece, whose rule says $x \leq 1$.
$\text{LHL} = \text{RHL} = f(1) = 1$ All three numbers are 1. So $f$ is continuous at 1.
$f(0.99) = 0.9801$ and $f(1.01) = 2(1.01) - 1 = 1.02$ Check with inputs near 1. Both values are near 1, and they get nearer as you move closer.
Check yourself
For $f(x) = x^2$ when $x \leq 1$ and $f(x) = 2x - 1$ when $x > 1$, what is $\lim_{x \to 1^-} f(x)$?
$1$, from $x^2$, the piece that applies for $x\leq 1$, evaluated at $x=1$
$2$, from the derivative of $x^2$, which is $2x$, evaluated at $x=1$ instead of the function itself
does not exist, because $x=1$ is the point where the two pieces switch
$0$, from $x^2$ evaluated at $x=0$ instead of $x=1$
Check your answer
✓ $1$, from $x^2$, the piece that applies for $x\leq 1$, evaluated at $x=1$ — (A) $1$, approaching $x=1$ from the left stays inside the $x\leq 1$ piece, so the limit uses $x^2$ evaluated at $x=1$, giving $1^2 = 1$.
$2$, from the derivative of $x^2$, which is $2x$, evaluated at $x=1$ instead of the function itself — $2x$ is the derivative of $x^2$, not the function value the limit asks for. The limit of $f(x)$ as $x$ approaches $1$ needs $x^2$ itself, evaluated near $x=1$, giving $1$, not $2$.
does not exist, because $x=1$ is the point where the two pieces switch — A piece switching at $x=1$ does not stop the left-hand limit from existing. Approaching from the left, $x$ stays inside the $x\leq 1$ piece the whole way, so $x^2$ applies smoothly right up to $1$.
$0$, from $x^2$ evaluated at $x=0$ instead of $x=1$ — The limit is taken as $x$ approaches $1$, not $0$, evaluating the correct formula, $x^2$, at the wrong point gives a value that answers a different question entirely.
For the same function, $f(x) = x^2$ when $x \leq 1$ and $f(x) = 2x - 1$ when $x > 1$, what is $\lim_{x \to 1^+} f(x)$?
$2$, from $2x-1$ but omitting the $-1$ term, giving $2x$ evaluated at $x=1$
$1$, from $2x-1$, the piece that applies for $x>1$, evaluated at $x=1$
does not exist, because the piece $2x-1$ applies only where $x>1$, which does not include $x=1$ itself
$3$, from $2x-1$ but with the sign flipped, giving $2x+1$ evaluated at $x=1$
Check your answer
$2$, from $2x-1$ but omitting the $-1$ term, giving $2x$ evaluated at $x=1$ — $2x-1$ evaluated at $x=1$ is $2(1)-1 = 1$, not $2$, dropping the $-1$ term leaves out part of the formula.
✓ $1$, from $2x-1$, the piece that applies for $x>1$, evaluated at $x=1$ — (B) $1$, approaching $x=1$ from the right stays inside the $x>1$ piece, so the limit uses $2x-1$ evaluated at $x=1$, $2(1)-1 = 1$.
does not exist, because the piece $2x-1$ applies only where $x>1$, which does not include $x=1$ itself — A limit as $x$ approaches $1$ from the right only needs $2x-1$ to be defined for values near $1$ with $x>1$, which it is, the limit does not require the formula to hold at $x=1$ itself, only arbitrarily close to it.
$3$, from $2x-1$ but with the sign flipped, giving $2x+1$ evaluated at $x=1$ — Flipping the sign turns $2x-1$ into $2x+1$, which is not the piece given, the formula for $x>1$ keeps the minus sign, giving $2(1)-1 = 1$.
Apply the same LHL/RHL/$f(a)$ test to a new piecewise function: let $g(x) = x^3$ for $x \leq 2$ and $g(x) = 6x - 8$ for $x > 2$. Is $g$ continuous at $x=2$?
yes, LHL $=$ RHL $=$ $g(2) = 8$, since all three quantities come out equal
no, evaluating $g(2)$ from the $x>2$ piece gives $g(2) = 4$, which disagrees with LHL $=8$
yes, both pieces are polynomials, and polynomials are always continuous, so the join point needs no separate check
no, LHL $= 8$ and $g(2) = 8$, but RHL $= 4$, so LHL and RHL disagree
Check your answer
yes, LHL $=$ RHL $=$ $g(2) = 8$, since all three quantities come out equal — Evaluating $6x-8$ at $x=2$ gives $6(2)-8 = 4$, not $8$, RHL does not match LHL here, which is exactly why this function is not continuous at $x=2$.
no, evaluating $g(2)$ from the $x>2$ piece gives $g(2) = 4$, which disagrees with LHL $=8$ — $g(2)$ is always read from the piece that actually defines the point, here, the $x\leq 2$ piece, $x^3$, giving $g(2) = 2^3 = 8$, not the $x>2$ piece.
yes, both pieces are polynomials, and polynomials are always continuous, so the join point needs no separate check — Each piece being individually a continuous polynomial says nothing about what happens exactly at the join, the two pieces still have to meet at the same value there, and here they do not.
✓ no, LHL $= 8$ and $g(2) = 8$, but RHL $= 4$, so LHL and RHL disagree — (D) No, LHL $= \lim_{x \to 2^-} x^3 = 8$ and $g(2) = 2^3 = 8$ agree, but RHL $= \lim_{x \to 2^+} (6x-8) = 6(2)-8 = 4$ does not match, so LHL and RHL disagree and $g$ is discontinuous at $x=2$.
Walk along the V. Left of 0, each unit step along the x-axis goes down 1; right of 0, up 1.
Go back to the V shape you drew in Class 11, the graph of the absolute value. Walk down the left arm towards the corner. The graph drops 1 for every unit step along the x-axis, so its slope is minus 1. Now walk up the right arm. It rises 1 for every unit step along the x-axis, so its slope is plus 1. At the corner the two arms meet with no gap. But they come in with different slopes. So at the corner there is no single slope.
Naming the one-sided derivatives
To measure a slope from one side, use the limit from Class 11. The chord from $a$ to $a + h$ has slope $(f(a + h) - f(a))/h$. Let $h$ get close to 0 through negative values. The limit is the LEFT-HAND DERIVATIVE, or LHD. Let $h$ get close to 0 through positive values. The limit is the RIGHT-HAND DERIVATIVE, or RHD.
A function is DIFFERENTIABLE at $a$ when the LHD and the RHD both exist and are equal. Their common value is the DERIVATIVE at $a$, written $f'(a)$. If $y = f(x)$, the derivative is also written $d y/d x$. Both are in use, and this book uses both.
A chord from (1, 1): to the left, slope 1.5; to the right, 2.5. Shrink h and each settles on one number.
Let us check a point where it works. Take $f(x) = x^2$ at $a = 1$.
Worked example
The slope of $x^2$ at 1, from both sides
$(f(1 + h) - f(1))/h = ((1 + h)^2 - 1)/h$ Write the chord slope from 1 to $1 + h$ for $f(x) = x^2$.
$((1 + h)^2 - 1)/h = (2h + h^2)/h$ Open the bracket: $(1 + h)^2 = 1 + 2h + h^2$, and the two 1s cancel.
$(2h + h^2)/h = 2 + h$ Divide the top and the bottom by $h$, which is allowed because $h \neq 0$.
$\text{LHD} = \text{RHD} = 2$ Let $h$ get close to 0 from either side: $2 + h$ gets close to 2.
*$f'(1) = 2$* The two sides agree, so $f$ is differentiable at 1.
Check: $d/d x (x^2) = 2x$, and $2 \cdot 1 = 2$ The Class 11 rule gives the same slope at $x = 1$.
Both sides settle on 2, so x² is differentiable at 1 and f′(1) = 2. The tangent has that slope.
chord slope
1.9
1.99
2
2.01
2.1
The chord slope is 2 + h: 1.9 and 1.99 from the left, 2.1 and 2.01 from the right. Both close on 2.
A function that is differentiable at every point of an interval is differentiable on that interval. Its graph then has one slope at each of those points, and no corner.
Differentiable at every point of an interval: x² has one slope at each point, here −2, 0, 2 and 4.
Your turn: take $f(x) = \left|x - 2\right|$ at $x = 2$. Find the LHD and the RHD. Is $f$ differentiable at 2?
A SCHOLAR INDIA REMEMBERS
You found $f′(1) = 2$ for $x^2$. Did you calculate that $2$, or only remember it? Put a bound on it. The chord slope is $2 + h$. With $h = 0.1$ it is $2.1$. With $h = -0.01$ it is $1.99$. The chord slope is never more than $\left|h\right|$ away from $2$. So however small a band you choose around $2$, a small enough step keeps the chord slope inside it. That is what the limit $2$ means.
Aryabhata · the hoopoe
Check yourself
At $x=a$, which expression correctly gives the right-hand derivative (RHD)?
$\lim_{h \to 0} (f(a+h)-f(a))/h$
$\lim_{x \to a^+} f(x)$
$\lim_{h \to 0^+} (f(a+h)-f(a))/h$
$\lim_{h \to 0^-} (f(a+h)-f(a))/h$
Check your answer
$\lim_{h \to 0} (f(a+h)-f(a))/h$ — This is the plain two-sided derivative limit, with no direction attached. RHD is specifically the one-sided version, defined before the two-sided limit is even known to exist.
$\lim_{x \to a^+} f(x)$ — This is the formula for the right-hand limit of $f(x)$ itself, the RHL, a separate limit defined earlier. RHD is built from the difference quotient $(f(a+h)-f(a))/h$, a different expression entirely.
✓ $\lim_{h \to 0^+} (f(a+h)-f(a))/h$ — (C) RHD is exactly this limit, the difference quotient $(f(a+h)-f(a))/h$, taken as $h$ approaches 0 from the positive side.
$\lim_{h \to 0^-} (f(a+h)-f(a))/h$ — It is tempting to pair the two labels with the two one-sided limits in whichever order feels natural, but the assignment is fixed, RHD is the limit as $h$ approaches 0 from the positive side, and the $h \to 0^-$ limit is LHD instead.
A function $f$ has LHD equal to 3 at $x=a$. What additional fact is needed before you can say $f$ is differentiable at $a$ with $f'(a)=3$?
nothing further: one one-sided derivative existing is already enough
RHD must also exist, but it need not equal LHD: as long as both exist, differentiability holds
RHD must also exist and equal 3
$f$ must be continuous at $a$; nothing about RHD is needed
Check your answer
nothing further: one one-sided derivative existing is already enough — A single one-sided derivative existing never settles the two-sided question, since the other side could easily disagree.
RHD must also exist, but it need not equal LHD: as long as both exist, differentiability holds — Two one-sided derivatives that both exist but disagree still leave $f'(a)$ undefined, exactly the way a corner point does, existence alone is not enough.
✓ RHD must also exist and equal 3 — (C) Differentiability at $a$ needs both one-sided derivatives to exist and agree, RHD has to exist and match LHD’s value of 3 before $f'(a)$ is defined at all.
$f$ must be continuous at $a$; nothing about RHD is needed — Continuity is a different property altogether. LHD and RHD both existing and matching is what defines $f'(a)$, and continuity alone says nothing about whether the two slopes agree.
LHD and RHD are one-sided limits of the difference quotient $(f(a+h)-f(a))/h$. What do they measure?
how the function’s slope approaches a limit as $h$ approaches 0
the average rate of change of $f$ over its whole domain
how the function’s value $f(x)$ behaves as $x$ approaches $a$ from one side
whether $f(a)$ is defined at all
Check your answer
✓ how the function’s slope approaches a limit as $h$ approaches 0 — (A) LHD and RHD track the slope’s one-sided approach, the difference quotient’s limiting behaviour as $h$ shrinks to 0 from one side, not the function’s value.
the average rate of change of $f$ over its whole domain — An average rate of change over an interval is a global fact. LHD/RHD are strictly local and one-sided, built from a limit as $h$ shrinks toward 0, not a fixed interval.
how the function’s value $f(x)$ behaves as $x$ approaches $a$ from one side — This describes LHL/RHL, the one-sided limits of $f(x)$ itself, a separate pair of limits from LHD/RHD; LHD/RHD ask a different question, about the slope, not the value.
whether $f(a)$ is defined at all — $f(a)$ being defined is a separate fact from what the resulting limit measures, this option would only matter if the difference quotient could not even be written down.
At some point $a$, LHD exists and equals 2, but RHD does not exist (the right-hand difference-quotient limit fails to settle). What can you conclude about $f$ at $a$?
$f$ is differentiable at $a$, and $f'(a)$ is undefined but the function is still smooth there
$f$ is differentiable at $a$ with $f'(a)=2$, since at least one one-sided derivative exists
nothing can be concluded without knowing $f(a)$
$f$ is not differentiable at $a$, since RHD fails to exist
Check your answer
$f$ is differentiable at $a$, and $f'(a)$ is undefined but the function is still smooth there — This tries to keep differentiable and $f'(a)$-undefined both true at once, which is a contradiction, differentiability at a point is the statement that $f'(a)$ exists there.
$f$ is differentiable at $a$ with $f'(a)=2$, since at least one one-sided derivative exists — The agreement condition needs both sides to exist, a missing RHD ends the question immediately, regardless of what LHD alone showed.
nothing can be concluded without knowing $f(a)$ — The LHD/RHD existence-and-agreement condition settles the question on its own, without needing $f(a)$’s value at all.
✓ $f$ is not differentiable at $a$, since RHD fails to exist — (D) $f'(a)$ is defined only when both LHD and RHD exist and agree, RHD failing to exist already breaks that condition, so $f$ is not differentiable at $a$ regardless of what LHD alone showed.
Left: both sides approach 1.5, but f of 1 is 2.6. Right: a V with no gap, but with slopes minus 1 and plus 1.
LHL and LHD look alike. Both are limits from one side, and the names differ by one letter. But they are limits of different things. The LHL is a limit of values, $f(x)$. The LHD is a limit of chord slopes, $(f(a + h) - f(a))/h$. So the LHL and the RHL test for a jump. The LHD and the RHD test for a corner.
Pair
Tests for
LHL and RHL
a jump in the value
LHD and RHD
a corner in the slope
Here is a working on $f(x) = \left|x\right|$ at 0. Which step is wrong?
From the left, $\left|x\right|$ gets close to 0. So the LHL is 0.
From the right, $\left|x\right|$ also gets close to 0. So the RHL is 0.
Also $f(0) = 0$, so $\text{LHL} = \text{RHL} = f(0)$.
So $f$ is differentiable at 0.
Keep your answer. The worked example on $\left|x\right|$, a little further on, finds its two slopes at 0.
Step 1
LHL at 0
0
Step 2
RHL at 0
0
Step 3
LHL = RHL = f(0)
0
Step 4
so differentiable
?
The working above, one step to a panel. The worked example on |x| shows which step fails.
the values
the slopes
Same function, two questions. The values at ±0.1 are both 0.1. The chord slopes are −1 and 1.
Before you test a point, name the question. Is it about the value, or about the slope? The value question is continuity. The slope question is differentiability.
Check yourself
A student says, “LHL/RHL and LHD/RHD are really the same kind of one-sided limit calculation, so if I have already checked one, I do not need to separately check the other.” What is wrong with this claim?
the claim is wrong because LHD/RHD are harder to compute than LHL/RHL, not because they test different things
the claim is wrong only because LHL/RHL come before LHD/RHD in the chapter, not because they test different things
nothing is wrong: checking one does cover the other
LHL/RHL track the value as $x$ approaches $a$; LHD/RHD track the slope as $h$ approaches 0
Check your answer
the claim is wrong because LHD/RHD are harder to compute than LHL/RHL, not because they test different things — This reaches for difficulty instead, LHD/RHD can be just as easy to compute as LHL/RHL for many functions; the reason checking one does not cover the other is that they measure different properties, not that one is harder.
the claim is wrong only because LHL/RHL come before LHD/RHD in the chapter, not because they test different things — This reaches for the order these two pairs happen to be presented in as the explanation, but the real boundary has nothing to do with sequence, it is about what each pair of limits is a limit of.
nothing is wrong: checking one does cover the other — This is the trap itself, restated as if it were true, a function can pass the LHL/RHL test and still fail the LHD/RHD test, exactly as $|x|$ does at 0.
✓ LHL/RHL track the value as $x$ approaches $a$; LHD/RHD track the slope as $h$ approaches 0 — (D) LHL/RHL and LHD/RHD look alike because both are one-sided limits, but they are limits of different quantities, the function’s value versus its slope, so one settling tells you nothing about whether the other does.
A function $g$ is found to be continuous at $x=3$: its LHL and RHL both agree with $g(3)$. A student concludes $g$ must also be differentiable at $x=3$. What does the LHL/RHL-versus-LHD/RHD distinction say about this conclusion?
the conclusion does not follow: continuity settles the value, but differentiability needs the slope test
the conclusion follows automatically: once LHL and RHL agree, the corresponding LHD and RHD must agree too
the conclusion cannot be evaluated without knowing $g$’s formula explicitly
the conclusion follows, but only because $g$ is continuous at every nearby point, not just at $x=3$
Check your answer
✓ the conclusion does not follow: continuity settles the value, but differentiability needs the slope test — (A) LHL and RHL agreeing only settles the value question. LHD and RHD are a separate, independent test on the slope, and nothing about the value test forces them to agree as well.
the conclusion follows automatically: once LHL and RHL agree, the corresponding LHD and RHD must agree too — This assumes the two tests are linked, but they are not, a function’s value can settle smoothly from both sides while its slope still jumps, exactly the pattern $|x|$ shows at 0.
the conclusion cannot be evaluated without knowing $g$’s formula explicitly — This treats the question as unanswerable without more information, but the general implication holds regardless of g’s specific formula, that is precisely why it needs a separate LHD/RHD check.
the conclusion follows, but only because $g$ is continuous at every nearby point, not just at $x=3$ — This adds a condition, continuity at nearby points, that the gap does not depend on at all, the gap can appear even when continuity holds only at the one point in question.
Which statement correctly matches each pair of one-sided limits to what it tracks?
LHL/RHL track the function’s slope; LHD/RHD track the function’s value
both pairs track the function’s value; only the direction of approach differs between them
LHL/RHL track the function’s value; LHD/RHD track the function’s slope
both pairs track the function’s slope; only the direction of approach differs between them
Check your answer
LHL/RHL track the function’s slope; LHD/RHD track the function’s value — This swaps the two pairings, it is LHL/RHL that track the value and LHD/RHD that track the slope, not the reverse.
both pairs track the function’s value; only the direction of approach differs between them — This collapses the two pairs into tracking the same quantity, differing only by direction, but the two pairs differ in what they are limits of, one built from $f(x)$, the other from the difference quotient.
✓ LHL/RHL track the function’s value; LHD/RHD track the function’s slope — (C) LHL/RHL are one-sided limits of $f(x)$, the value; LHD/RHD are one-sided limits of the difference quotient, the slope.
both pairs track the function’s slope; only the direction of approach differs between them — This also collapses the two pairs into tracking the same quantity, but LHL/RHL are limits of $f(x)$, the value, while only LHD/RHD are limits of the difference quotient, the slope.
Why can a function pass the LHL/RHL test (be continuous at a point) yet still fail the LHD/RHD test at that same point?
because LHL/RHL require only the output values to converge from both sides, while LHD/RHD also require the rate of approach to match
because the difference quotient equals $f(a)$ exactly when $h=0$, so LHD/RHD reduce to nothing more than LHL/RHL evaluated at the single point $h=0$
because LHL/RHL and LHD/RHD are computed using different values of a, so they are testing entirely different points
because LHL/RHL is a stronger condition than LHD/RHD, so passing the stronger test does not guarantee passing the weaker one
Check your answer
✓ because LHL/RHL require only the output values to converge from both sides, while LHD/RHD also require the rate of approach to match — (A) LHL/RHL only require the output values to converge; LHD/RHD go further and require the rate of approach, the slope, to also match from both sides, which is a strictly extra condition value agreement never guarantees.
because the difference quotient equals $f(a)$ exactly when $h=0$, so LHD/RHD reduce to nothing more than LHL/RHL evaluated at the single point $h=0$ — This tries to collapse the difference quotient at $h=0$ into the definition of LHL/RHL, but the quotient is not even defined at $h=0$ since it divides by $h$, the limit is taken as $h$ approaches 0, never evaluated there.
because LHL/RHL and LHD/RHD are computed using different values of a, so they are testing entirely different points — This is simply false, both pairs of limits are taken at the same point a; the difference is in what expression is being limited, not which point is used.
because LHL/RHL is a stronger condition than LHD/RHD, so passing the stronger test does not guarantee passing the weaker one — This gets the relationship backwards, neither test is simply stronger or weaker in a way that guarantees one from the other; $|x|$ at 0 is exactly the case where the value test passes while the slope test fails.
Here is a working on a function with two pieces. Take $f(x) = 2x$ for $x \leq 1$ and $f(x) = 2x + 3$ for $x > 1$, and test it at $x = 1$. Which step is wrong?
Step 1
left slope
2
Step 2
right slope
2
Step 3
so differentiable
?
Both pieces have slope 2. Does that settle the slope at x = 1?
To the left of 1 the rule is $2x$, whose slope is 2. So the LHD is 2.
To the right of 1 the rule is $2x + 3$, whose slope is also 2. So the RHD is 2.
The LHD equals the RHD, so $f$ is differentiable at 1.
Start with the values: from the left 2, at the point 2, from the right 5. The graph jumps by 3.
Start with the values, not the slopes. The LHL at 1 is 2, and so is $f(1)$. But the RHL is 5. So the graph jumps at 1, and $f$ is not even continuous there.
The graph jumps by 3 at x = 1. Measured from (1, 2), the chord slopes 8, 14 and 26 keep growing, never settling.
chord slope
8
14
26
50
Halve h and the chord slope nearly doubles: 8, 14, 26, 50. It settles on no number, so there is no RHD.
Now look at the RHD again. It is not the slope of the right-hand piece. It is measured from the point $(1, f(1))$, which is $(1, 2)$. For a small $h > 0$, the chord slope is:
$(f(1 + h) - f(1))/h = (2(1 + h) + 3 - 2)/h$.
Open the bracket and split the fraction:
$(2(1 + h) + 3 - 2)/h = (3 + 2h)/h = 3/h + 2$.
Now let $h$ get close to 0. Then $3/h$ grows without limit. So the RHD does not exist, and the second step is wrong.
That is why you check for a jump first. Where a graph breaks, the chord slopes on at least one side do not settle on a number. So a function with a break at a point has no derivative there. The earlier section, “Differentiable implies continuous”, proves it.
So when a question asks about the slope at a point, test the three numbers for continuity first. If they fail, the answer is already no.
Check yourself
A student says, “$f$ has a jump at $x=a$, but its LHD and RHD might still agree, so I will compute both before I decide whether $f$ is differentiable there.” What is wrong with this?
the approach is fine, but only works for polynomial functions
the approach is backwards: RHD should always be computed before LHD, not continuity before differentiability
nothing is wrong: at a jump, LHD and RHD can both exist and agree
at a jump at least one of LHD and RHD fails to exist, so $f$ is not differentiable there
Check your answer
the approach is fine, but only works for polynomial functions — This invents a restriction to polynomials that has nothing to do with the actual issue, the precondition holds for any function.
the approach is backwards: RHD should always be computed before LHD, not continuity before differentiability — This redirects the ordering problem onto LHD versus RHD, a different pair entirely, the ordering that matters is continuity before differentiability, not which one-sided derivative gets computed first.
nothing is wrong: at a jump, LHD and RHD can both exist and agree — If LHD and RHD both existed as finite numbers, $f$ would be continuous at $a$, so it could have no jump there. At a jump at least one of them fails to exist.
✓ at a jump at least one of LHD and RHD fails to exist, so $f$ is not differentiable there — (D) If both one-sided slopes were finite, $f(a+h) - f(a)$ would tend to $0$ from both sides and $f$ would have no jump at $a$. So at a jump at least one of LHD and RHD fails to exist, and $f$ is not differentiable there. Spotting the jump first saves computing them.
If $f$ is discontinuous at $x=a$, what can you say about LHD and RHD at $a$?
they are unrelated to continuity, so they may still exist and agree at $a$
they cannot even be written down, because the difference quotient needs $f$ to be continuous
they always agree, by coincidence, so the slope test gives a false yes
at least one of them fails to exist, so $f$ is not differentiable at $a$
Check your answer
they are unrelated to continuity, so they may still exist and agree at $a$ — They are not unrelated. If LHD and RHD both existed as finite numbers, then $f(a+h) - f(a) = h \cdot (f(a+h) - f(a))/h$ would tend to $0$ from both sides, and $f$ would be continuous at $a$. A break rules that out.
they cannot even be written down, because the difference quotient needs $f$ to be continuous — The difference quotient $(f(a+h) - f(a))/h$ can be written for every $h \neq 0$. What fails is its limit: on at least one side it does not settle on a finite number.
they always agree, by coincidence, so the slope test gives a false yes — They cannot both exist, let alone agree: if both one-sided limits were finite numbers, $f$ would be continuous at $a$. At a break at least one of them fails to exist.
✓ at least one of them fails to exist, so $f$ is not differentiable at $a$ — (D) If LHD and RHD both existed as finite numbers, then $f(a+h) - f(a) = h \cdot (f(a+h) - f(a))/h$ would tend to $0$ from both sides, so $f$ would be continuous at $a$. At a break, at least one of them fails to exist.
What is the efficient order for testing whether $f$ is differentiable at a point $a$?
check LHD and RHD first; only check continuity if they disagree
check RHD first, then LHD, then continuity last, since continuity follows automatically once both one-sided derivatives are known
check continuity and LHD/RHD simultaneously: the order does not matter, since they test independent facts
check continuity first; only compute LHD and RHD if continuity already holds
Check your answer
check LHD and RHD first; only check continuity if they disagree — This reverses the efficient order. At a break at least one of LHD and RHD fails to exist, so computing them first is limit work that checking continuity would have saved.
check RHD first, then LHD, then continuity last, since continuity follows automatically once both one-sided derivatives are known — If RHD and LHD exist and agree, $f$ is differentiable and so continuous, so the last check adds nothing. But this order does all the limit work first, even where a break would have settled the question at once.
check continuity and LHD/RHD simultaneously: the order does not matter, since they test independent facts — This treats the two checks as independent, but continuity is a precondition for differentiability, not a separate fact that can be checked in any order.
✓ check continuity first; only compute LHD and RHD if continuity already holds — (D) Differentiability needs continuity. If $f$ is broken at $a$, at least one of LHD and RHD fails to exist, so checking continuity first means you compute them only when they can both exist.
A problem gives a piecewise function $h$ and asks you to test differentiability at $x=5$. Before computing anything, you notice the two pieces of $h$ do not match up at $x=5$, so $h$ is discontinuous there. What should you conclude, and why?
$h$ could still be differentiable at $x=5$ as long as one of LHD or RHD exists
the discontinuity only affects the value of $h'(5)$, not whether $h'(5)$ exists in the first place
$h$ is not differentiable at $x=5$, and there is no need to compute LHD or RHD: continuity already fails
you should still compute LHD and RHD, since they might happen to agree despite the discontinuity, which would make $h$ differentiable anyway
Check your answer
$h$ could still be differentiable at $x=5$ as long as one of LHD or RHD exists — One existing one-sided derivative is never enough: differentiability needs both to exist and agree. And at a break they cannot both exist.
the discontinuity only affects the value of $h'(5)$, not whether $h'(5)$ exists in the first place — This tries to separate whether $h'(5)$ exists from what it equals, but a discontinuity blocks existence itself, there is no derivative value left to compute, correct or otherwise.
✓ $h$ is not differentiable at $x=5$, and there is no need to compute LHD or RHD: continuity already fails — (C) If both one-sided slopes at $x=5$ were finite, the two pieces would have to meet there. They do not, so at least one of LHD and RHD fails to exist and $h$ is not differentiable at $x=5$. Seeing the break first saves the limit work.
you should still compute LHD and RHD, since they might happen to agree despite the discontinuity, which would make $h$ differentiable anyway — They cannot both exist at a break: if both one-sided slopes were finite, the two pieces would meet at $x=5$. So at least one of them fails to exist, and there is nothing left to agree.
If $f$ is differentiable at $x = a$, then $f$ is continuous at $a$. That is why you can stop testing for a jump once you know a function is differentiable: differentiability already rules one out. Here is the reason, line by line.
Rise = slope × step along the x-axis. From (1, 1) with h = 0.5: 2.5 × 0.5 = 1.25, the gap in height.
Why it works
Differentiable at $a$ gives continuous at $a$
$f(a + h) - f(a) = (f(a + h) - f(a))/h \cdot h$ For $h \neq 0$, divide by $h$ and multiply by $h$: the two cancel.
$\lim_{h \to 0} (f(a + h) - f(a))/h = f'(a)$ The first factor is the chord slope, and $f$ is differentiable at $a$.
$\lim_{h \to 0} (f(a + h) - f(a)) = f'(a) \cdot 0 = 0$ The limit of a product is the product of the limits, and $h$ gets close to 0.
$\lim_{h \to 0} f(a + h) = f(a)$ Add $f(a)$ to both sides.
*$\lim_{x \to a} f(x) = f(a)$* Write $x = a + h$: as $h$ gets close to 0, $x$ gets close to $a$.
That is continuity at $a$. The height at $a + h$ gets close to the height at $a$, so the graph has no jump there.
Chords from (1, 1) have slopes 3, 2.5 and 2.25, closing on tangent slope 2. Height gaps 3, 1.25 and 0.5625 close on 0.
height gap
1.25
0.5625
0.21
0.0201
The gap is 2h + h²: 1.25, 0.5625, 0.21, 0.0201. It closes on 0, so x² has no jump at 1.
Check it on $f(x) = x^2$ at 1, where the slope is 2.
Worked example
Check it on $x^2$ at 1
$f(1 + h) - f(1) = (1 + h)^2 - 1$ Write the change in height from 1 to $1 + h$.
$(1 + h)^2 - 1 = 2h + h^2$ Open the bracket, and the two 1s cancel.
$\lim_{h \to 0} (2h + h^2) = 0$ Both terms get close to 0 as $h$ does.
Check: $\lim_{h \to 0} f(1 + h) = f(1) = 1$ Add $f(1)$ to both sides, so $x^2$ has no jump at 1.
The other direction is false. A graph with no jump can still turn a corner, as the V of $\left|x\right|$ does at 0.
Check yourself
Between continuity and differentiability at a point, which is the stronger requirement?
neither is stronger: the two properties are logically independent of each other
they are equally strong: a function is continuous at a point exactly when it is differentiable there
continuity: every continuous function is differentiable, but not every differentiable function is continuous
differentiability: every differentiable function is continuous, but the reverse does not hold
Check your answer
neither is stronger: the two properties are logically independent of each other — This denies that any implication holds at all, but one direction, differentiable implies continuous, always holds and can be proved directly.
they are equally strong: a function is continuous at a point exactly when it is differentiable there — This claims the two properties always coincide, but $|x|$ at $x=0$ is continuous without being differentiable, so the two are not equivalent.
continuity: every continuous function is differentiable, but not every differentiable function is continuous — This reverses the actual direction, continuity is the weaker property, implied by differentiability, not the other way round.
✓ differentiability: every differentiable function is continuous, but the reverse does not hold — (D) Differentiability is the stronger requirement, it always forces continuity, while continuity never forces differentiability back.
In the proof that differentiability implies continuity, $f(a+h)-f(a)$ is written as $[(f(a+h)-f(a))/h] \cdot h$ for $h \neq 0$. As $h$ approaches 0, why does this product tend to 0?
because the bracket tends to the finite number $f'(a)$, while $h$ tends to 0, so their product shrinks to 0
because the bracket $(f(a+h)-f(a))/h$ tends to 0 as well, so 0 times 0 gives 0
because $h$ appears in both the numerator and denominator of the bracket, so it cancels out algebraically before any limit is taken
because differentiability guarantees $f(a+h) = f(a)$ for all $h$ near 0
Check your answer
✓ because the bracket tends to the finite number $f'(a)$, while $h$ tends to 0, so their product shrinks to 0 — (A) The bracket tends to the finite value $f'(a)$ because $f$ is assumed differentiable, and the separate factor $h$ shrinks to 0, a finite number times something shrinking to 0 shrinks the whole product to 0.
because the bracket $(f(a+h)-f(a))/h$ tends to 0 as well, so 0 times 0 gives 0 — This mistakes what the bracket tends to, it is defined to tend to $f'(a)$, a finite number, precisely because $f$ is assumed differentiable at a; it is the outer factor $h$, not the bracket, that shrinks to 0.
because $h$ appears in both the numerator and denominator of the bracket, so it cancels out algebraically before any limit is taken — This tries to cancel $h$ algebraically inside the bracket, but $f(a+h)$ is a function evaluated at a shifted point, not a literal multiple of $h$, there is no $h$ to cancel before the limit is taken.
because differentiability guarantees $f(a+h) = f(a)$ for all $h$ near 0 — This would make $f$ constant near a, which is far stronger than differentiability and not what the identity assumes or needs.
What is the purpose of writing $f(a+h)-f(a)$ as $[(f(a+h)-f(a))/h] \cdot h$ in this proof?
to simplify the formula for $f'(a)$ into a form that is easier to compute numerically
to replace the definition of the derivative with the definition of continuity, since the two are proved to be identical
to split the change in $f$ into a factor whose limit is known, $f'(a)$, and a factor, $h$, that visibly tends to 0
to show that $f(a+h)-f(a)$ can only be zero when $h$ is zero
Check your answer
to simplify the formula for $f'(a)$ into a form that is easier to compute numerically — This misreads the purpose as computational convenience for $f'(a)$ itself, but $f'(a)$ is already assumed known here, the identity is used to prove continuity, not to compute the derivative.
to replace the definition of the derivative with the definition of continuity, since the two are proved to be identical — This overstates the conclusion, the proof shows differentiability forces continuity, not that the two definitions are the same thing; continuity remains a strictly weaker, separate property.
✓ to split the change in $f$ into a factor whose limit is known, $f'(a)$, and a factor, $h$, that visibly tends to 0 — (C) The identity splits the change in $f$ into a piece with a known limit and a piece that visibly vanishes, so the product’s limit, and therefore continuity, can be read off directly without any extra work.
to show that $f(a+h)-f(a)$ can only be zero when $h$ is zero — This claims something the identity does not establish, $f(a+h)-f(a)$ can certainly be zero for other reasons too; the identity is not making a claim about when the difference is zero.
A textbook problem states $f$ is differentiable at $x=7$ and separately asks you to verify $f$ is continuous at $x=7$ as an additional step. Given the differentiable-implies-continuous result, is that separate verification necessary?
no, continuity comes automatically once differentiability holds; it is not a separate fact to prove
yes, because differentiability only implies continuity for polynomial functions, not in general
yes, because differentiability and continuity must always be checked independently with no shortcut between them
no, but only because $x=7$ happens to be a convenient integer value
Check your answer
✓ no, continuity comes automatically once differentiability holds; it is not a separate fact to prove — (A) Once differentiability at a point is established, continuity there comes for free, it is a necessary precondition already built into differentiability, not a separate fact that needs its own proof.
yes, because differentiability only implies continuity for polynomial functions, not in general — This invents a restriction to polynomials that the general proof does not make, the identity $[(f(a+h)-f(a))/h] \cdot h$ works for any differentiable function, not just polynomials.
yes, because differentiability and continuity must always be checked independently with no shortcut between them — This denies the implication proved in this section, differentiability at a point always forces continuity there, so treating the two as needing fully independent checks ignores that proof.
no, but only because $x=7$ happens to be a convenient integer value — This invents a dependence on the specific value of $x$ that the result does not have, the implication holds for any point a, integer or not.
✕MISCONCEPTION
Here is an argument about $f(x) = \left|x\right|$ at 0. Which step is wrong?
The LHL, the RHL and $f(0)$ are all 0. So $\left|x\right|$ is continuous at 0.
A function that is differentiable at a point is continuous there.
So $\left|x\right|$ is differentiable at 0.
Line 1
continuous at 0
yes
Line 2
differentiable ⇒ continuous
always
Line 3
so differentiable
?
The first two lines are true. Does the third follow from them?
The first two lines are true. Keep your answer, then test it on the graph. The V comes into the corner with slope $-1$ and leaves it with slope $1$.
The V comes into the corner with slope −1 and leaves with slope 1.
The second line runs in one direction only. Differentiable gives continuous. Continuous does not give differentiable. A function with no jump can still have a corner, and at a corner the two one-sided slopes differ.
continuous and differentiable
x²2x + 1eˣ
continuous, not differentiable
|x|1 − |x|
not continuous
the rate cardthe jump function
Differentiable gives continuous, never the other way. No function is differentiable but not continuous, so there is no third box for it.
The worked example below finds the LHD and the RHD of $\left|x\right|$ at 0 from the limit.
First try
I looked at the graph of $f(x) = \left|x\right|$ and saw no break anywhere, not even at $x=0$, so I wrote that $f$ must be differentiable there too, since nothing seemed to go wrong.
Second look
A graph with no break only tests continuity, not whether one slope fits at that point. At $x=0$, $\text{LHD} = -1$ and $\text{RHD} = 1$ for $f(x)=\left|x\right|$, two different one-sided slopes meeting at the same point, so $f'(0)$ does not exist even though $f$ is continuous there.
Continuous asks whether the graph breaks. Differentiable asks whether one slope fits. $\left|x\right|$ answers yes to the first question and no to the second.
Check yourself
A student says, “If a function’s graph has no break at a point, then it must be continuous there, and since it is continuous, the tangent line (and so the derivative) must exist there too.” What is wrong with this reasoning?
the reasoning is correct for polynomial functions but breaks down for functions involving absolute value
the reasoning is backwards: differentiability should be checked before continuity, not after
continuity does not imply differentiability; $f(x)=|x|$ is continuous at $0$ but its LHD, RHD disagree
nothing is wrong: an unbroken graph always has a well-defined tangent line at every point
Check your answer
the reasoning is correct for polynomial functions but breaks down for functions involving absolute value — This invents a restriction to functions involving absolute value, but the failure has nothing to do with algebraic form specifically, it is about corner points in general, and the boundary is continuity-versus-differentiability.
the reasoning is backwards: differentiability should be checked before continuity, not after — This confuses this trap with the separate check-order issue, the reasoning here is about which property implies which, not about which property to test first.
✓ continuity does not imply differentiability; $f(x)=|x|$ is continuous at $0$ but its LHD, RHD disagree — (C) Continuity never guarantees differentiability, only the reverse direction always holds, and $|x|$ at $x=0$ is the standing counterexample where the graph is unbroken yet the two one-sided slopes disagree.
nothing is wrong: an unbroken graph always has a well-defined tangent line at every point — This restates the trap as if it were true, an unbroken graph can still meet at a corner, where no single tangent line exists, exactly as $|x|$ shows at 0.
Which of the following correctly states the logical relationship between continuity and differentiability at a point?
neither property ever implies the other: they must always be checked completely independently
continuous implies differentiable, and differentiable implies continuous: the two are logically equivalent
continuous implies differentiable, but differentiable does not imply continuous
differentiable implies continuous, but continuous does not imply differentiable
Check your answer
neither property ever implies the other: they must always be checked completely independently — This denies any implication at all, but differentiable implies continuous is proved to hold in general, for every differentiable function, it is not merely a coincidence to be checked case by case.
continuous implies differentiable, and differentiable implies continuous: the two are logically equivalent — This claims full equivalence, but $|x|$ at 0 breaks it, continuous there, not differentiable.
continuous implies differentiable, but differentiable does not imply continuous — This states the one direction that does not hold as if it did, and drops the direction that actually always holds.
✓ differentiable implies continuous, but continuous does not imply differentiable — (D) Only one direction always holds, differentiable implies continuous. The reverse, continuous implies differentiable, fails, as $|x|$ at $x=0$ shows.
For $f(x)=|x|$ at $x=0$, what specifically breaks differentiability even though the function is continuous there?
$f$ is not defined for $x<0$, so the left-hand derivative cannot be computed
LHD $=-1$, RHD $=+1$ at $x=0$: they disagree, so no derivative exists
$f(0)$ is undefined, so the derivative has nothing to be based on
the left-hand and right-hand limits of $f(x)$ itself disagree at $x=0$
Check your answer
$f$ is not defined for $x<0$, so the left-hand derivative cannot be computed — This is false, $|x|$ is defined for every real $x$, negative values included, so the left-hand derivative can be computed just as normally as the right-hand one.
✓ LHD $=-1$, RHD $=+1$ at $x=0$: they disagree, so no derivative exists — (B) LHD and RHD come out to $-1$ and $+1$ respectively, two different numbers, so $f'(0)$ has no single value to settle on, even though $f$ itself is continuous at 0.
$f(0)$ is undefined, so the derivative has nothing to be based on — This is false, $f(0)=|0|=0$ is perfectly well defined, so the derivative has a starting point; the problem is not that $f(0)$ is missing.
the left-hand and right-hand limits of $f(x)$ itself disagree at $x=0$ — This describes exactly what would break continuity, not differentiability, but LHL and RHL of $|x|$ do agree at 0, both equal to 0, which is why the function is continuous there in the first place.
Define $g(x) = |x-2|$, which has a corner at $x=2$ shaped just like $|x|$’s corner at $x=0$. Given that continuity never guarantees differentiability, what should you expect about $g$ at $x=2$?
$g$’s differentiability at $x=2$ cannot be predicted at all without a formal LHD/RHD computation, since continuity guarantees nothing either way
$g$ must be differentiable at $x=2$, since $g$ is continuous everywhere
$g$ is continuous at $x=2$ but, by the same reasoning as $|x|$ at $0$, its one-sided derivatives should disagree there
$g$ is not even continuous at $x=2$, since it has a corner there
Check your answer
$g$’s differentiability at $x=2$ cannot be predicted at all without a formal LHD/RHD computation, since continuity guarantees nothing either way — This is overly cautious, while a formal computation is needed for full certainty, the structural parallel to $|x|$ at 0 gives strong reason to anticipate the same disagreement in LHD and RHD.
$g$ must be differentiable at $x=2$, since $g$ is continuous everywhere — This repeats the very implication that fails here, continuity everywhere never guarantees differentiability everywhere, and a corner is precisely where that gap tends to show up.
✓ $g$ is continuous at $x=2$ but, by the same reasoning as $|x|$ at $0$, its one-sided derivatives should disagree there — (C) $g$ is continuous at its corner, but by the same structural reasoning as $|x|$ at 0, the one-sided slopes should be expected to disagree there, the corner shape is exactly what signals a likely differentiability failure.
$g$ is not even continuous at $x=2$, since it has a corner there — This confuses a corner with a break, g’s two pieces meet at the same height at $x=2$ (both give $g(2) = 0$), so continuity holds there even though the graph has a corner rather than a smooth turn.
The V of the absolute value has no gap at 0. The LHD is minus 1 and the RHD is plus 1.
The quotient |h|/h is −1 for every step left and 1 for every step right. It never settles on one number.
Worked example
Compute LHD and RHD of $f(x) = |x|$ at $x = 0$
$f(x) = |x|$ at $x=0$ Here $f(0) = 0$. Find both one-sided slopes from the limit.
$\text{LHD} = \lim_{h \to 0^-} (|h|-0)/h$ The LHD uses steps to the left of 0, so $h < 0$.
$\text{LHD} = \lim_{h \to 0^-} -h / h = -1$ When $h < 0$, $\left|h\right| = -h$. So the quotient is $-1$ for every such $h$.
$\text{RHD} = \lim_{h \to 0^+} (|h|-0)/h$ The RHD uses steps to the right of 0, so $h > 0$.
$\text{RHD} = \lim_{h \to 0^+} h/h = 1$ When $h > 0$, $\left|h\right| = h$. So the quotient is 1 for every such $h$.
$\text{LHD} = -1 \neq 1 = \text{RHD}$ The two slopes differ. So $f'(0)$ does not exist, though $f$ is continuous at 0.
$h = 0.01$: $\left|0.01\right|/0.01 = 1$, and $h = -0.01$: $\left|-0.01\right|/(-0.01) = -1$ Check with one small step on each side. The right arm gives 1 and the left arm gives $-1$.
Worked example
The same test, on a ball thrown straight up
A ball thrown straight up has velocity $v(t) = u - g t$, with $u = 20$ m/s and $g = 10$ m/s² Physics 11, Chapter 2, Motion in a Straight Line, gives this as the velocity under constant acceleration $-g$.
Speed is $\left|v(t)\right|$ exactly, with no interval to blur it Instantaneous speed is the magnitude of instantaneous velocity, true at a single instant.
$v(t) = 0$ at $t = u/g = 2$ s, the top of the flight Solve $20 - 10 t = 0$ for the moment the ball turns around.
For $t < 2$, $v(t) > 0$, so $\text{speed}(t) = v(t) = 20 - 10 t$ Before the top, the ball still rises, so speed equals velocity directly.
For $t > 2$, $v(t) < 0$, so $\text{speed}(t) = -v(t) = 10 t - 20$ After the top, the ball falls, so speed is the negative of velocity.
$\text{LHD} = \lim_{h \to 0^-} (\text{speed}(2+h)-\text{speed}(2))/h = -10$ Just before $t=2$, speed follows $20-10t$, whose slope is $-10$.
$\text{RHD} = \lim_{h \to 0^+} (\text{speed}(2+h)-\text{speed}(2))/h = 10$ Just after $t=2$, speed follows $10t-20$, whose slope is $10$.
*$\text{LHD} = -10 \neq 10 = \text{RHD}$, so speed is continuous at the top of the flight but not differentiable there* The same corner as $\left|x\right|$ at $0$, now on a thrown ball.
Check yourself
Computing LHD for $f(x)=|x|$ at $x=0$: $\lim_{h \to 0^-} (|h|-0)/h$. For $h<0$, what does $|h|$ equal, and what does this make the limit?
for $h<0$, the quotient $(|h|-0)/h$ is undefined, since dividing by a negative number is not permitted inside a limit
for $h<0$, $|h|=-h$, so the quotient is $(-h)/h=-1$ for every such $h$, giving LHD $=-1$
for $h<0$, $|h|=-h$, so the quotient is $(-h)/h=-1$, but since $h$ approaches 0, the limit itself must be 0, giving LHD $=0$
for $h<0$, $|h|=h$, so the quotient is $h/h=1$, giving LHD $=1$
Check your answer
for $h<0$, the quotient $(|h|-0)/h$ is undefined, since dividing by a negative number is not permitted inside a limit — This invents a restriction that does not exist, dividing by a negative number is perfectly well defined; the quotient $(-h)/h$ simplifies cleanly to $-1$ for any nonzero negative $h$.
✓ for $h<0$, $|h|=-h$, so the quotient is $(-h)/h=-1$ for every such $h$, giving LHD $=-1$ — (B) For $h$ negative, $|h|$ equals $-h$, so the quotient $(-h)/h$ simplifies to the constant $-1$ for every such $h$, the limit is just that constant, LHD $=-1$.
for $h<0$, $|h|=-h$, so the quotient is $(-h)/h=-1$, but since $h$ approaches 0, the limit itself must be 0, giving LHD $=0$ — This correctly simplifies the quotient to $-1$ but then confuses the quotient’s constant value with $h$ itself shrinking to 0, since the quotient equals $-1$ for every negative $h$, not just near 0, its limit is $-1$, not 0.
for $h<0$, $|h|=h$, so the quotient is $h/h=1$, giving LHD $=1$ — This gets the sign of $|h|$ backwards for negative $h$, $|h|=-h$ when $h<0$, not $|h|=h$; that identity is only true for $h>0$.
Computing RHD for $f(x)=|x|$ at $x=0$: $\lim_{h \to 0^+} (|h|-0)/h$. What is the value of this limit, and why?
$1$, because for $h>0$, $|h|=h$, so the quotient is $h/h=1$ for every such $h$, giving RHD $=1$
$-1$, because $|h|=-h$ regardless of the sign of $h$
$0$, because as $h \to 0^+$, both $|h|$ and $h$ shrink to 0, so their quotient must shrink to 0 too
undefined, because $|h|$ is not a differentiable expression
Check your answer
✓ $1$, because for $h>0$, $|h|=h$, so the quotient is $h/h=1$ for every such $h$, giving RHD $=1$ — (A) For $h$ positive, $|h|$ equals $h$, so the quotient $h/h$ simplifies to the constant $1$ for every such $h$, the limit is that constant, RHD $=1$.
$-1$, because $|h|=-h$ regardless of the sign of $h$ — This applies the negative-h identity, $|h|=-h$, to the positive-h case, but $|h|=h$ is the correct identity once $h$ is positive.
$0$, because as $h \to 0^+$, both $|h|$ and $h$ shrink to 0, so their quotient must shrink to 0 too — This misapplies limit reasoning by taking the limits of numerator and denominator separately and dividing the results, but the quotient $h/h$ is already the constant $1$ for every positive $h$, so there is no separate shrinking to track.
undefined, because $|h|$ is not a differentiable expression — This confuses differentiability of $|x|$ as a whole, which does fail at 0, with whether this particular one-sided limit can be computed, it can, and gives $1$ cleanly.
Apply the same LHD/RHD method to $f(x)=|x-2|$ at $x=2$. Using $h$ as the small offset so the point becomes $2+h$, what are LHD and RHD there, and is $f$ differentiable at $x=2$?
LHD $=1$ and RHD $=-1$; since they disagree, $f$ is not differentiable at $x=2$
LHD $=-1$ and RHD $=1$; since they agree in magnitude, $f$ is differentiable at $x=2$ with $f'(2)=1$
LHD $=-1$ and RHD $=1$, as for $|x|$ at $0$; they disagree, so $f$ is not differentiable at $x=2$
LHD $=1$ and RHD $=1$; since they agree, $f$ is differentiable at $x=2$ with $f'(2)=1$
Check your answer
LHD $=1$ and RHD $=-1$; since they disagree, $f$ is not differentiable at $x=2$ — This computes the correct disagreement but swaps which side gives which sign, LHD, from $h$ approaching 0 from below, gives $-1$, not $+1$, matching the same sign pattern as $|x|$ at 0.
LHD $=-1$ and RHD $=1$; since they agree in magnitude, $f$ is differentiable at $x=2$ with $f'(2)=1$ — This notices the two values have the same magnitude but treats matching magnitude as if it were agreement, LHD and RHD must be equal, not just equal in size, before $f'(2)$ is defined; $-1$ and $1$ are not the same number.
✓ LHD $=-1$ and RHD $=1$, as for $|x|$ at $0$; they disagree, so $f$ is not differentiable at $x=2$ — (C) The shifted point $2+h$ makes $f(2+h)-f(2) = |h|-0 = |h|$, exactly the same difference quotient as $|x|$ at $0$, so LHD $=-1$ and RHD $=1$ again, and $f$ is not differentiable at $x=2$ despite being continuous there.
LHD $=1$ and RHD $=1$; since they agree, $f$ is differentiable at $x=2$ with $f'(2)=1$ — This miscomputes LHD as $1$ instead of $-1$, the left-hand offset still gives $|h|=-h$ for $h<0$, producing $-1$, not $1$.
Two functions in a row: 4 goes in, the square root gives 2, the sine gives 0.909.
Chapter 1’s two functions in a row are back. Feed a number into the first function, and its output goes straight into the second. Say the first function takes the square root and the second takes the sine. Feed in 4. The first function gives 2, and the second gives the sine of 2. Together the two make one function.
0.01 nudge
→
√
slope 1/4
→
sin
slope cos 2 ≈ −0.416
→
-0.00104
Nudge 4 by 0.01. The first move is 0.0025, a quarter of it; the second is −0.00104, cos 2 times that.
Now nudge the input a little. The first output moves by about its own slope times the nudge. That move is the nudge for the second function. So the final output moves by about the second slope times that. The two slopes multiply.
√x
sin u
sin √x
Three slopes: 0.25 for the inner function, −0.416 for the outer, and −0.104, their product, for the two in a row.
That is the CHAIN RULE. Call the inner function $u = g(x)$ and the outer one $y = f(u)$. Then
$d y/d x = d y/d u \cdot d u/d x$.
g(x) = 2
→
cos
→
-0.416
x = 4
→
cos
→
-0.654
The outer slope is taken at the inner output, 2, not at 4: cos 2 = −0.416, while cos 4 = −0.654.
In the other notation, $d y/d x = f'(g(x)) \cdot g'(x)$. The outer slope is taken at the inner output, $g(x)$, not at $x$. The rule needs $g$ differentiable at $x$ and $f$ differentiable at $g(x)$.
Both ways give 8x + 4. At x = 1 the tangent to (2x + 1)² has slope 12.
Let us check it on a function you can also expand. Take $y = (2x + 1)^2$.
Worked example
Two ways to the slope of $(2x + 1)^2$
$u = 2x + 1$ Name the inner function, the one applied first.
$d u/d x = 2$ Differentiate the inner function with respect to $x$.
$y = u^2$ The outer function squares $u$.
$d y/d u = 2u$ Differentiate the outer function with respect to $u$, not $x$.
$d y/d x = d y/d u \cdot d u/d x = 2u \cdot 2$ Multiply the two slopes: this is the chain rule.
*$d y/d x = 4(2x + 1) = 8x + 4$* Put $u = 2x + 1$ back, so the answer is in $x$, then open the bracket.
Check: $y = 4x^2 + 4x + 1$, so $d y/d x = 8x + 4$ Expand first and differentiate term by term: the two answers match.
With the chain rule, you can find the slope of any function built from functions whose slopes you know.
Your turn: differentiate $y = \sin(3x)$. Name the inner function first.
A SCHOLAR INDIA REMEMBERS
A method is worth having when it works for the next case too. Try three functions in a row: double, then square root, then sine. Feed in $x = 2$. Doubling gives $4$, the root gives $2$, and the sine gives $\sin 2$. The three slopes are $2$, then $1/(2 \sqrt{4}) = 1/4$, then $\cos 2$. Multiply all three. You get $2 \cdot 1/4 \cdot \cos 2$, about $-0.208$. The rule just takes one more factor.
Aryabhata · the hoopoe
Check yourself
For a composite function $y = f(g(x))$, what does the chain rule state for $d y/d x$?
$d y/d x = f'(g(x)) + g'(x)$
$d y/d x = f'(x) \cdot g'(x)$
$d y/d x = f(g'(x))$
$d y/d x = f'(g(x)) g'(x)$
Check your answer
$d y/d x = f'(g(x)) + g'(x)$ — This adds the two derivatives instead of multiplying them. The chain rule is a product of two rates, the outer function’s slope at the inner value, times the inner function’s own slope, never a sum.
$d y/d x = f'(x) \cdot g'(x)$ — This evaluates $f'$ at $x$, the composite’s own input, but the chain rule needs $f'$ evaluated at $g(x)$, the inner function’s output, since the outer function only ever acts on the inner function’s output, not on $x$ directly.
$d y/d x = f(g'(x))$ — This leaves the outer function $f$ entirely undifferentiated, differentiating only $g$ inside its argument, the chain rule requires differentiating both layers, $f'$ evaluated at $g(x)$, then multiplied by $g'(x)$.
✓ $d y/d x = f'(g(x)) g'(x)$ — (D) The chain rule differentiates the outer function $f$ at the inner value $g(x)$, keeping $g(x)$ intact as the input to $f$, then multiplies by the inner function’s own derivative $g'(x)$.
In the chain rule for $y = f(g(x))$, what happens to the inner function $g$ when computing the composite’s derivative?
$g(x)$ is plugged into the outer function’s derivative unchanged, while $g$’s own derivative is multiplied in afterward
$g(x)$ is dropped once the outer function’s derivative is found, since that alone determines the composite’s rate of change
the inner function’s derivative is plugged into the outer function’s derivative in place of $g(x)$ itself
$g$ is differentiated twice: once to produce the value plugged into the outer function’s derivative, and again as the multiplying factor
Check your answer
✓ $g(x)$ is plugged into the outer function’s derivative unchanged, while $g$’s own derivative is multiplied in afterward — (A) $g(x)$ stays intact as the value fed into the outer function’s derivative, while the inner function’s own derivative is computed separately and multiplied in, the two roles never merge.
$g(x)$ is dropped once the outer function’s derivative is found, since that alone determines the composite’s rate of change — Dropping $g(x)$ after finding the outer function’s derivative throws away the multiplication by the inner function’s own derivative that the chain rule requires, the outer derivative alone gives only part of the answer.
the inner function’s derivative is plugged into the outer function’s derivative in place of $g(x)$ itself — The outer function’s derivative must be evaluated at $g(x)$, the inner function’s actual value, plugging in its derivative instead evaluates the outer function’s slope at the wrong input entirely.
$g$ is differentiated twice: once to produce the value plugged into the outer function’s derivative, and again as the multiplying factor — $g$ is differentiated only once, to produce the multiplying factor. The value plugged into the outer function’s derivative is $g(x)$ itself, the original inner function, not a second differentiation of it.
Writing $u = g(x)$ so $y = f(u)$, the chain rule can be rewritten in a layered form using $u$ and $y$. Which is the correct layered form for $d y/d x$?
$d y/d x = ((d y)/(d u)) \cdot ((d u)/(d x))$
$d y/d x = ((d y)/(d u)) \cdot ((d x)/(d u))$
$d y/d x = ((d y)/(d u))/((d u)/(d x))$
$d y/d x = d y/d u$
Check your answer
✓ $d y/d x = ((d y)/(d u)) \cdot ((d u)/(d x))$ — (A) The layered chain rule multiplies the outer rate $d y/d u$ by the inner rate $d u/d x$, the same product structure as before, just relabeled with $u$ standing in for $g(x)$.
$d y/d x = ((d y)/(d u)) \cdot ((d x)/(d u))$ — $d x/d u$ is the reciprocal of $d u/d x$, not $d u/d x$ itself. Substituting the reciprocal changes the value the formula computes.
$d y/d x = ((d y)/(d u))/((d u)/(d x))$ — Dividing $d y/d u$ by $d u/d x$ is not the chain rule, which multiplies. It is not the parametric formula either: that one divides $d y/d t$ by $d x/d t$, two rates against the same parameter $t$.
$d y/d x = d y/d u$ — This keeps only the outer rate $d y/d u$ and drops $d u/d x$ entirely. Without multiplying by how fast $u$ itself changes with $x$, the chain through to $x$ is broken.
The chain rule’s layered form $d y/d x = ((d y)/(d u)) \cdot ((d u)/(d x))$ reappears repeatedly in what follows. Which two later techniques rely on it directly?
testing continuity via LHL/RHL and testing differentiability via LHD/RHD
implicit differentiation and logarithmic differentiation
Rolle’s theorem and the Mean Value Theorem
the derivative of $e^x$ and the derivative of $\ln x$
Check your answer
testing continuity via LHL/RHL and testing differentiability via LHD/RHD — LHL/RHL and LHD/RHD test whether limits or one-sided derivatives exist and agree, they come from the definitions of continuity and differentiability, not from applying the chain rule’s layered form.
✓ implicit differentiation and logarithmic differentiation — (B) Implicit differentiation applies the chain rule to every $y$-term, since $y$ is itself a function of $x$; logarithmic differentiation applies it after taking logarithms of both sides. Both lean directly on the layered $d y/d u$ times $d u/d x$ form.
Rolle’s theorem and the Mean Value Theorem — Rolle’s theorem and the Mean Value Theorem are about the existence of a point with a certain derivative value, not about differentiating a composite function, they do not use the layered chain-rule form at all.
the derivative of $e^x$ and the derivative of $\ln x$ — $d/d x (e^x) = e^x$ and $d/d x (\ln x) = 1/x$ are standard results stated on their own. The chain rule uses them, as in $d/d x (e^{2x}) = 2 e^{2x}$, but they do not rest on its layered form.
The curve y equals the sine of the square root of x. At x equals 1 its height is 0.841 and its slope 0.270.
1
→
√
slope 1/2
→
sin
slope cos 1 ≈ 0.540
→
0.841
At x = 1 the slopes are 1/2 and 0.540. Multiplied, 0.270, the slope of the tangent.
Worked example
Differentiate $y = \sin(\sqrt{x})$ using the chain rule
$y = \sin(\sqrt{x})$ Two functions in a row: the square root first, then the sine. Take $x > 0$, where the square root has a slope.
$u = \sqrt{x}$, so $y = \sin(u)$ Call the inner function’s output $u$. Then the outer function is the sine of $u$.
$d y/d u = \cos(u)$ The slope of the outer function, the sine, with respect to $u$.
$d u/d x = 1/(2 \sqrt{x})$ The slope of the inner function. The square root is $x^{1/2}$, so its slope is $1/2 x^{-1/2}$.
$d y/d x = ((d y)/(d u)) \cdot ((d u)/(d x)) = \cos(u) \cdot 1/(2 \sqrt{x})$ The chain rule multiplies the two slopes.
$d y/d x = \cos(\sqrt{x}) / (2 \sqrt{x})$ Put $u = \sqrt{x}$ back, so the answer is in $x$ alone.
$x = 1$: $\cos(1)/(2 \sqrt{1}) \approx 0.5403/2 \approx 0.270$ Check at $x = 1$. The tangent in the figure rises 0.270 for every unit step along the x-axis.
Check yourself
To differentiate $y = \sin(\sqrt{x})$ with the chain rule, which substitution correctly splits the composite into two layers?
$u = \sin(\sqrt{x})$, so $y = u$
$u = x$, so $y = \sin(\sqrt{u})$
$u = \sqrt{x}$, so $y = \sin(u)$
$u = \sin(x)$, so $y = \sqrt{u}$
Check your answer
$u = \sin(\sqrt{x})$, so $y = u$ — Setting $u$ equal to the entire expression leaves nothing left to differentiate as an outer function, $y = u$ is a relabeling, not a split into two layers the chain rule can act on.
$u = x$, so $y = \sin(\sqrt{u})$ — Setting $u = x$ changes nothing; $y = \sin(\sqrt{u})$ is still a composite of two functions, not the single-layer form the chain rule needs before differentiating.
✓ $u = \sqrt{x}$, so $y = \sin(u)$ — (C) $\sqrt{x}$ is computed first, so it is the inner function; $\sin$ is then applied to that result, making it the outer function. Setting $u = \sqrt{x}$ gives $y = \sin(u)$, a genuine single outer function applied to a single inner value.
$u = \sin(x)$, so $y = \sqrt{u}$ — This swaps which function is applied first. In $\sin(\sqrt{x})$, $\sqrt{x}$ is computed first and fed into $\sin$, not the other way round, so the inner layer is $\sqrt{x}$, not $\sin(x)$.
Differentiating $y = \sin(\sqrt{x})$ with $u = \sqrt{x}$, $d y/d u = \cos(u)$ and $d u/d x = 1/(2 \sqrt{x})$. What is $d y/d x$?
$\cos(x)/(2 \sqrt{x})$
$\sin(\sqrt{x})/(2 \sqrt{x})$
$\cos(\sqrt{x}) \cdot 2 \sqrt{x}$
$\cos(\sqrt{x})/(2 \sqrt{x})$
Check your answer
$\cos(x)/(2 \sqrt{x})$ — $\cos(u)$ means $\cos$ evaluated at $u = \sqrt{x}$, which is $\cos(\sqrt{x})$, not $\cos(x)$. Forgetting to substitute $u$ back in leaves the wrong argument inside cosine.
$\sin(\sqrt{x})/(2 \sqrt{x})$ — $\sin(\sqrt{x})$ is $y$ itself, the original function, not its derivative. Differentiating the outer function $\sin$ gives $\cos$, not a repeat of $\sin$.
$\cos(\sqrt{x}) \cdot 2 \sqrt{x}$ — $d u/d x = 1/(2 \sqrt{x})$, not $2 \sqrt{x}$, its reciprocal. Multiplying by the reciprocal instead of the actual derivative of $\sqrt{x}$ inverts the second factor.
✓ $\cos(\sqrt{x})/(2 \sqrt{x})$ — (D) $d y/d x = ((d y)/(d u)) \cdot ((d u)/(d x)) = \cos(u) \cdot 1/(2 \sqrt{x})$; substituting $u = \sqrt{x}$ back in gives $\cos(\sqrt{x})/(2 \sqrt{x})$.
Using the same two-layer method as $y = \sin(\sqrt{x})$, what is $d y/d x$ for $y = \cos(\sqrt{x})$?
$-\sin(\sqrt{x}) / (2 \sqrt{x})$
$-\cos(\sqrt{x}) / (2 \sqrt{x})$
$-\sin(\sqrt{x}) \cdot 2 \sqrt{x}$
$\sin(\sqrt{x})/(2 \sqrt{x})$
Check your answer
✓ $-\sin(\sqrt{x}) / (2 \sqrt{x})$ — (A) With $u = \sqrt{x}$, $d y/d u = -\sin(u)$ and $d u/d x = 1/(2 \sqrt{x})$; multiplying gives $d y/d x = -\sin(\sqrt{x}) / (2 \sqrt{x})$, the same method as $\sin(\sqrt{x})$ with cosine’s derivative carrying its negative sign through.
$-\cos(\sqrt{x}) / (2 \sqrt{x})$ — Differentiating $\cos(u)$ gives $-\sin(u)$, not $\cos(u)$ again, $\cos(u)$ is the original function being differentiated, not its own derivative.
$-\sin(\sqrt{x}) \cdot 2 \sqrt{x}$ — $d u/d x = 1/(2 \sqrt{x})$ here too, not $2 \sqrt{x}$, multiplying by the reciprocal inverts the inner derivative’s fraction.
$\sin(\sqrt{x})/(2 \sqrt{x})$ — $d/d u (\cos(u)) = -\sin(u)$, with a negative sign attached. Dropping the sign treats cosine’s derivative as if it matched sine’s, but the two differ by exactly this sign.
The inverse sine key, given 0.5, shows 30: one angle, though many angles have a sine of 0.5.
The level 0.5 crosses the sine curve many times but the principal branch once, at π/6, which is 30 degrees.
Press the inverse sine key on your calculator and enter 0.5. In degree mode it shows 30, an angle in degrees. Many angles have a sine of 0.5, but the key gives just one. That is Chapter 2’s inverse sine: the angle whose sine is the number, taken from the principal branch. Now ask how fast that angle changes as the number changes. You do not need a new limit for it.
y = sin x
y = sin⁻¹x
The slope of sin⁻¹x at 0.5 is 1 over the slope of sin at π/6: 1/0.866 = 1.155.
Write it as an equation. Take $y = \sin^{-1} x$, for $x$ between $-1$ and $1$. Then $y$ is on the principal branch, from $-\pi/2$ to $\pi/2$.
Why it works
The slope of $\sin^{-1} x$
$\sin y = x$ Take the sine of both sides: on the principal branch, sine undoes $\sin^{-1}$.
$d/d x (\sin y) = d/d x (x)$ Differentiate both sides with respect to $x$, keeping $y$ as a function of $x$.
$\cos y \cdot d y/d x = 1$ Use the chain rule on the left, since $x$ gives $y$ and then sine gives $\sin y$.
*$d y/d x = 1/(\cos y)$* Divide both sides by $\cos y$, which needs $\cos y \neq 0$.
On the principal branch, from −π/2 to π/2, cos y is never negative. So it is the positive square root.
On the principal branch, $\cos y$ is never negative. So $\cos y$ is the positive square root of $1 - \sin^2 y$, and $\sin y = x$:
$\cos y = \sqrt{1 - \sin^2 y} = \sqrt{1 - x^2}$.
Put this into $d y/d x = 1/(\cos y)$. That gives
$d/d x (\sin^{-1} x) = 1/\sqrt{1 - x^2}$, for $-1 < x < 1$.
The slope 1/√(1 − x²): 1 at 0, 1.155 at 0.5, 2.294 at 0.9. It grows without limit toward the ends.
has a value
has a value
sin⁻¹x has a value from −1 to 1. Its slope has a value only strictly between them.
At $x = 1$ and at $x = -1$ the bottom is 0, so the formula has no value there.
at 0.2: sin⁻¹x = 0.201, x = 0.2
Near 0, sin⁻¹x is close to x: at 0.2 it is 0.201. Both have slope 1 at 0.
Check it at $x = 0$. The formula gives $1/\sqrt{1} = 1$. Near 0, $\sin y$ is close to $y$. So $\sin^{-1} x$ is close to $x$, whose slope is 1.
The six inverse trig functions, each derivative with the values of x where it holds. Each pair adds to pi over 2, so its derivatives differ in sign.
The other five come from the same three moves: write the function as an equation, differentiate both sides, then swap back to $x$. Here they are for $\cos^{-1} x$, for $x$ between $-1$ and $1$.
Why it works
The slope of $\cos^{-1} x$
$\cos y = x$ Write the inverse as an equation, with $y$ from $0$ to $\pi$.
$-\sin y \cdot d y/d x = 1$ Differentiate both sides with respect to $x$, using the chain rule on the left.
$d y/d x = -1/(\sin y)$ Divide both sides by $-\sin y$, which needs $\sin y \neq 0$.
$\sin y = \sqrt{1 - \cos^2 y} = \sqrt{1 - x^2}$ On this branch $\sin y$ is never negative, so take the positive root.
*$d/d x (\cos^{-1} x) = -1/\sqrt{1 - x^2}$* Swap back to $x$: the slope of $\sin^{-1} x$ with a minus sign.
The figure below shows why the sign flips.
Two more hints.
For $\tan^{-1} x$, start from $\tan y = x$ and use $\sec^2 y = 1 + \tan^2 y = 1 + x^2$.
For $\sec^{-1} x$, start from $\sec y = x$. Differentiate to get $d y/d x = 1/(\sec y \tan y)$.
On the principal branch $\tan y$ has the same sign as $x$, so $\sec y \tan y = \left|x\right| \sqrt{x^2 - 1}$: that is where the $\left|x\right|$ in the answer comes from.
The other two are the partners in each pair, each the negative of the one above it.
sin⁻¹x + cos⁻¹x = π/2, a constant. So their slopes add to 0: at 0.5, 1.155 and −1.155.
$d/d x (\cos^{-1} x) = -1/\sqrt{1 - x^2}$, for $-1 < x < 1$.
$d/d x (\tan^{-1} x) = 1/(1 + x^2)$, for every real $x$.
$d/d x (\cot^{-1} x) = -1/(1 + x^2)$, for every real $x$.
$d/d x (\sec^{-1} x) = 1/(\left|x\right| \sqrt{x^2 - 1})$, for $\left|x\right| > 1$.
$d/d x (\text{cosec}^{-1} x) = -1/(\left|x\right| \sqrt{x^2 - 1})$, for $\left|x\right| > 1$.
Why do the pairs differ only in sign? From Chapter 2, $\sin^{-1} x + \cos^{-1} x = \pi/2$. A constant has slope 0, so the two slopes add to 0. The same holds for $\tan^{-1}$ and $\cot^{-1}$, and for $\sec^{-1}$ and $\text{cosec}^{-1}$.
Your turn: find the slope of $\tan^{-1} x$ the same way. Start from $\tan y = x$.
Check yourself
What is the key difference between $d/d x (\sin^{-1} x)$ and $d/d x (\cos^{-1} x)$, both valid for $|x|<1$?
they are negatives of each other: $1/\sqrt{1-x^2}$ and $-1/\sqrt{1-x^2}$
they are identical, since $\sin^{-1} x + \cos^{-1} x$ is constant
they hold on different domains: $|x|<1$ for sine, $|x|>1$ for cosine
they have different denominators: $\sqrt{1-x^2}$ for sine, $1+x^2$ for cosine
Check your answer
✓ they are negatives of each other: $1/\sqrt{1-x^2}$ and $-1/\sqrt{1-x^2}$ — (A) $d/d x (\sin^{-1} x) = 1/\sqrt{1-x^2}$ and $d/d x (\cos^{-1} x) = -1/\sqrt{1-x^2}$, same denominator, opposite sign, both for $|x|<1$.
they are identical, since $\sin^{-1} x + \cos^{-1} x$ is constant — $\sin^{-1} x + \cos^{-1} x$ is constant, so differentiating both sides gives $0$, but that means the two derivatives are negatives of each other, summing to zero, not equal to each other.
they hold on different domains: $|x|<1$ for sine, $|x|>1$ for cosine — $|x|>1$ is the domain for $\sec^{-1} x$ and $\text{cosec}^{-1} x$, not $\cos^{-1} x$, sine and cosine inverse share the same domain, $|x|<1$.
they have different denominators: $\sqrt{1-x^2}$ for sine, $1+x^2$ for cosine — $1+x^2$ is the denominator pattern for $\tan^{-1} x$ and $\cot^{-1} x$, not for sine or cosine inverse, $\sin^{-1} x$ and $\cos^{-1} x$ both use $\sqrt{1-x^2}$.
What are $d/d x (\tan^{-1} x)$ and $d/d x (\cot^{-1} x)$, and over what domain do they hold?
$1/(1+x^2)$ and $-1/(1+x^2)$, for all $x \in \mathbb{R}$
$1/\sqrt{1-x^2}$ and $-1/\sqrt{1-x^2}$, for $|x|<1$
$1/(1+x^2)$ and $-1/(1+x^2)$, but only for $|x|<1$
$1/(|x| \sqrt{x^2-1})$ and $-1/(|x| \sqrt{x^2-1})$, for $|x|>1$
Check your answer
✓ $1/(1+x^2)$ and $-1/(1+x^2)$, for all $x \in \mathbb{R}$ — (A) $d/d x (\tan^{-1} x) = 1/(1+x^2)$ and $d/d x (\cot^{-1} x) = -1/(1+x^2)$, and since $1+x^2$ is never zero or negative, both hold for every real $x$.
$1/\sqrt{1-x^2}$ and $-1/\sqrt{1-x^2}$, for $|x|<1$ — $\sqrt{1-x^2}$ in the denominator, with domain $|x|<1$, belongs to $\sin^{-1} x$ and $\cos^{-1} x$, $\tan^{-1} x$ and $\cot^{-1} x$ use $1+x^2$ instead, an expression that is never zero or negative for any real $x$.
$1/(1+x^2)$ and $-1/(1+x^2)$, but only for $|x|<1$ — $1+x^2$ is always positive for every real $x$, so there is no domain to restrict, $\tan^{-1} x$ and $\cot^{-1} x$ are differentiable for all $x \in \mathbb{R}$, unlike the other two pairs.
$1/(|x| \sqrt{x^2-1})$ and $-1/(|x| \sqrt{x^2-1})$, for $|x|>1$ — $|x| \sqrt{x^2-1}$ in the denominator, with domain $|x|>1$, belongs to $\sec^{-1} x$ and $\text{cosec}^{-1} x$, $\tan^{-1} x$ and $\cot^{-1} x$ use the simpler $1+x^2$, valid everywhere.
Why do $d/d x (\sec^{-1} x) = 1/(|x| \sqrt{x^2-1})$ and $d/d x (\text{cosec}^{-1} x) = -1/(|x| \sqrt{x^2-1})$ require $|x|>1$?
because $|x|$ in the denominator must be positive, which fails only at $x=0$
because $\sqrt{x^2-1}$ sits in the denominator and must be nonzero, which requires $|x|>1$
because the derivative must match the domain of $\tan^{-1} x$, which $\sec^{-1} x$ is built from
because $\sec^{-1} x$ and $\text{cosec}^{-1} x$ are undefined for negative $x$
Check your answer
because $|x|$ in the denominator must be positive, which fails only at $x=0$ — $|x|$ is zero only at $x=0$, a single point, that alone would restrict just one value, not the whole band $-1<x<1$ that is actually excluded.
✓ because $\sqrt{x^2-1}$ sits in the denominator and must be nonzero, which requires $|x|>1$ — (B) $\sqrt{x^2-1}$ is in the denominator, so it must be defined and nonzero: $x^2-1 > 0$, that is $x^2 > 1$, which holds exactly when $|x|>1$.
because the derivative must match the domain of $\tan^{-1} x$, which $\sec^{-1} x$ is built from — This formula stands on its own; nothing about a different function’s domain enters into it. $\sqrt{x^2-1}$ sits in the denominator, so it needs $x^2-1 > 0$, independent of any other function.
because $\sec^{-1} x$ and $\text{cosec}^{-1} x$ are undefined for negative $x$ — $\sec^{-1} x$ and $\text{cosec}^{-1} x$ are defined for $x \leq -1$ as well as $x \geq 1$, negative $x$ is not the problem. The band $-1<x<1$ is excluded, not all negative numbers.
Which domain correctly matches its inverse-trig derivative formula?
$|x|<1$ for $\tan^{-1} x$ and $\cot^{-1} x$; all $x \in \mathbb{R}$ for $\sin^{-1} x$ and $\cos^{-1} x$; $|x|>1$ for $\sec^{-1} x$ and $\text{cosec}^{-1} x$
$|x|<1$ for $\sin^{-1} x$ and $\cos^{-1} x$; all $x \in \mathbb{R}$ for $\sec^{-1} x$ and $\text{cosec}^{-1} x$; $|x|>1$ for $\tan^{-1} x$ and $\cot^{-1} x$
$|x|<1$ for $\sin^{-1} x$ and $\cos^{-1} x$; all $x \in \mathbb{R}$ for $\tan^{-1} x$ and $\cot^{-1} x$; $|x|>1$ for $\sec^{-1} x$ and $\text{cosec}^{-1} x$
$|x|>1$ for $\sin^{-1} x$ and $\cos^{-1} x$; all $x \in \mathbb{R}$ for $\tan^{-1} x$ and $\cot^{-1} x$; $|x|<1$ for $\sec^{-1} x$ and $\text{cosec}^{-1} x$
Check your answer
$|x|<1$ for $\tan^{-1} x$ and $\cot^{-1} x$; all $x \in \mathbb{R}$ for $\sin^{-1} x$ and $\cos^{-1} x$; $|x|>1$ for $\sec^{-1} x$ and $\text{cosec}^{-1} x$ — $\sin^{-1} x$ and $\cos^{-1} x$ need $|x|<1$, since their denominator $\sqrt{1-x^2}$ requires $1-x^2 \geq 0$. $\tan^{-1} x$ and $\cot^{-1} x$ face no such restriction, since their denominator $1+x^2$ is always positive, that pair is the one that holds for all real $x$, not sine and cosine.
$|x|<1$ for $\sin^{-1} x$ and $\cos^{-1} x$; all $x \in \mathbb{R}$ for $\sec^{-1} x$ and $\text{cosec}^{-1} x$; $|x|>1$ for $\tan^{-1} x$ and $\cot^{-1} x$ — $\sec^{-1} x$ and $\text{cosec}^{-1} x$ need $|x|>1$, since their denominator involves $\sqrt{x^2-1}$, which requires $x^2 \geq 1$. $\tan^{-1} x$ and $\cot^{-1} x$ have no such restriction, $1+x^2$ is never zero or negative.
✓ $|x|<1$ for $\sin^{-1} x$ and $\cos^{-1} x$; all $x \in \mathbb{R}$ for $\tan^{-1} x$ and $\cot^{-1} x$; $|x|>1$ for $\sec^{-1} x$ and $\text{cosec}^{-1} x$ — (C) Each pair’s domain traces back to its denominator, $\sqrt{1-x^2}$ forces $|x|<1$ for sine and cosine inverse, $1+x^2$ is always positive so tangent and cotangent inverse need no restriction, and $\sqrt{x^2-1}$ forces $|x|>1$ for secant and cosecant inverse.
$|x|>1$ for $\sin^{-1} x$ and $\cos^{-1} x$; all $x \in \mathbb{R}$ for $\tan^{-1} x$ and $\cot^{-1} x$; $|x|<1$ for $\sec^{-1} x$ and $\text{cosec}^{-1} x$ — $\sin^{-1} x$ and $\cos^{-1} x$ need $|x|<1$, not $|x|>1$, since $\sqrt{1-x^2}$ requires $1-x^2 \geq 0$. $\sec^{-1} x$ and $\text{cosec}^{-1} x$ need the reverse, $|x|>1$, since their expression is $\sqrt{x^2-1}$.
The inverse sine of 2x exists only between minus one half and one half. At x equals one quarter its slope is 2.309.
1/4
→
×2
slope 2
→
sin⁻¹
slope 1.155
→
π/6
Two functions in a row: ×2 then sin⁻¹. At x = 1/4 the slopes are 2 and 1.155, so the slope is 2.309.
(−1/2, 1/2)
(−1, 1)
2x lies strictly between −1 and 1 exactly when x lies strictly between −1/2 and 1/2.
Worked example
Differentiate $y = \sin^{-1}(2x)$
$y = \sin^{-1}(2x)$ Two functions in a row: double $x$, then take the angle whose sine is that number.
$u = 2x$ Call the inner output $u = 2x$. Then $y = \sin^{-1} u$.
$d y/d u = 1/\sqrt{1-u^2}$ The slope of $\sin^{-1} u$, from the formula above. It needs $-1 < u < 1$.
$d u/d x = 2$ The inner function doubles, so its slope is 2.
$d y/d x = d y/d u \cdot d u/d x = 2/\sqrt{1-u^2}$ Multiply the two slopes, as the chain rule says.
$d y/d x = 2/\sqrt{1-4x^2}$ Put $u = 2x$ back. The condition $-1 < 2x < 1$ becomes $\left|x\right| < 1/2$.
$x = 1/4$: $2/\sqrt{1 - 4 \cdot 1/16} = 2/\sqrt{3/4} = 4/\sqrt{3} \approx 2.309$ Check at $x = 1/4$. The answer matches the slope the figure draws there.
Worked example
Differentiate $y = \tan^{-1}(2x / (1 - x^2))$, for $-1 < x < 1$
$y = \tan^{-1}(2x / (1 - x^2))$, $-1 < x < 1$ The inside has the shape of $\tan 2 \theta = 2 \tan \theta / (1 - \tan^2 \theta)$. So put $x = \tan \theta$.
$x = \tan \theta$, with $-\pi/4 < \theta < \pi/4$ For $-1 < x < 1$, the angle $\theta = \tan^{-1} x$ lies between $-\pi/4$ and $\pi/4$.
$2x / (1 - x^2) = 2 \tan \theta / (1 - \tan^2 \theta) = \tan 2 \theta$ The double-angle formula for tangent.
$y = \tan^{-1}(\tan 2 \theta) = 2 \theta$ $2 \theta$ lies between $-\pi/2$ and $\pi/2$, the principal range of $\tan^{-1}$ from Chapter 2, so $\tan^{-1}$ undoes $\tan$ here.
$d y/d x = 2/(1 + x^2)$ The slope of $\tan^{-1} x$ is $1/(1 + x^2)$, so twice it is $2/(1 + x^2)$.
$x = 1/2$: $\tan^{-1}(4/3) \approx 0.9273$ and $2 \tan^{-1}(1/2) \approx 0.9273$ Check the substitution at $x = 1/2$, where $2x / (1 - x^2) = 1 / (3/4) = 4/3$. The two sides agree.
Check yourself
To differentiate $y = \sin^{-1}(2x)$ using the chain rule, which substitution and pair of derivatives correctly set up the problem?
$u = 2x$, so $d y/d u = 1/\sqrt{1-u^2}$ and $d u/d x = 2x$
$u = 2x$, so $d y/d u = 1/\sqrt{1-u^2}$ and $d u/d x = 2$
$u = \sin^{-1}(2x)$, so $d y/d u = 1$ and $d u/d x = 2/\sqrt{1-4x^2}$
$u = 2x$, so $d y/d u = 2/\sqrt{1-u^2}$ and $d u/d x = 1$
Check your answer
$u = 2x$, so $d y/d u = 1/\sqrt{1-u^2}$ and $d u/d x = 2x$ — $d u/d x$ is the derivative of $2x$ with respect to $x$, which is the constant $2$, not $2x$. Differentiating $2x$ should remove the $x$ entirely, not leave it in place.
✓ $u = 2x$, so $d y/d u = 1/\sqrt{1-u^2}$ and $d u/d x = 2$ — (B) Setting $u = 2x$ splits the composite cleanly, $d y/d u = 1/\sqrt{1-u^2}$ is the standard arcsine-derivative pattern applied to $u$, and $d u/d x = 2$ is the derivative of the inner linear function $2x$.
$u = \sin^{-1}(2x)$, so $d y/d u = 1$ and $d u/d x = 2/\sqrt{1-4x^2}$ — Setting $u$ equal to the whole answer skips the substitution’s purpose entirely, $y = u$ becomes a trivial relabeling with nothing left for the chain rule to act on.
$u = 2x$, so $d y/d u = 2/\sqrt{1-u^2}$ and $d u/d x = 1$ — The factor of $2$ comes from differentiating the inner function $2x$, so it belongs in $d u/d x$, not in $d y/d u$. $d y/d u = 1/\sqrt{1-u^2}$ is just the standard arcsine-derivative pattern, with no extra factor attached.
Multiplying $d y/d u = 1/\sqrt{1-u^2}$ by $d u/d x = 2$ and substituting $u = 2x$ back in, what is $d y/d x$ for $y = \sin^{-1}(2x)$, and over what domain is it valid?
$d y/d x = 1/\sqrt{1-4x^2}$, valid for $|x| < 1/2$
$d y/d x = 2/\sqrt{1-4x^2}$, valid for $|x| < 1/2$
$d y/d x = 2/\sqrt{1-2x^2}$, valid for $|x| < 1/2$
$d y/d x = 2/\sqrt{1-4x^2}$, valid for $|x| < 1$
Check your answer
$d y/d x = 1/\sqrt{1-4x^2}$, valid for $|x| < 1/2$ — The multiplication by $d u/d x = 2$ is part of the chain rule, not optional, dropping it leaves only the outer derivative’s factor, understating the true rate of change by half.
✓ $d y/d x = 2/\sqrt{1-4x^2}$, valid for $|x| < 1/2$ — (B) $d y/d x = 1/\sqrt{1-u^2} \cdot 2 = 2/\sqrt{1-4x^2}$ after substituting $u = 2x$, and since $1-4x^2>0$ requires $x^2<1/4$, the domain is $|x| < 1/2$.
$d y/d x = 2/\sqrt{1-2x^2}$, valid for $|x| < 1/2$ — $u = 2x$ means $u^2 = (2x)^2 = 4x^2$, not $2x^2$, squaring the coefficient along with the variable is required when substituting back in.
$d y/d x = 2/\sqrt{1-4x^2}$, valid for $|x| < 1$ — The domain scales along with the substitution, $u = 2x$ must satisfy $|u|<1$, i.e. $|2x|<1$, which gives $|x| < 1/2$, not the unscaled $|x| < 1$ that applies to plain $\sin^{-1} x$.
Using the same substitution method as $\sin^{-1}(2x)$, what is $d y/d x$ for $y = \tan^{-1}(2x)$?
$d y/d x = 2/(1+4x^2)$, valid for all $x \in \mathbb{R}$
$d y/d x = 2/(1+4x^2)$, valid only for $|x| < 1/2$
$d y/d x = 2/\sqrt{1-4x^2}$, valid for $|x| < 1/2$
$d y/d x = 1/(1+4x^2)$, valid for all $x \in \mathbb{R}$
Check your answer
✓ $d y/d x = 2/(1+4x^2)$, valid for all $x \in \mathbb{R}$ — (A) With $u = 2x$, $d y/d u = 1/(1+u^2)$ and $d u/d x = 2$, giving $d y/d x = 2/(1+4x^2)$; since $1+u^2$ is never zero or negative, no domain restriction applies, the formula holds for all $x \in \mathbb{R}$.
$d y/d x = 2/(1+4x^2)$, valid only for $|x| < 1/2$ — $\tan^{-1}(u)$ uses denominator $1+u^2$, which is always positive, unlike $\sin^{-1}(u)$, it carries no domain restriction to inherit. The $|x| < 1/2$ restriction belongs to the arcsine version, not this one.
$d y/d x = 2/\sqrt{1-4x^2}$, valid for $|x| < 1/2$ — $\sqrt{1-4x^2}$ is the arcsine-derivative pattern, not the arctangent one. $\tan^{-1}(u)$ differentiates to $1/(1+u^2)$, giving $1+4x^2$ in the denominator after substitution, not a square root at all.
$d y/d x = 1/(1+4x^2)$, valid for all $x \in \mathbb{R}$ — $d u/d x = 2$, from differentiating the inner function $2x$, and that factor must still multiply in, dropping it understates the derivative by half, the same error as in the arcsine case.
e to the x is 2.718 high at x equals 1. The tangent there rises 2.718 per unit step along the x-axis: slope equals height.
(1 + 1/n)ⁿ
2
2.25
2.441
2.613
2.715
2.718
Interest added more and more often: 2, 2.25, 2.441, 2.613, 2.715. The amounts close on e ≈ 2.718.
$e^x$ is the function whose slope at every point equals its own value. In symbols, $d/d x (e^x) = e^x$. The number $e$ is about 2.718. You met compound interest in Class 8. Put one rupee in at 100 percent a year, and add the interest more and more often. In a year it grows towards $e$ rupees.
Here is where the slope comes from. For a base $a > 0$, the limit from Class 11 gives
The slope of aˣ at 0 is the limit of (aʰ − 1)/h. For 2, e and 4 it is 0.69, 1 and 1.39.
Naming the natural base
The last limit depends only on $a$. The number $e$ is the base that makes it exactly 1. So for $e$, the slope is $e^x$ itself.
At every point the slope of eˣ equals its height: 0.368, 1, 2.718 and 7.389.
Check it with numbers. At $x = 0$, $e^x$ is 1, and its slope is 1. At $x = 1$, the height is about 2.718. The tangent there rises about 2.718 for every unit step along the x-axis.
Now compare bases. $2^x$, $e^x$ and $4^x$ all pass through $(0, 1)$, because any base to the power 0 is 1. But their slopes there differ: 0.69, 1 and 1.39, the three limits you just saw. Only $e$ gives slope 1 at $(0, 1)$, and with it a slope equal to the height everywhere. That is what makes $e$ the natural base.
Three bases, one point (0, 1), three slopes: 0.69 for 2, 1.39 for 4, exactly 1 for e. So e is the natural base.
Every other base needs a correction. Write $a^x$ as $e^{x \ln a}$, and the chain rule gives it at once: $d/d x (a^x) = e^{x \ln a} \cdot \ln a = a^x \ln a$, where $\ln$ is the natural log of the next section. Since $\ln e = 1$, the base $e$ needs none. At $x = 0$, $2^x$ has slope $\ln 2 \approx 0.69$, and $4^x$ has slope $\ln 4 \approx 1.39$.
That is why growth and decay are written with $e$ in later chapters. Its slope needs no correction.
Your turn: find the slope of $e^x$ at $x = 2$. Then use the chain rule to find the slope of $e^{3x}$ at $x = 0$.
A SCHOLAR INDIA REMEMBERS
How big is $e$? Do not take $2.718$ on trust. Bound it first. The slope of $a^x$ at $0$ is close to $(a^h - 1)/h$ for a small $h$. Take $h = 0.001$. For $a = 2$ you get about $0.693$, too small. For $a = 3$, about $1.099$, too big. So $e$ is between $2$ and $3$. For $2.7$ you get $0.994$, and for $2.8$ you get $1.030$. So $e$ is between $2.7$ and $2.8$.
Aryabhata · the hoopoe
Check yourself
What single property makes $e^x$ special among exponential functions, according to this fact?
$e^x$ is the only exponential function with an inverse function
$e^x$ is its own derivative: $d/d x (e^x) = e^x$
$e^x$ is the only exponential function whose derivative is a constant
$e^x$ is the only exponential function that is always increasing
Check your answer
$e^x$ is the only exponential function with an inverse function — Every exponential $a^x$ with $a>0$ and $a \neq 1$ has a well-defined inverse, invertibility is common to the whole family, not something that singles out $e^x$.
✓ $e^x$ is its own derivative: $d/d x (e^x) = e^x$ — (B) $d/d x (e^x) = e^x$, differentiating $e^x$ returns $e^x$ unchanged. No other base does this without an extra multiplying factor.
$e^x$ is the only exponential function whose derivative is a constant — $d/d x (e^x) = e^x$ is a function of $x$, not a constant, the derivative changes value exactly as $e^x$ itself does; it never settles to a fixed number.
$e^x$ is the only exponential function that is always increasing — Every exponential function $a^x$ with base $a>1$ is always increasing, that property is shared broadly, not unique to $e^x$, and it is not the fact being described here.
What is the general formula for $d/d x (a^x)$, for an arbitrary positive base $a$?
$x \cdot a^{x-1}$
$a^x \cdot \ln(x)$
$a^x$
$a^x \cdot \ln(a)$
Check your answer
$x \cdot a^{x-1}$ — $x \cdot a^{x-1}$ is the power rule, which applies to $x^n$ where the variable is the base and a fixed number is the exponent, here it is reversed, with $a$ the fixed base and $x$ the variable exponent, so the power rule does not apply.
$a^x \cdot \ln(x)$ — $\ln(x)$ is the natural log of the variable $x$, not of the base $a$, the correction factor must be $\ln(a)$, a constant, since $a$ is what is fixed here, not $x$.
$a^x$ — Dropping $\ln(a)$ treats every base as if it behaved like $e$, but only $e^x$ differentiates back to itself with no factor attached. Any other base needs the $\ln(a)$ multiplier.
✓ $a^x \cdot \ln(a)$ — (D) $d/d x (a^x) = a^x \cdot \ln(a)$, the original function multiplied by the natural log of the base, a constant correction factor.
Why does the general formula $d/d x (a^x) = a^x \cdot \ln(a)$ reduce to $d/d x (e^x) = e^x$ exactly when $a=e$?
because $e$ is the only base for which $a^x$ itself equals $x$
because $\ln(e) = 1$, so the correction factor vanishes
because $\ln(a)$ is undefined for every base except $a=e$
because $e^x$ is the only exponential function with a derivative at all
Check your answer
because $e$ is the only base for which $a^x$ itself equals $x$ — $e^x$ never equals $x$ itself except at isolated points that vary by coincidence, not by identity, this is not the mechanism behind the formula reducing at $a=e$.
✓ because $\ln(e) = 1$, so the correction factor vanishes — (B) $\ln(e) = 1$ by definition of the natural logarithm, so $a^x \cdot \ln(a)$ becomes $e^x \cdot 1 = e^x$ exactly when $a=e$, the correction factor disappears because it equals $1$, not for any other reason.
because $\ln(a)$ is undefined for every base except $a=e$ — $\ln(a)$ is defined for every positive base $a$, including $a=e$, where it happens to equal exactly $1$, it is not undefined anywhere in this formula’s domain.
because $e^x$ is the only exponential function with a derivative at all — Every exponential function $a^x$ with $a>0$ is differentiable, not just $e^x$, the general formula $a^x \cdot \ln(a)$ applies to all of them; $e$ is not singled out by differentiability.
Using the general formula $d/d x (a^x) = a^x \cdot \ln(a)$, what is $d/d x (2^x)$?
$2^x \cdot \ln(x)$
$2^x$
$2^x \cdot \ln(2)$
$x \cdot 2^{x-1}$
Check your answer
$2^x \cdot \ln(x)$ — $\ln(x)$ varies with $x$; the correction factor here must be $\ln(2)$, a fixed constant, since $2$ is the base that is fixed in this formula, not $x$.
$2^x$ — Only $e^x$ differentiates to itself with no extra factor. Base $2$ needs its own correction factor, $\ln(2)$, a fixed number close to $0.693$, not simply dropped.
✓ $2^x \cdot \ln(2)$ — (C) Substituting $a=2$ into $a^x \cdot \ln(a)$ gives $2^x \cdot \ln(2)$ directly, the same general formula, with the constant correction factor that belongs to base $2$.
$x \cdot 2^{x-1}$ — $x \cdot 2^{x-1}$ is the power rule, for functions like $x^n$ where $x$ is the base, here $2$ is the fixed base and $x$ is the variable exponent, so the exponential rule applies instead.
ln undoes eˣ: 1 goes in, eˣ gives 2.718, and ln gives 1 again.
domain of ln x
ln x has a value only for x greater than 0.
The NATURAL LOG, $\ln x$, undoes $e^x$. If $e^y = x$, then $y = \ln x$. It is defined only for $x > 0$, because $e^y$ is always positive. NCERT writes this function as $\log x$, with the base $e$ understood. This book writes $\ln x$, and the two mean the same thing.
ln 1 = 0 and ln e = 1: the curve crosses the x-axis at 1 and reaches height 1 at e.
Check it with numbers. Each fact about $e$ reads backwards as a fact about $\ln$:
$e^0 = 1$, so $\ln 1 = 0$.
$e^1 = e$, so $\ln e = 1$.
2
→
ln
slope 1/2
→
eʸ
slope 2
→
2
0.693
→
eʸ
slope 2
→
ln
slope 1/2
→
0.693
The two slopes multiply to 1: at x = 2, ln has slope 1/2 and eʸ has slope 2.
Finding the slope of ln
Here is the result first. Its slope is $d/d x (\ln x) = 1/x$, for $x > 0$. The lines below show why.
Why it works
The slope of $\ln x$
$y = \ln x$ Call the function $y$.
$x = e^y$ Undo the log: raise $e$ to the power of both sides, since $e^{\ln x} = x$.
$1 = e^y \cdot d y/d x$ Differentiate both sides with respect to $x$, using the chain rule on the right.
$d y/d x = 1/e^y$ Divide both sides by $e^y$, which is never 0.
*$d y/d x = 1/x$* Put $e^y = x$ back, so the answer is in $x$.
The slope of ln x is 1/x: 1 at 1, 1/2 at 2, 1/10 at 10.
Look at what the slope does. At $x = 1$ it is 1. At $x = 2$ it is $1/2$, and at $x = 10$ it is $1/10$. So the graph of $\ln x$ keeps rising, but more and more gently.
The curves e to the x and ln x mirror each other in y equals x. Mirrored points have slopes 2.014 and 1 over 2.014.
The graph of $\ln x$ is the graph of $e^x$ reflected in the line $y = x$. Their slopes at mirrored points are reciprocals. At $x = 0.7$ the slope of $e^x$ is about 2.014. At the mirrored point, $x = 2.014$, the slope of $\ln x$ is $1/2.014$, about 0.497.
8
→
ln
→
2.079
2 × 4
→
ln 2 + ln 4
0.693 + 1.386
→
2.079
ln turns a product into a sum: ln 8 = 2.079, and ln 2 + ln 4 = 0.693 + 1.386 = 2.079 too.
The log turns a product into a sum and a power into a multiple. That makes it the tool for the hardest slopes, a few sections on.
Your turn: find the slope of $\ln(2x)$ for $x > 0$. Compare it with the slope of $\ln x$.
Check yourself
What is $d/d x (\ln x)$, and over what domain does it hold?
$1/x$, for all $x \neq 0$
$x$, for $x>0$
$1/x$, for $x>0$
$\ln x$, for $x>0$
Check your answer
$1/x$, for all $x \neq 0$ — $\ln x$ itself is only defined for $x>0$, it has no value at all for $x \leq 0$, so its derivative cannot hold anywhere $\ln x$ does not exist, even though $1/x$ is defined for negative $x$ too.
$x$, for $x>0$ — Differentiating $\ln x$ gives $1/x$, a reciprocal function, not $x$ itself, that would be the derivative of a different function entirely.
✓ $1/x$, for $x>0$ — (C) $d/d x (\ln x) = 1/x$, and since $\ln x$ itself is defined only for $x>0$, the derivative formula inherits that same restriction.
$\ln x$, for $x>0$ — $\ln x$ differentiating back to itself is a property of $e^x$, not $\ln x$, the natural log’s derivative is the reciprocal function $1/x$, a genuinely different function from $\ln x$.
Why is the domain of $d/d x (\ln x) = 1/x$ restricted to $x>0$, rather than all $x \neq 0$?
because $1/x$ is undefined at negative values of $x$
because differentiation itself is only valid for positive numbers
because $\ln x$ is only defined for $x>0$, matching its domain
because $\ln x$ is decreasing for $x<0$
Check your answer
because $1/x$ is undefined at negative values of $x$ — $1/x$ is defined for every $x$ except $x=0$, including all negative numbers, the restriction to $x>0$ comes from $\ln x$ itself, not from any limitation of $1/x$.
because differentiation itself is only valid for positive numbers — Differentiation works perfectly well for negative numbers in general, $d/d x (x^2) = 2x$ holds for negative $x$ too. The restriction here is specific to the domain of $\ln x$, not to differentiation as a process.
✓ because $\ln x$ is only defined for $x>0$, matching its domain — (C) A derivative can only be evaluated where the original function is defined, and $\ln x$ itself has no value for $x \leq 0$, so the domain restriction traces back to $\ln x$, not to $1/x$.
because $\ln x$ is decreasing for $x<0$ — $\ln x$ has no value whatsoever for $x<0$, it cannot be called decreasing, increasing, or anything else there, since there is nothing to describe. The function simply does not exist on that side.
The formula $d/d x (\ln x) = 1/x$ follows from combining which two facts?
$e^x$ being its own derivative, and its inverse relation to $\ln x$
the power rule, and $\ln x$ being equivalent to $x^{-1}$
the chain rule, and $\ln x$ being a composite function of two layers
the product rule, and $\ln x$ being a product of two simpler functions
Check your answer
✓ $e^x$ being its own derivative, and its inverse relation to $\ln x$ — (A) Because $e^x$ is its own derivative and $\ln x$ is defined as the inverse function of $e^x$, the inverse-function rule combines the two facts to produce $d/d x (\ln x) = 1/x$.
the power rule, and $\ln x$ being equivalent to $x^{-1}$ — $\ln x$ and $x^{-1}$ are entirely different functions with different graphs and different values, $\ln x$ is never equal to $1/x$ as a function, even though $1/x$ happens to be its derivative.
the chain rule, and $\ln x$ being a composite function of two layers — $\ln x$ on its own is a single function, not two layers composed together, there is no inner function to identify the way $\sin(\sqrt{x})$ has $\sqrt{x}$ inside $\sin$. The chain rule is not what produces this formula.
the product rule, and $\ln x$ being a product of two simpler functions — $\ln x$ is not built as a product of two separate functions multiplied together, it is the inverse function of $e^x$, a relationship the product rule has nothing to do with.
Why does the formula $d/d x (\ln x) = 1/x$ matter for logarithmic differentiation, a technique used later in this chapter?
logarithmic differentiation is a shortcut for finding $d/d x (\ln x)$ itself, faster than the direct formula
logarithmic differentiation takes $\ln$ of both sides, and differentiating $\ln y$ needs this formula via the chain rule
logarithmic differentiation only applies to functions that already contain $\ln x$ somewhere in their formula
logarithmic differentiation replaces every derivative in a problem with $1/x$, regardless of what function is being differentiated
Check your answer
logarithmic differentiation is a shortcut for finding $d/d x (\ln x)$ itself, faster than the direct formula — The relationship runs the other way, $d/d x (\ln x) = 1/x$ is a known building-block formula that logarithmic differentiation uses, via the chain rule on $\ln y$, not something the technique exists to compute.
✓ logarithmic differentiation takes $\ln$ of both sides, and differentiating $\ln y$ needs this formula via the chain rule — (B) Logarithmic differentiation takes $\ln$ of both sides of $y =$ some function, and differentiating $\ln y$ with respect to $x$ uses the chain rule together with $d/d x (\ln x) = 1/x$, producing $(1/y) \cdot d y/d x$ on that side.
logarithmic differentiation only applies to functions that already contain $\ln x$ somewhere in their formula — Logarithmic differentiation is most useful exactly where $\ln$ does not already appear, functions like $x^x$ or long products, where taking $\ln$ of both sides first is what makes the derivative tractable at all.
logarithmic differentiation replaces every derivative in a problem with $1/x$, regardless of what function is being differentiated — $1/x$ is specifically the derivative of $\ln x$, not a universal replacement for every derivative, it only appears where $\ln$ of something is actually being differentiated, via the chain rule.
x equals 3 meets both arcs, at heights 4 and minus 4. The slopes there are minus 3 over 4 and plus 3 over 4.
Think of the circle of radius 5 with its centre at the origin. You met its equation in Class 11. Every point on it is 5 units from the centre. The point $(3, 4)$ is on it. So is the point $(3, -4)$. One input gives two heights, so no single formula for the height covers the whole circle.
Near (3, 4) the circle is the graph of one function, even though no single formula gives all of it.
The equation is $x^2 + y^2 = 25$. It ties $x$ and $y$ together without giving $y$ alone. You can still find the slope at a point. Near most points, the circle is a curve whose height $y$ depends on $x$. Differentiate both sides of the equation with respect to $x$, and treat $y$ as a function of $x$. This is IMPLICIT DIFFERENTIATION.
x
→
y(x)
slope dy/dx
→
square
slope 2y
→
y²
y² is two functions in a row: x to y, then squaring. So its slope is 2y times dy/dx.
Every term with $y$ in it needs the chain rule: $y^2$ is two functions in a row, $x$ gives $y$, and then $y$ is squared.
So the slope of $y^2$ is $2y \cdot d y/d x$, not $2y$.
After you differentiate, collect the $d y/d x$ terms on one side and solve. The answer usually has both $x$ and $y$ in it.
With y = 2x: 2y × dy/dx = 2 × 2 × 2 = 8 at x = 1. 4x² has slope 8 there.
Let us check the $y^2$ step on a line, $y = 2x$. First put $y = 2x$ into $y^2$ and differentiate directly:
$d/d x (y^2) = d/d x (4x^2) = 8x$.
Now use the chain rule, with $d y/d x = 2$:
$d/d x (y^2) = 2y \cdot d y/d x = 2(2x) \cdot 2 = 8x$.
Both ways give $8x$, so the $y^2$ step is right.
So you can find slopes on curves where $y$ cannot be written alone, or only with effort.
Your turn: differentiate both sides of $y^2 = x$ with respect to $x$. What is $d y/d x$ at the point $(4, 2)$?
Check yourself
An equation like $x^2+y^2=25$ tangles $x$ and $y$ together, with no easy way to solve for $y$ alone. What does implicit differentiation do about this?
it eliminates $y$ from the equation before differentiating, leaving only $x$
it differentiates both sides directly with respect to $x$, treating $y$ as a function of $x$
it estimates $d y/d x$ numerically at a point without finding an algebraic formula
it first solves the equation for $y$ as an explicit formula in $x$, then differentiates that
Check your answer
it eliminates $y$ from the equation before differentiating, leaving only $x$ — The method keeps $y$ in the equation throughout and differentiates both sides as they stand, it never eliminates $y$ before differentiating.
✓ it differentiates both sides directly with respect to $x$, treating $y$ as a function of $x$ — (B) Implicit differentiation works on the equation as given, differentiating both sides with respect to $x$ and treating $y$ as a function of $x$ throughout.
it estimates $d y/d x$ numerically at a point without finding an algebraic formula — The method gives an exact algebraic formula for $d y/d x$, not a numerical or graphical approximation.
it first solves the equation for $y$ as an explicit formula in $x$, then differentiates that — Implicit differentiation is needed precisely when $y$ cannot easily be isolated on one side, an equation like $x^2+y^2=25$ never gets solved for a single formula $y=f(x)$ first.
While differentiating $x^2+y^2=25$ with respect to $x$, what does differentiating the $y^2$ term correctly give?
$2y \cdot ((d y)/(d x))$
$2y \cdot x$
$0$, since $y^2$ is treated as a constant
$2y$
Check your answer
✓ $2y \cdot ((d y)/(d x))$ — (A) Since $y$ is itself a function of $x$, differentiating $y^2$ needs the chain rule, giving $2y \cdot ((d y)/(d x))$, not just $2y$.
$2y \cdot x$ — There is no product of two separate $x$-functions here to invoke the product rule on, $y^2$ is a single function of $y$, composed with $y$’s own dependence on $x$, which is exactly what the chain rule handles.
$0$, since $y^2$ is treated as a constant — Treating $y$ as a constant gives a derivative of $0$ for the $y^2$ term, but $y$ is itself a function of $x$, so it cannot be held fixed while differentiating with respect to $x$.
$2y$ — Applying the power rule alone, as if $y$ were the variable being differentiated with respect to itself, drops the required $d y/d x$ factor that the chain rule supplies.
After differentiating both sides of an equation with respect to $x$, what is the next step in implicit differentiation?
the process is finished: the differentiated equation is the final answer
collect every $d y/d x$ term on one side and solve algebraically for $d y/d x$
integrate both sides again to recover the original equation
substitute a specific numerical value for $x$ and $y$ before solving for $d y/d x$
Check your answer
the process is finished: the differentiated equation is the final answer — Differentiating both sides is only the first half of the procedure, the $d y/d x$ terms are usually scattered across the equation and still need to be gathered and solved for.
✓ collect every $d y/d x$ term on one side and solve algebraically for $d y/d x$ — (B) The differentiated equation still has $d y/d x$ terms scattered through it, collecting them on one side and solving algebraically gives the formula for $d y/d x$.
integrate both sides again to recover the original equation — This describes finding an antiderivative, the reverse of differentiation, implicit differentiation differentiates the given equation, it does not integrate it.
substitute a specific numerical value for $x$ and $y$ before solving for $d y/d x$ — The procedure solves for $d y/d x$ algebraically as a formula, before any particular point’s coordinates are ever substituted in.
After solving algebraically for $d y/d x$ from an implicitly differentiated equation, what form does the result usually take?
a formula in $x$ alone, with $y$ eliminated
no usable formula: implicit differentiation only gives a value at one point, not a general expression
a formula in $y$ alone, with $x$ eliminated
a formula still in terms of both $x$ and $y$
Check your answer
a formula in $x$ alone, with $y$ eliminated — A formula in $x$ alone would require every $y$ to cancel out of the solved expression, which is a special case, not the general outcome of the method.
no usable formula: implicit differentiation only gives a value at one point, not a general expression — The algebraic collect-and-solve step reliably produces a usable formula for $d y/d x$, it is just ordinarily a formula in both $x$ and $y$, not a formula that fails to exist.
a formula in $y$ alone, with $x$ eliminated — A formula in $y$ alone would require every $x$ to cancel out of the solved expression, which is just as special a case as the $x$-alone outcome.
✓ a formula still in terms of both $x$ and $y$ — (D) The collect-and-solve step produces a formula for $d y/d x$ that is usually still expressed in terms of both $x$ and $y$, not one variable alone.
At (3, 4) on the circle, the radius has slope 4 over 3. The tangent, at right angles, has slope minus 3 over 4.
dy/dx = −x/y: −3/4 at (3, 4), −4/3 at (4, 3), 0 at the top. At (5, 0), y = 0: the tangent is upright.
Worked example
Find $d y/d x$ for the circle $x^2+y^2=25$
$x^2 + y^2 = 25$ The circle of radius 5. Near most points it is a curve $y$ of $x$, though no one formula gives $y$.
$2x + 2y \: d y/d x = 0$ Differentiate each term with respect to $x$. The term $y^2$ needs the chain rule, and the constant 25 has slope 0.
$2y \: d y/d x = -2x$ Move $2x$ to the other side.
$d y/d x = -x/y$ Divide by $2y$. This needs $y \neq 0$, so it fails at $(5, 0)$ and $(-5, 0)$.
At $(3, 4)$: $-x/y = -3/4$, and $(-3/4) \cdot (4/3) = -1$ Check at $(3, 4)$. The radius there has slope $4/3$. The two slopes multiply to $-1$, so the tangent meets the radius at a right angle.
Check yourself
Differentiating $x^2+y^2=25$ with respect to $x$, term by term, gives which equation?
✓ $2x + 2y \cdot ((d y)/(d x)) = 0$ — (A) The $x^2$ term differentiates by the ordinary power rule to $2x$, the $y^2$ term needs the chain rule to give $2y \cdot ((d y)/(d x))$, and the constant $25$ differentiates to $0$.
$2x + 2y \cdot ((d y)/(d x)) = 25$ — The right side is the constant $25$, and a constant’s derivative is $0$, it never produces a leftover term.
$2x + 2y = 0$ — The $y^2$ term needs the chain rule because $y$ is a function of $x$, dropping the $d y/d x$ factor leaves just $2y$, missing the very piece that makes the equation solvable for $d y/d x$.
$2x \cdot ((d y)/(d x)) + 2y \cdot ((d y)/(d x)) = 0$ — $x$ is the variable of differentiation itself, so $x^2$ differentiates by the ordinary power rule to $2x$, with no chain-rule factor needed.
Solving $2x + 2y \cdot ((d y)/(d x)) = 0$ for $d y/d x$ gives which result?
$d y/d x = -y/x$
$d y/d x = x/y$
$d y/d x = -2x$
$d y/d x = -x/y$
Check your answer
$d y/d x = -y/x$ — Solving $2y \cdot ((d y)/(d x))=-2x$ for $d y/d x$ divides by $2y$, putting $x$ on top and $y$ on the bottom, not the other way round.
$d y/d x = x/y$ — Moving $2x$ to the right side of $2x+2y \cdot ((d y)/(d x))=0$ gives $2y \cdot ((d y)/(d x))=-2x$, and the negative sign carries through to the final ratio, it does not disappear.
$d y/d x = -2x$ — Moving $2x$ across gives $2y \cdot d y/d x = -2x$, and $-2x$ is what $2y$ times the slope equals, not the slope. The last step, dividing both sides by $2y$, is missing.
✓ $d y/d x = -x/y$ — (D) Moving $2x$ across and dividing by $2y$ gives $d y/d x = -x/y$, with the factor of $2$ cancelling and the negative sign carried through.
$x^2+y^2=100$ is a circle of radius $10$, structurally the same as $x^2+y^2=25$. Applying the same implicit differentiation method, what is $d y/d x$?
$d y/d x = -y/x$
$d y/d x$ cannot be solved for without first finding a chain-rule factor on the $x^2$ term
$d y/d x = -x/y$
$d y/d x = -x/y$, with an extra constant term from the $100$
Check your answer
$d y/d x = -y/x$ — Solving $2y \cdot ((d y)/(d x))=-2x$ for $d y/d x$ divides by $2y$, putting $x$ on top and $y$ on the bottom, the same ratio orientation as the radius-$5$ case, unaffected by the different radius.
$d y/d x$ cannot be solved for without first finding a chain-rule factor on the $x^2$ term — The $y^2$ term still needs the chain rule here, exactly as it did for $x^2+y^2=25$, skipping it leaves no $d y/d x$ term to solve for at all.
✓ $d y/d x = -x/y$ — (C) The same method applies regardless of the radius, the constant on the right always differentiates to $0$, so $d y/d x = -x/y$ for this circle too.
$d y/d x = -x/y$, with an extra constant term from the $100$ — The right side is a constant, $100$, and a constant’s derivative is $0$, it plays no role in the slope formula, exactly as $25$ played none for the radius-$5$ circle.
x³ has a fixed power, 3ˣ a fixed base. xˣ has neither, so neither rule applies.
Each rule alone holds one part of x to the x fixed, so its line cuts across the curve. Adding the two slopes gives the line that lies along it.
$x^x$ is neither a power nor an exponential. In a power, such as $x^3$, the exponent is fixed. In an exponential, such as $3^x$, the base is fixed. In $x^x$ both change, so neither rule fits.
2² = 4
→
ln
→
1.386
2
→
ln, then ×2
0.693 × 2
→
1.386
ln brings a power down to a product: ln 4 = ln 2² = 2 × 0.693 = 1.386.
LOGARITHMIC DIFFERENTIATION handles it. Take $\ln$ of both sides first, and then differentiate. The log turns a power into a product, so $\ln(x^x) = x \ln x$. You can differentiate a product with the product rule.
y = xˣ > 0
Taking ln needs y > 0. For xˣ that holds when x > 0, the domain we work on.
The left side needs the chain rule: $\ln y$ is two functions in a row, so its slope is $(1/y) \: d y/d x$.
Once both sides are differentiated, multiply through by $y$ to get $d y/d x$ alone.
Logs need positive inputs, so the method works only where $y > 0$. For $x^x$ that means $x > 0$.
1 × 2 × 3 = 6
→
ln
→
1.792
1, 2, 3
→
ln each, add
0 + 0.693 + 1.099
→
1.792
A long product becomes a sum: ln 6 = 1.792, and 0 + 0.693 + 1.099 = 1.792.
The same move helps with a long product or quotient. The log turns a product into a sum and a quotient into a difference. Then you differentiate each piece on its own.
The log way on x³ gives y × 3/x. At x = 1 that is 3, the slope the power rule gives.
Let us check the method on a slope you know.
Worked example
Check logs on $y = x^3$
$y = x^3, \quad x > 0$ Start from a slope you know, with $x > 0$ so that $y$ is positive.
$\ln y = 3 \ln x$ Take $\ln$ of both sides: the log turns the power 3 into a factor 3.
$(1/y) \: d y/d x = 3/x$ Differentiate both sides with respect to $x$, using the chain rule on the left.
$d y/d x = y \cdot 3/x$ Multiply both sides by $y$.
Check: $d y/d x = x^3 \cdot 3/x = 3x^2$ Put $y = x^3$ back: this is the power rule’s answer.
Your turn: use logs to differentiate $y = 2^x$. Compare your answer with the rule for $a^x$.
Check yourself
$y = x^x$ has both a variable base and a variable exponent. Why does this call for logarithmic differentiation?
because the ordinary power rule handles it, once $\ln$ makes the exponent easier to read
because logarithmic differentiation is only for equations where $y$ is tangled implicitly with $x$
because the exponential rule for $a^x$ already covers this case directly
because neither the power nor exponential rule applies when base and exponent are both variable
Check your answer
because the ordinary power rule handles it, once $\ln$ makes the exponent easier to read — The ordinary power rule needs a fixed constant exponent, like $x^3$, here the exponent is $x$ itself, a variable, so the power rule does not apply.
because logarithmic differentiation is only for equations where $y$ is tangled implicitly with $x$ — Logarithmic differentiation is not limited to implicit equations, $y=x^x$ is already solved explicitly for $y$, and still needs the method because neither ordinary rule reaches a variable base raised to a variable exponent.
because the exponential rule for $a^x$ already covers this case directly — The exponential rule for $a^x$ needs a fixed constant base, like $2^x$, here the base is $x$ itself, a variable, so that rule does not apply either.
✓ because neither the power nor exponential rule applies when base and exponent are both variable — (D) With both the base and the exponent variable, neither the power rule nor the exponential rule for a constant base applies, this is exactly the gap logarithmic differentiation fills.
For a long product of several factors, each possibly raised to a power, why does taking $\ln$ of both sides before differentiating help?
because differentiation always turns a sum of factors into a simpler expression
because $\ln$ turns a sum of factors into a single product, which is easier to differentiate
because $\ln$ turns a product into a sum, and powers into multiples, easier to differentiate
because $\ln$ turns the whole expression into a single quotient of two terms
Check your answer
because differentiation always turns a sum of factors into a simpler expression — This describes what differentiating a sum term by term does, not what taking $\ln$ does to a product of factors before any differentiation happens.
because $\ln$ turns a sum of factors into a single product, which is easier to differentiate — This is the reverse of what $\ln$ actually does, $\ln$ turns a product into a sum, not a sum into a product.
✓ because $\ln$ turns a product into a sum, and powers into multiples, easier to differentiate — (C) $\ln$ converts a product of several factors into a sum of $\ln$ terms, and any exponents on those factors into simple multiples out front, both far easier to differentiate than the original product.
because $\ln$ turns the whole expression into a single quotient of two terms — $\ln$ of a single quotient does turn it into a subtraction, but the question is about a long product of several factors, which $\ln$ turns into a sum of terms, one per factor.
A student differentiates $\ln(y)=x \cdot \ln x$ with respect to $x$ and writes the left side as just $1/y$, with no other factor. What has gone wrong?
the product rule was needed instead, since $\ln(y)$ is a product of $\ln$ and $y$
the student should have gotten $0$, since $y$ is held constant while differentiating with respect to $x$
nothing has gone wrong: $1/y$ is the correct derivative of $\ln(y)$ with respect to $x$
the chain rule was skipped: $\ln(y)$ depends on $x$ through $y$, needing a factor of $d y/d x$
Check your answer
the product rule was needed instead, since $\ln(y)$ is a product of $\ln$ and $y$ — $\ln(y)$ is a single composite function, not a product of two separate factors, so the product rule does not apply here, the chain rule does.
the student should have gotten $0$, since $y$ is held constant while differentiating with respect to $x$ — Treating $y$ as a fixed constant inside $\ln(y)$ gives a derivative of $0$, but $y$ is a function of $x$ and cannot be held fixed while differentiating with respect to $x$.
nothing has gone wrong: $1/y$ is the correct derivative of $\ln(y)$ with respect to $x$ — This drops the chain-rule factor. $\ln(y)$ is a function of $x$ only through $y$, so differentiating it needs $1/y$ multiplied by $d y/d x$, not $1/y$ alone.
✓ the chain rule was skipped: $\ln(y)$ depends on $x$ through $y$, needing a factor of $d y/d x$ — (D) $\ln(y)$ is a function of $x$ only through $y$, so the chain rule applies, the correct left side is $(1/y) \cdot ((d y)/(d x))$, not the bare $1/y$ the student wrote.
After differentiating $\ln(y) =$ (some expression in $x$) to get $(1/y) \cdot ((d y)/(d x)) =$ (that expression’s derivative), what is the final step to solve for $d y/d x$?
multiply both sides by $x$
divide both sides by $y$
the equation is already solved for $d y/d x$, with nothing further needed
multiply both sides by $y$
Check your answer
multiply both sides by $x$ — The variable being cleared from the denominator is $y$, not $x$, $x$ does not appear as a divisor on the left side at all.
divide both sides by $y$ — The left side already has $y$ in the denominator as $1/y$; isolating $d y/d x$ requires multiplying both sides by $y$ to clear that denominator, not dividing by it again.
the equation is already solved for $d y/d x$, with nothing further needed — After differentiating, the left side is $(1/y) \cdot ((d y)/(d x))$, a formula for the derivative of $\ln(y)$, not yet $d y/d x$ itself, one more algebraic step is needed.
✓ multiply both sides by $y$ — (D) The left side has $y$ in the denominator as $1/y$; multiplying both sides by $y$ clears it and isolates $d y/d x$.
x to the x is lowest at 1 over e, about 0.37, with a flat tangent. At x equals 1.5 its slope is 2.582.
The slope xˣ(ln x + 1) is 0 at x = 1/e, 1 at x = 1 and 2.582 at 1.5.
Worked example
Find $d y/d x$ for $y = x^x$
$y = x^x, \quad x > 0$ $x^x$ is neither a power nor an exponential, so neither rule fits. The log needs $x > 0$.
$\ln y = x \ln x$ Take $\ln$ of both sides. The log turns the power into a product.
$(1/y) \: d y/d x = \ln x + x \cdot 1/x$ Differentiate both sides. The left side needs the chain rule, and the right side needs the product rule.
$(1/y) \: d y/d x = \ln x + 1$ Simplify $x \cdot 1/x$ to 1.
$d y/d x = y(\ln x + 1)$ Multiply both sides by $y$.
$d y/d x = x^x (\ln x + 1)$ Put $y = x^x$ back.
$x = 1/e$: $x^x (\ln(1/e) + 1) = x^x (-1 + 1) = 0$ Check against the graph. The slope is 0 at $x = 1/e$, about 0.37, where the curve is lowest.
Check yourself
Taking $\ln$ of both sides of $y = x^x$ gives which equation?
$\ln y = x \ln x$
$\ln y = (\ln x)^x$
$\ln y = \ln x$
$\ln y = x^{\ln x}$
Check your answer
✓ $\ln y = x \ln x$ — (A) The log identity $\ln(a^b) = b \cdot \ln a$ pulls the exponent $x$ out as a multiplying factor, giving $\ln y = x \cdot \ln x$.
$\ln y = (\ln x)^x$ — There is no log identity that raises $\ln x$ to a power, the correct identity multiplies $x$ by $\ln x$, it does not exponentiate $\ln x$.
$\ln y = \ln x$ — The exponent $x$ must appear in the result as a multiplying factor; dropping it and taking $\ln$ of the base alone loses the exponent’s contribution entirely.
$\ln y = x^{\ln x}$ — The identity $\ln(x^x) = x \cdot \ln x$ pulls the exponent out as a multiplying factor, it does not leave the exponent attached in any other form.
Differentiating $\ln y = x \cdot \ln x$ with respect to $x$, using the chain rule on the left and the product rule on the right, gives which equation?
$(1/y) \cdot ((d y)/(d x)) = \ln x$
$(1/y) \cdot ((d y)/(d x)) = \ln x + x \cdot (1/x)$
$(1/y) \cdot ((d y)/(d x)) = \ln x + x \cdot (1/x^2)$
$1/y = \ln x + x \cdot (1/x)$
Check your answer
$(1/y) \cdot ((d y)/(d x)) = \ln x$ — The right side, $x \cdot \ln x$, is a product of two functions of $x$, so it needs the product rule, differentiating just one factor and leaving the other untouched misses half the terms.
✓ $(1/y) \cdot ((d y)/(d x)) = \ln x + x \cdot (1/x)$ — (B) The left side needs the chain rule, giving $(1/y) \cdot ((d y)/(d x))$; the right side needs the product rule on $x \cdot \ln x$, giving $\ln x + x \cdot (1/x)$, which simplifies to $\ln x + 1$.
$(1/y) \cdot ((d y)/(d x)) = \ln x + x \cdot (1/x^2)$ — The chain rule is applied correctly on the left. But the right side’s second term needs $d/d x (x) = 1$ multiplied by $1/x$, giving $x \cdot (1/x)$, this instead differentiates $1/x$ itself, giving $x \cdot (1/x^2)$, which is not what the product rule calls for.
$1/y = \ln x + x \cdot (1/x)$ — The left side is missing its chain-rule factor here even though the right side’s product rule is done correctly, both sides must be handled correctly together.
$y = (\sin x)^x$ has a variable base and a variable exponent, structurally the same as $y=x^x$. Applying the same logarithmic differentiation method, taking $\ln$ of both sides gives which equation?
$\ln y = \ln(\sin x)$
$y = x \cdot \sin x$, found directly by the ordinary power rule
$\ln y = \ln(x \cdot \sin x)$
$\ln y = x \cdot \ln(\sin x)$
Check your answer
$\ln y = \ln(\sin x)$ — The exponent $x$ must appear as a multiplying factor in the result; dropping it and taking $\ln$ of $\sin x$ alone loses the exponent’s contribution, the same error as on $x^x$.
$y = x \cdot \sin x$, found directly by the ordinary power rule — $x$ is a variable exponent here too, so the ordinary power rule does not apply, this is exactly why logarithmic differentiation is needed, the same reason as for $x^x$.
$\ln y = \ln(x \cdot \sin x)$ — The identity $\ln(a^b)=b \cdot \ln a$ pulls the exponent out as a multiplying factor, exactly as it did for $x^x$, it does not leave the exponent attached inside the log.
✓ $\ln y = x \cdot \ln(\sin x)$ — (D) Applying $\ln(a^b)=b \cdot \ln a$ with $a=\sin x$ and $b=x$ pulls the exponent out as a multiplying factor, giving $\ln y = x \cdot \ln(\sin x)$, exactly parallel to $\ln y = x \cdot \ln x$ for $y=x^x$.
A ball thrown up and forward. Its position at every moment is given by two numbers.
In Physics in Class 11 you studied a ball thrown forward. Its forward distance and its height both change with time. You worked out each one from the time, on its own. The path is a curve, but you never needed the height as a formula in the distance. You wrote both in terms of time.
Each dot is one moment: at t = 0.5 the ball is at (5, 3.75); at t = 1, at the top, (10, 5).
Naming the parametric form
Say a ball is thrown at 10 metres a second forward and 10 metres a second up. Take $g$ as 10 metres per second per second. After $t$ seconds it is $x = 10t$ forward and $y = 10t - 5t^2$ up. A curve given like this is in PARAMETRIC FORM, and $t$ is the PARAMETER.
At t = 0.5 the ball moves 10 forward and 5 up per second. The path’s slope is 5/10 = 0.5.
The slope of the path is still $d y/d x$. By the chain rule:
$d y/d t = d y/d x \cdot d x/d t$.
Divide both sides by $d x/d t$, as long as it is not 0:
$d y/d x = ((d y)/(d t))/((d x)/(d t))$
Check it on the ball. Its two rates are:
$d x/d t = 10, \quad d y/d t = 10 - 10t$.
Now divide them, the $y$ rate over the $x$ rate:
$d y/d x = (10 - 10t)/10 = 1 - t$.
Read the slope. At $t = 0$ the slope is 1, so the ball leaves at 45 degrees. At $t = 1$ the slope is 0, so the ball is at the top of its path.
dy/dx = 1 − t: 1 at the throw, at 45°; 0 at the top; −0.5 on the way down.
You get the slope without ever writing $y$ in terms of $x$. For many curves that is the only easy way.
Your turn: for $x = t^2$ and $y = t^3$, find $d y/d x$ in terms of $t$.
Check yourself
For a curve given parametrically by $x=f(t)$ and $y=g(t)$, what is $d y/d x$ in terms of the two rates with respect to $t$?
$((d y)/(d t))/((d x)/(d t))$
$((d y)/(d t)) \cdot ((d x)/(d t))$
$((d x)/(d t))/((d y)/(d t))$
$((d y)/(d t)) - ((d x)/(d t))$
Check your answer
✓ $((d y)/(d t))/((d x)/(d t))$ — (A) Dividing the chain-rule identity $d y/d t = ((d y)/(d x)) \cdot ((d x)/(d t))$ through by $d x/d t$ isolates $d y/d x = ((d y)/(d t))/((d x)/(d t))$.
$((d y)/(d t)) \cdot ((d x)/(d t))$ — The chain-rule identity $d y/d t = ((d y)/(d x)) \cdot ((d x)/(d t))$ does involve a product, but isolating $d y/d x$ requires dividing through by $d x/d t$, not stopping at the product of the two rates.
$((d x)/(d t))/((d y)/(d t))$ — Inverting numerator and denominator gives $d x/d y$, the reciprocal slope, not $d y/d x$.
$((d y)/(d t)) - ((d x)/(d t))$ — The two rates only ever combine by division in this formula, there is no subtraction step anywhere in the derivation.
The chain rule gives $d y/d t = ((d y)/(d x)) \cdot ((d x)/(d t))$. Why does dividing both sides by $d x/d t$ correctly isolate $d y/d x$?
because $d y/d x$ is already known before the division, so dividing only confirms it
because $d x/d t$ and $d y/d t$ are always equal for a parametric curve, so either one may be divided out
because $d x/d t$ multiplies $d y/d x$ on the right side, so dividing by it undoes that multiplication
because $t$ is the same variable on both sides and therefore cancels first, before any division
Check your answer
because $d y/d x$ is already known before the division, so dividing only confirms it — The division is what produces $d y/d x$ in the first place, nothing about it is known beforehand to be confirmed.
because $d x/d t$ and $d y/d t$ are always equal for a parametric curve, so either one may be divided out — The two rates $d x/d t$ and $d y/d t$ depend on the curve and are almost never equal, the identity holds for any values they take, not because they match.
✓ because $d x/d t$ multiplies $d y/d x$ on the right side, so dividing by it undoes that multiplication — (C) On the right side of $d y/d t = ((d y)/(d x)) \cdot ((d x)/(d t))$, $d x/d t$ multiplies $d y/d x$; dividing both sides by $d x/d t$ undoes that multiplication and leaves $d y/d x$ alone.
because $t$ is the same variable on both sides and therefore cancels first, before any division — $t$ does not appear as a separate factor to cancel, it is the variable both derivatives are taken with respect to; the identity is an algebraic relation between the two rates themselves.
If $d x/d t = 0$ at some value of $t$, what goes wrong with using $d y/d x = ((d y)/(d t))/((d x)/(d t))$ there?
the ratio still works, giving $d y/d x = 0$ automatically
it means the curve itself is discontinuous there
the formula is undefined there: the denominator is zero
it means $y$ is not a function of $t$ at that point
Check your answer
the ratio still works, giving $d y/d x = 0$ automatically — A zero denominator makes a fraction undefined, not equal to zero, the mistake here treats “denominator is $0$” as if it meant “the fraction is $0$”, which is the opposite of how division behaves.
it means the curve itself is discontinuous there — A curve can be perfectly smooth and continuous at a point where $d x/d t=0$. The leftmost and rightmost points of a circle, where the tangent is vertical, are exactly this case, and the curve does not break there.
✓ the formula is undefined there: the denominator is zero — (C) $d y/d x = ((d y)/(d t))/((d x)/(d t))$ is a fraction, and a zero denominator makes any fraction undefined at that point, regardless of the numerator.
it means $y$ is not a function of $t$ at that point — $d x/d t = 0$ says only that $x$ is momentarily not changing with $t$, it says nothing about whether $y$ depends on $t$, which it still does.
What is the main practical advantage of computing $d y/d x$ via $((d y)/(d t))/((d x)/(d t))$ for a parametric curve, rather than first eliminating $t$?
it always gives a simpler final expression than eliminating $t$ would
it works even when $x$ and $y$ do not both depend on the same parameter $t$
it avoids ever needing an explicit formula for $y$ in terms of $x$
it avoids the chain rule entirely
Check your answer
it always gives a simpler final expression than eliminating $t$ would — The parametric route can be more or less simple than eliminating $t$ depending on the curve, its real advantage is sidestepping elimination, not guaranteeing simplicity.
it works even when $x$ and $y$ do not both depend on the same parameter $t$ — The method assumes $x=f(t)$ and $y=g(t)$ share the same parameter $t$, if they depended on different parameters, this ratio would not even make sense.
✓ it avoids ever needing an explicit formula for $y$ in terms of $x$ — (C) Solving for $y$ explicitly in terms of $x$ is often impossible or ugly for a parametric curve; dividing the two rates with respect to $t$ sidesteps that step entirely.
it avoids the chain rule entirely — The formula $d y/d x = ((d y)/(d t))/((d x)/(d t))$ is derived directly from the chain rule identity $d y/d t = ((d y)/(d x)) \cdot ((d x)/(d t))$, it relies on the chain rule rather than avoiding it.
At t equals pi over 4, the rates are minus 1.414 for x and 1.414 for y. Their ratio, minus 1, is the slope.
dy/dx = −cot t: −1.732, −1, −0.577, then 0 at the top. At t = 0, dx/dt = 0 and the formula has no value.
Worked example
Find $d y/d x$ for the parametric circle $x = a \cos t$, $y = a \sin t$
$x = a \cos t, \quad y = a \sin t$ Both coordinates come from one parameter, the angle $t$. The circle has radius $a$.
$d x/d t = -a \sin t$ Differentiate $x$ with respect to $t$. The slope of $\cos t$ is $-\sin t$.
$d y/d t = a \cos t$ Differentiate $y$ with respect to $t$.
$d y/d x = ((d y)/(d t))/((d x)/(d t)) = a \cos t / (-a \sin t)$ Divide the two rates. This needs $d x/d t \neq 0$, so $\sin t \neq 0$.
$d y/d x = -\cot t$ The $a$ cancels, and $(\cos t)/(\sin t)$ is $\cot t$.
$t = \pi/4$: $-\cot(\pi/4) = -1$, and $-x/y = -1$ because $x = y$ there Check against implicit differentiation. On $x^2 + y^2 = a^2$ the slope is $-x/y$, and both give $-1$.
Check yourself
For the parametric circle $x = a \cos t$, $y = a \sin t$, what are $d x/d t$ and $d y/d t$?
$d x/d t = -a \sin t$ and $d y/d t = a \cos t$
$d x/d t = a \sin t$ and $d y/d t = -a \cos t$
$d x/d t = -a \cos t$ and $d y/d t = -a \sin t$
$d x/d t = a \cos t$ and $d y/d t = a \sin t$
Check your answer
✓ $d x/d t = -a \sin t$ and $d y/d t = a \cos t$ — (A) $d/d t (\cos t) = -\sin t$ and $d/d t (\sin t) = \cos t$, so $d x/d t = -a \sin t$ and $d y/d t = a \cos t$.
$d x/d t = a \sin t$ and $d y/d t = -a \cos t$ — $d/d t (\cos t) = -\sin t$ carries the negative sign, but $d/d t (\sin t) = \cos t$ does not, this option puts a negative sign on both, when only one belongs there.
$d x/d t = -a \cos t$ and $d y/d t = -a \sin t$ — Differentiating a trig function swaps it for the other trig function, not just its own sign, this keeps $\cos t$ and $\sin t$ in place and only negates them, missing the swap.
$d x/d t = a \cos t$ and $d y/d t = a \sin t$ — This lists $\cos t$ and $\sin t$ unchanged, differentiating actually swaps them to $-\sin t$ and $\cos t$ respectively, it does not leave them as they were.
Simplifying $(a \cos t)/(-a \sin t)$ for the parametric circle, what does $d y/d x$ equal?
$-\cot t$
$\tan t$
$\cot t$
$-\tan t$
Check your answer
✓ $-\cot t$ — (A) The two $a$’s cancel and $\cos t/\sin t = \cot t$, with the denominator’s negative sign carrying through, giving $-\cot t$.
$\tan t$ — This combines two separate slips, inverting $\cot t$ to $\tan t$ and dropping the negative sign that the $-a \sin t$ denominator carries.
$\cot t$ — The two $a$’s cancel and the leftover negative sign survives, $\cos t/(-\sin t) = -\cot t$, not $\cot t$; dropping the sign loses information the negative denominator carries.
$-\tan t$ — $\cos t/\sin t = \cot t$, not $\tan t$, this inverts which trig ratio the quotient actually forms, on top of missing the sign.
Using the same method, find $d y/d x$ for $x = 2 \cos t$, $y = 3 \sin t$.
$-(3/2) \cot t$
$-(3/2) \tan t$
$(3/2) \cot t$
$-(2/3) \cot t$
Check your answer
✓ $-(3/2) \cot t$ — (A) $d x/d t = -2 \sin t$ and $d y/d t = 3 \cos t$, so $d y/d x = 3 \cos t / (-2 \sin t) = -(3/2) \cot t$.
$-(3/2) \tan t$ — $\cos t/\sin t = \cot t$, the same relationship as the plain circle, using $\tan t$ here inverts which trig ratio the quotient forms.
$(3/2) \cot t$ — The denominator $-2 \sin t$ is negative, so that sign carries through to the final ratio, dropping it loses that.
$-(2/3) \cot t$ — This computes $d x/d t$ over $d y/d t$ instead of $d y/d t$ over $d x/d t$, the coefficient ends up upside down as $2/3$ because the whole ratio is upside down.
Position, velocity, acceleration: each is the slope of the one above. For the dropped ball, s = 5t², v = 10t, a = 10.
In Physics in Class 11 you met three ways to describe a moving object. Its position says where it is. Its velocity says how fast the position changes. Its acceleration says how fast the velocity changes. So acceleration is a rate of change of a rate of change. You find it by taking a rate twice.
x⁴
→
d/dx
→
d/dx
→
12x²
Differentiate twice: x⁴ gives 4x³, then 12x². That is the second derivative, written d²y/dx² or y″.
In maths this is the SECOND DERIVATIVE. Differentiate $y = f(x)$ once to get $d y/d x$. Differentiate that again to get the second derivative. It is written $d^2 y / d x^2$, or $f''(x)$, or $y''$. All three mean the same thing.
5t²
→
d/dt
at t = 2: 20 m/s
→
d/dt
10 m/s² at every t
→
10
At t = 2 the ball has fallen 20 m and moves at 20 m/s. Its speed grows by 10 m/s every second.
Let us try it on a ball dropped from rest, taking $g$ as 10. After $t$ seconds it has fallen this far, in metres:
$s = 5t^2$.
Differentiate once for the velocity:
$d s/d t = 10t$.
Differentiate again for the acceleration:
$d^2 s / d t^2 = 10$.
It is the same at every moment. That is the pull of gravity, taken as 10 metres per second per second.
Both tangents touch at x equals 1. The cup-shaped curve stays above its tangent, and the cap-shaped curve stays below.
sin x bends like a cap from 0 to π, where y″ < 0, and like a cup from π to 2π.
On a graph, $d y/d x$ is the slope. The second derivative says how that slope is changing. Where it is positive, the slope is increasing, and the curve bends up like a cup. Where it is negative, the curve bends down like a cap. The next chapter uses this to tell a highest point from a lowest one.
x⁴
→
d/dx
→
d/dx
→
d/dx
→
24x
One more step gives the third derivative: x⁴, 4x³, 12x², 24x.
You can keep going. Differentiate a third time to get the third derivative, $d^3 y / d x^3$.
Your turn: find $d^2 y / d x^2$ for $y = x^3$. What is its value at $x = 2$?
Check yourself
What is $d^2 y/d x^2$, in terms of $d y/d x$?
the value of $d y/d x$ evaluated twice at the same point
the derivative of $d y/d x$ with respect to $x$
the reciprocal of $d y/d x$
the square of $d y/d x$
Check your answer
the value of $d y/d x$ evaluated twice at the same point — Evaluating the same function twice at one point gives the same number both times, the second derivative instead requires actually differentiating $d y/d x$ again, producing a new function.
✓ the derivative of $d y/d x$ with respect to $x$ — (B) $d^2 y/d x^2$ is just $d y/d x$ differentiated one more time, with respect to $x$.
the reciprocal of $d y/d x$ — Nothing in the second-order derivative involves flipping a fraction, it is a fresh derivative taken of the slope function, not its reciprocal.
the square of $d y/d x$ — “Second” here names how many times differentiation is applied, not an exponent, $d^2 y/d x^2$ means differentiating twice, never $((d y)/(d x))^2$.
Which three notations all name the same thing, the second-order derivative of $y$ with respect to $x$?
$d^2 y/d x^2$, $f''(x)$, $y''$
$((d y)/(d x))^2$, $f''(x)$, $y''$
$(d y^2)/d x$, $f''(x)$, $y''$
$d^2 y/d x^2$, $f'(x)$, $y'$
Check your answer
✓ $d^2 y/d x^2$, $f''(x)$, $y''$ — (A) $d^2 y/d x^2$, $f''(x)$, and $y''$ are three interchangeable ways of writing the same second-order derivative.
$((d y)/(d x))^2$, $f''(x)$, $y''$ — $((d y)/(d x))^2$ is the slope function squared, a different quantity from differentiating it again, it shares no notation with the genuine second derivative.
$(d y^2)/d x$, $f''(x)$, $y''$ — $(d y^2)/d x$ places the exponent on $y$ instead of marking that the differentiation is applied twice, the correct notation is $d^2 y/d x^2$, with the $2$ on both $d$ and $d x$.
$d^2 y/d x^2$, $f'(x)$, $y'$ — $f'(x)$ and $y'$ name the first derivative, $d y/d x$, one differentiation short of $d^2 y/d x^2$.
Geometrically, what does $d^2 y/d x^2$ track about a curve?
how the curve’s slope itself is changing
how steep the curve is at a point
how far the curve is from the x-axis
how many times the curve crosses the x-axis
Check your answer
✓ how the curve’s slope itself is changing — (A) $d y/d x$ tracks the slope; $d^2 y/d x^2$ tracks how that slope itself is changing, a first step toward describing concavity.
how steep the curve is at a point — Steepness at a point is exactly what $d y/d x$, the first derivative, tracks, the second derivative tracks how that steepness is changing, one step further.
how far the curve is from the x-axis — Distance from the x-axis is the value of $y$ itself, tracked by neither derivative, that is what the original function measures.
how many times the curve crosses the x-axis — Crossing count is a property of the function’s own graph, unrelated to either derivative.
A curve’s slope is positive and getting steeper as $x$ increases. What does this say about $d y/d x$ and $d^2 y/d x^2$?
$d y/d x$ is negative and $d^2 y/d x^2$ is positive
$d y/d x$ is positive and $d^2 y/d x^2$ is zero
$d y/d x$ is positive and $d^2 y/d x^2$ is positive
$d y/d x$ is positive and $d^2 y/d x^2$ is negative
Check your answer
$d y/d x$ is negative and $d^2 y/d x^2$ is positive — The slope itself is described as positive, not negative, flipping that sign contradicts the given information before the second derivative is even considered.
$d y/d x$ is positive and $d^2 y/d x^2$ is zero — A second derivative of zero means the slope is not changing at all, staying constant, that contradicts a slope described as getting steeper.
✓ $d y/d x$ is positive and $d^2 y/d x^2$ is positive — (C) A positive, increasing slope means $d y/d x$ is positive, and since that slope is itself increasing, $d^2 y/d x^2$ is positive too.
$d y/d x$ is positive and $d^2 y/d x^2$ is negative — “Getting steeper” means the slope is increasing as $x$ increases, which is exactly what a positive second derivative records, a negative second derivative would mean the slope is decreasing instead.
At x = −2: height 0.541, slope 0, second derivative −0.271, so the curve is a cap there.
Worked example
Find $d^2 y / d x^2$ for $y = x^2 e^x$
$y = x^2 e^x$ A product of $x^2$ and $e^x$.
$d y/d x = 2x e^x + x^2 e^x$ Use the product rule. The slope of $x^2$ is $2x$, and the slope of $e^x$ is $e^x$.
$d y/d x = e^x (x^2 + 2x)$ Take out the common factor $e^x$.
$d^2 y / d x^2 = e^x (x^2+2x) + e^x (2x+2)$ Differentiate again with the product rule. The slope of $x^2 + 2x$ is $2x + 2$.
$d^2 y / d x^2 = e^x (x^2 + 4x + 2)$ Add the like terms, $2x + 2x = 4x$.
$x = -2$: $e^{-2} (4 - 8 + 2) = -2 e^{-2} \approx -0.271$ Check at $x = -2$. The answer is the value the figure marks on its bottom curve.
Check yourself
For $y = x^2 \cdot e^x$, what is $d y/d x$ by the product rule?
$2x + e^x$
$e^x \cdot (x^2+2x)$
$2x \cdot e^x + x^2 \cdot 0$
$x^2 \cdot e^x$
Check your answer
$2x + e^x$ — $x^2 \cdot e^x$ is a product, not a sum, the sum rule, differentiate each piece and add, does not apply here; the product rule’s two-term structure is required instead.
✓ $e^x \cdot (x^2+2x)$ — (B) By the product rule, $d y/d x = 2x \cdot e^x + x^2 \cdot e^x = e^x \cdot (x^2+2x)$.
$2x \cdot e^x + x^2 \cdot 0$ — The product rule needs both pieces, $2x \cdot e^x$ from differentiating $x^2$, plus $x^2 \cdot e^x$ from differentiating $e^x$, this keeps only the first and drops the second.
$x^2 \cdot e^x$ — This keeps only the term from differentiating $e^x$ and drops the term from differentiating $x^2$, the product rule needs a contribution from each factor.
Continuing from $d y/d x = e^x \cdot (x^2+2x)$, what is $d^2 y/d x^2$?
$e^x \cdot (2x+2)$
$e^x \cdot (x^2+2x+2)$
$e^x \cdot (x^2+2x)$
$e^x \cdot (x^2+4x+2)$
Check your answer
$e^x \cdot (2x+2)$ — This keeps only the term from differentiating the bracket $(x^2+2x)$ and drops the term from differentiating $e^x$ itself, the product rule needs both, $e^x \cdot (x^2+2x) + e^x \cdot (2x+2)$.
$e^x \cdot (x^2+2x+2)$ — Combining $(x^2+2x)$ and $(2x+2)$ term by term gives $x^2+4x+2$, this drops one of the two $2x$ terms along the way, landing on $x^2+2x+2$ instead.
$e^x \cdot (x^2+2x)$ — This repeats the first derivative unchanged, the second derivative requires differentiating $d y/d x$ again, not restating it.
Using the same product-rule-twice method, find $d^2 y/d x^2$ for $y = x \cdot e^x$.
$e^x \cdot (x+1)^2$
$e^x \cdot x$
$e^x \cdot (x+1)$
$e^x \cdot (x+2)$
Check your answer
$e^x \cdot (x+1)^2$ — Squaring the first derivative’s bracket is not the same operation as differentiating it again, the second derivative comes from applying the product rule a second time, not from squaring.
$e^x \cdot x$ — Differentiating $e^x \cdot (x+1)$ needs both product-rule terms, $e^x \cdot (x+1) + e^x \cdot 1$, this keeps only part of that sum and loses the constant term.
$e^x \cdot (x+1)$ — $e^x \cdot (x+1)$ is the first derivative, $d y/d x$, reached correctly, but the question asks for $d^2 y/d x^2$, one differentiation step further.
✓ $e^x \cdot (x+2)$ — (D) $d y/d x = e^x + x \cdot e^x = e^x \cdot (x+1)$; differentiating again by the product rule gives $d^2 y/d x^2 = e^x \cdot (x+1) + e^x \cdot 1 = e^x \cdot (x+2)$.
A ball thrown straight up, at the top of its flight. For that moment its velocity is 0.
The board does not examine Rolle’s theorem or the mean value theorem in 2026-27. They explain why the next chapter’s slope test for increasing and decreasing functions works.
Thrown at t = 0, caught at t = 2 at the same height. Between, at t = 1, the slope is 0.
Throw a ball straight up and catch it at the same height. On the way up, its velocity points up. On the way down, it points down. The velocity changes without a jump, so at the top it passes through zero. For one instant the ball is still.
Both ends sit at height 2, at x equals 1 and 5. The tangent is flat at the top, at x equals 3.
ROLLE’S THEOREM says the same about a graph. Let $f$ be continuous on the closed interval $[a, b]$ and differentiable on the open interval $(a, b)$. If $f(a) = f(b)$, then there is at least one $c$ in $(a, b)$ with $f'(c) = 0$. At that point the tangent is flat.
x² − 4x is 0 at both ends of [0, 4]. Its slope 2x − 4 is 0 at c = 2.
Check it on a polynomial whose two ends are equal.
Worked example
Check Rolle’s theorem on $x^2 - 4x$
$f(x) = x^2 - 4x$ on $[0, 4]$ A polynomial is continuous and differentiable everywhere, so both conditions hold.
$f(0) = 0, \quad f(4) = 16 - 16 = 0$ Work out the values at the two ends: they are equal.
$f'(x) = 2x - 4$ Differentiate to get the slope at any point.
$2c - 4 = 0 \to c = 2$ Set the slope at $c$ equal to 0, and solve.
Check: $0 < 2 < 4$ The point $c = 2$ lies inside $(0, 4)$, as the theorem says.
a break at the end
a corner
unequal ends
Drop one condition and the flat tangent can vanish: a break at the end, a corner in the middle, or unequal ends.
Three graphs: a break, a corner, unequal ends. Each drops one condition of Rolle’s theorem, and none has a flat tangent inside.
A break: $f(x) = x$ on $[0, 1)$ with $f(1) = 0$ has equal ends, but its slope is always 1, so no point has slope 0.
A corner: $1 - \left|x\right|$ on $[-1, 1]$ has equal ends, but its top is a corner with no slope, so no point has slope 0.
Unequal ends: $y = x$ on $[0, 1]$ climbs with slope 1 all the way, so no point has slope 0.
So Rolle’s theorem turns two equal values into a flat tangent between them.
Check yourself
Rolle’s theorem concludes that some $c \in (a,b)$ satisfies which condition?
$f'(c) = 0$
$f(c) = f(a)$
$f''(c) = 0$
$f(c) = 0$
Check your answer
✓ $f'(c) = 0$ — (A) Rolle’s theorem concludes that some $c \in (a,b)$ has $f'(c)=0$.
$f(c) = f(a)$ — Matching an endpoint value is the hypothesis, $f(a)=f(b)$, not the conclusion, the conclusion is about the derivative vanishing at some interior point.
$f''(c) = 0$ — Rolle’s theorem says nothing about the second derivative. Its conclusion is about the first: the tangent at $c$ is flat, $f'(c)=0$.
$f(c) = 0$ — Rolle’s theorem is a statement about the derivative at $c$, not the function value, $f(c)=0$ is not part of the conclusion and need not hold.
What does Rolle’s theorem say happens geometrically between two points of equal height on a smooth curve?
the curve must be a straight line between them
the curve must cross the x-axis somewhere between them
the curve’s tangent must point toward one of the two endpoints
the curve’s tangent must be horizontal somewhere between them
Check your answer
the curve must be a straight line between them — The curve can bend arbitrarily between the two endpoints, Rolle’s theorem only guarantees one horizontal tangent somewhere in between, not that the whole arc is straight.
the curve must cross the x-axis somewhere between them — Rolle’s theorem says nothing about crossing the x-axis, that would require the curve’s height to hit zero, not its slope.
the curve’s tangent must point toward one of the two endpoints — A tangent line has a slope, not a direction “toward” a point, the theorem’s claim is specifically that this slope is zero, horizontal, at some interior point.
✓ the curve’s tangent must be horizontal somewhere between them — (D) If a smooth curve starts and ends at the same height, somewhere between the two ends its tangent line must be horizontal.
$f(x) = |x|$ on $[-1,1]$ has $f(-1)=f(1)=1$, but there is no $c \in (-1,1)$ with $f'(c)=0$. Which hypothesis of Rolle’s theorem fails?
differentiability on open interval $(-1,1)$: $f$ is not differentiable at $x=0$
none of the three hypotheses fail; Rolle’s theorem is simply wrong here
continuity on the closed interval $[-1,1]$: $|x|$ is discontinuous somewhere on it
the equal-endpoint condition: $f(-1)$ does not actually equal $f(1)$
Check your answer
✓ differentiability on open interval $(-1,1)$: $f$ is not differentiable at $x=0$ — (A) $|x|$ is not differentiable at $x=0$, which lies in the open interval $(-1,1)$, the differentiability hypothesis fails, so the theorem does not apply, and no contradiction occurs.
none of the three hypotheses fail; Rolle’s theorem is simply wrong here — Rolle’s theorem is not violated, it simply does not apply here, because one of its three hypotheses, differentiability on the open interval, fails first.
continuity on the closed interval $[-1,1]$: $|x|$ is discontinuous somewhere on it — $|x|$ is continuous everywhere, including on $[-1,1]$, the break is in its slope at $x=0$, not in the function’s value.
the equal-endpoint condition: $f(-1)$ does not actually equal $f(1)$ — The endpoints are given as equal, $f(-1)=f(1)=1$, that hypothesis holds; the failure is elsewhere.
$f(x) = x$ on $[0,1]$ has $f(0)=0 \neq f(1)=1$, so the equal-endpoints hypothesis fails, yet $f'(x)=1$ everywhere and is never zero on $(0,1)$. What does this show?
continuity must also have failed for this to happen
differentiability must also have failed for this to happen
dropping equal endpoints can remove that guarantee
Rolle’s theorem is contradicted by this example
Check your answer
continuity must also have failed for this to happen — $f(x)=x$ is continuous everywhere, including on $[0,1]$, continuity is not the hypothesis that failed here.
differentiability must also have failed for this to happen — $f(x)=x$ is differentiable everywhere, including on $(0,1)$, differentiability is not the hypothesis that failed here either.
✓ dropping equal endpoints can remove that guarantee — (C) With only the equal-endpoints hypothesis dropped, and $f'(x)=1$ never zero, this shows that hypothesis genuinely matters, losing it can genuinely remove the guarantee.
Rolle’s theorem is contradicted by this example — Rolle’s theorem only promises a zero derivative when all three hypotheses hold, here one is missing, so the theorem simply makes no promise; the missing zero derivative is not a contradiction.
To cover 120 km in 2 hours, the car must average 60 km/h. At 50 or 70 it would end at 100 or 140.
Say a car covers 120 kilometres in 2 hours. Its average speed is 60 kilometres an hour. It went faster at some times and slower at others. But at some moment its speedometer read exactly 60. If it had stayed below 60 the whole time, it would have covered less than 120 kilometres. If it had stayed above 60, it would have covered more.
The chord from (1, 1.25) to (5, 7.25) has slope 1.5. The tangent at x equals 3 has the same slope and runs parallel.
The MEAN VALUE THEOREM says this for a graph. Let $f$ be continuous on $[a, b]$ and differentiable on $(a, b)$. Then there is at least one $c$ in $(a, b)$ with $f'(c) = (f(b) - f(a))/(b - a)$.
Rolle: y = x² − 4x
MVT: y = x² − 3x
Rolle is the flat case: the chord has slope 0. Tilt the ends and some tangent still matches the chord.
The right side is the slope of the chord from $(a, f(a))$ to $(b, f(b))$. So at some point inside, the tangent is parallel to the chord. Rolle’s theorem is the case $f(a) = f(b)$, where the chord is flat.
s = 30t² covers 120 km in 2 hours. At t = 1 the speedometer reads 60, the mean speed.
Check it with the car from the start, which covered 120 kilometres in 2 hours.
Worked example
Check it with the car
$s = 30t^2$ Suppose this is the car’s distance in kilometres after $t$ hours.
$t = 2$: $s = 30 \cdot 2^2 = 120$ Work out the distance after 2 hours: 120 kilometres.
$(120 - 0)/(2 - 0) = 60$ The slope of the chord is the average speed, 60 kilometres an hour.
$d s/d t = 60t$ Differentiate to get the speed at each moment.
$60t = 60 \to t = 1$ Set the speed equal to the average speed, and solve.
Check: $0 < 1 < 2$ The moment $t = 1$ lies inside $(0, 2)$, as the theorem says.
c = √(7/3) ≈ 1.528
Slope 3x² is positive everywhere on [1, 2], so the chord slope, 7, is positive too: f(2) > f(1).
Here is why it matters for the next chapter. Suppose $f$ is differentiable on an interval, with $f'(x) > 0$ at every point of it. Take any two points $x_1 < x_2$ in it. By the mean value theorem, there is a $c$ between them with $f(x_2) - f(x_1) = f'(c)(x_2 - x_1)$. Both factors on the right are positive, so $f(x_2) > f(x_1)$. So $f$ is increasing there, and that is the next chapter’s test.
Check yourself
The Mean Value Theorem concludes that some $c \in (a,b)$ satisfies which equation?
$f(c) = (f(b)-f(a))/(b-a)$
$f'(c) = 0$
$f'(c) = (f(b)-f(a))/(b-a)$
$f'(c) = (f(a)-f(b))/(b-a)$
Check your answer
$f(c) = (f(b)-f(a))/(b-a)$ — The left side must be a derivative, $f'(c)$, matched against a slope, writing $f(c)$ instead reports a function value, not a rate of change.
$f'(c) = 0$ — $f'(c)=0$ is Rolle’s conclusion, the special case where $f(a)=f(b)$, the MVT’s general conclusion equates $f'(c)$ to the chord’s slope, not to zero.
✓ $f'(c) = (f(b)-f(a))/(b-a)$ — (C) The Mean Value Theorem concludes that some $c \in (a,b)$ has $f'(c) = (f(b)-f(a))/(b-a)$.
$f'(c) = (f(a)-f(b))/(b-a)$ — $(f(a)-f(b))/(b-a)$ flips the sign of the numerator relative to the correct chord slope, $(f(b)-f(a))/(b-a)$.
Geometrically, what does the Mean Value Theorem guarantee about the tangent line at $c$?
it passes through both endpoints $(a,f(a))$ and $(b,f(b))$
it is perpendicular to the chord joining the two endpoints
it is parallel to the chord joining the two endpoints
it is horizontal
Check your answer
it passes through both endpoints $(a,f(a))$ and $(b,f(b))$ — The tangent line at $c$ touches the curve at $c$ alone, it matches the chord’s slope, but there is no requirement that it pass through the two endpoints themselves.
it is perpendicular to the chord joining the two endpoints — The theorem matches the tangent’s slope to the chord’s slope, matching slopes means parallel, not a right-angle relationship.
✓ it is parallel to the chord joining the two endpoints — (C) The tangent line at $c$ has the same slope as the chord joining $(a,f(a))$ and $(b,f(b))$, so the two lines are parallel.
it is horizontal — A horizontal tangent is Rolle’s special case, where the chord itself is horizontal because $f(a)=f(b)$, the general MVT allows any chord slope.
How does Rolle’s theorem relate to the Mean Value Theorem?
Rolle’s theorem is the special case of the MVT where $f(a)=f(b)$, making the chord horizontal
the MVT is the special case of Rolle’s theorem where $f(a)=f(b)$
the two theorems are unrelated and require entirely different hypotheses
Rolle’s theorem replaces the MVT’s differentiability hypothesis with a continuity-only requirement
Check your answer
✓ Rolle’s theorem is the special case of the MVT where $f(a)=f(b)$, making the chord horizontal — (A) Rolle’s theorem is the MVT with the extra condition $f(a)=f(b)$, which makes the chord horizontal and collapses the conclusion to $f'(c)=0$.
the MVT is the special case of Rolle’s theorem where $f(a)=f(b)$ — The direction is the other way round, Rolle’s theorem is the narrower, special case; the MVT is the more general statement that Rolle’s sits inside.
the two theorems are unrelated and require entirely different hypotheses — The two share the same continuity-on-$[a,b]$ and differentiability-on-$(a,b)$ hypotheses, Rolle’s only adds one extra condition, $f(a)=f(b)$, on top.
Rolle’s theorem replaces the MVT’s differentiability hypothesis with a continuity-only requirement — Both theorems require differentiability on the open interval, Rolle’s does not trade that requirement away; it only adds the equal-endpoints condition.
If a function satisfies the MVT’s hypotheses on $[a,b]$ and also has $f(a)=f(b)$, what does the MVT’s conclusion simplify to?
$f'(c) = b-a$
$f'(c) = 0$, exactly Rolle’s conclusion
$f'(c) = 1$
the MVT no longer applies once $f(a)=f(b)$
Check your answer
$f'(c) = b-a$ — The denominator $(b-a)$ is just the interval width, it is the numerator, $f(b)-f(a)$, that becomes $0$ when $f(a)=f(b)$, driving the whole fraction to $0$, not to $b-a$.
✓ $f'(c) = 0$, exactly Rolle’s conclusion — (B) With $f(a)=f(b)$, the numerator $f(b)-f(a)$ becomes $0$, so the MVT’s conclusion collapses to $f'(c)=0$, exactly Rolle’s theorem.
$f'(c) = 1$ — Setting $f(a)=f(b)$ makes the numerator $f(b)-f(a)$ equal to $0$, not the whole fraction equal to $1$, treating it as if numerator equalled denominator conflates two unrelated quantities.
the MVT no longer applies once $f(a)=f(b)$ — The MVT’s hypotheses, continuity and differentiability, are unaffected by the endpoint values, it still applies, and its conclusion simply simplifies to Rolle’s.
The chord from (1, 1) to (4, 16) has slope 5. The tangent at c equals 5 over 2 is parallel to it.
[1, 4]
For x², the c the theorem finds is always the midpoint: here 5/2, halfway from 1 to 4.
Worked example
Verify the Mean Value Theorem for $f(x)=x^2$ on $[1,4]$
$f(x) = x^2$ A polynomial is continuous and differentiable everywhere. So both conditions hold on $[1, 4]$.
$f(1) = 1, \quad f(4) = 16$ Work out the values at the two ends.
$(f(4)-f(1))/(4-1) = (16-1)/3 = 5$ This is the slope of the chord from $(1, 1)$ to $(4, 16)$. The theorem says some tangent inside has this slope.
$f'(x) = 2x$ Differentiate $f$ to get the slope at any point.
$2c = 5 \to c = 5/2$ Set the slope at $c$ equal to the chord’s slope, and solve.
$1 < 5/2 < 4$ The point $c = 5/2$ lies inside $(1, 4)$, as the theorem needs.
$f'(5/2) = 2 \cdot 5/2 = 5$ Check that the slope at $c$ is the chord’s slope, 5.
Check yourself
For $f(x)=x^2$ on $[1,4]$, why are the MVT’s continuity and differentiability hypotheses automatically satisfied?
because $[1,4]$ contains only positive numbers
because $f(1)$ and $f(4)$ happen to be different values
because the interval $[1,4]$ is closed
because polynomials are continuous and differentiable
Check your answer
because $[1,4]$ contains only positive numbers — Whether the interval contains positive or negative numbers has nothing to do with continuity or differentiability, $x^2$ is continuous and differentiable on every interval, not just positive ones.
because $f(1)$ and $f(4)$ happen to be different values — $f(1) \neq f(4)$ is a fact about the endpoint values, not about why $f$ is continuous or differentiable, that comes from $f$ being a polynomial, full stop.
because the interval $[1,4]$ is closed — The interval being closed is a fact about $[1,4]$ itself, not about the function $f$, it does not by itself make $f$ continuous or differentiable; $f$ being a polynomial does.
✓ because polynomials are continuous and differentiable — (D) Every polynomial, including $f(x)=x^2$, is continuous and differentiable at every real number, so both hypotheses hold on $[1,4]$ with nothing further to check.
For $f(x)=x^2$ on $[1,4]$, what is the MVT slope $(f(4)-f(1))/(4-1)$?
$3$
$15$
$17$
$5$
Check your answer
$3$ — $3$ is just the interval width $4-1$ on its own, the slope needs the value difference $16-1=15$ divided by that width, not the width alone.
$15$ — $f(4)-f(1) = 16-1 = 15$ is only the numerator, the MVT slope still needs dividing by the interval width, $4-1=3$, giving $5$.
$17$ — $f(4)+f(1) = 16+1=17$ adds the two endpoint values; the slope formula subtracts them, $16-1=15$, then divides by $3$.
✓ $5$ — (D) $f(1)=1$ and $f(4)=16$, so the MVT slope is $(16-1)/3 = 5$.
Using the same method, verify the MVT for $f(x)=x^2$ on $[2,5]$: what are the MVT slope and the resulting $c$?
slope $=7$, $c = 7$
slope $=21$, $c = 21/2$
slope $=3$, $c = 3/2$
slope $=7$, $c = 7/2$
Check your answer
slope $=7$, $c = 7$ — The slope $7$ is correct, but solving $2c=7$ gives $c=7/2$, not $c=7$, dividing by $2$, the derivative’s coefficient, is the step this skips.
slope $=21$, $c = 21/2$ — $f(5)-f(2)=25-4=21$ is only the numerator, the MVT slope still needs dividing by the interval width $5-2=3$, giving $7$, not $21$.
slope $=3$, $c = 3/2$ — $5-2=3$ is the interval width, not the slope, the slope also needs the value difference $f(5)-f(2)=21$ divided by that width.
✓ slope $=7$, $c = 7/2$ — (D) $f(2)=4$ and $f(5)=25$, giving slope $(25-4)/3=7$; solving $f'(c)=2c=7$ gives $c=7/2$, which lies in $(2,5)$.
Fourteen idea cards joined by arrows, from continuity down to the mean value theorem. Each arrow means the later idea uses the earlier one.
The question at the start had three parts. Which graphs can you draw without lifting the pen? Which of those have one slope at every point? And how do you find the slope of the new functions of Class 12?
x² at 1
the card at 500 g
|x| at 0
The opening question answered: x² is both, the rate card breaks at 500 g, and the V of |x| is unbroken but not smooth.
CONTINUOUS at $a$: $\text{LHL} = \text{RHL} = f(a)$. For $2x + 1$ at 3, all three are 7.
DIFFERENTIABLE at $a$: the LHD and the RHD exist and are equal. For $x^2$ at 1, both are 2.
ONE WAY ONLY: differentiable gives continuous. $\left|x\right|$ is continuous at 0, but its LHD is $-1$ and its RHD is $1$.
CHAIN RULE: $d y/d x = d y/d u \cdot d u/d x$. For $(2x + 1)^2$ it gives $8x + 4$.
NEW SLOPES: $\sin^{-1} x$ has slope $1/\sqrt{1 - x^2}$, and $e^x$ has slope $e^x$. At $x = 1$, $\ln x$ has slope $1/x = 1$.
PARAMETRIC: $d y/d x = ((d y)/(d t))/((d x)/(d t))$. For the ball, $x = 10t$ and $y = 10t - 5t^2$ give $1 - t$.
Answering the three questions
Here is the answer. A graph you can draw without lifting the pen has no jump, so it is continuous. A graph with one slope at each point has no corner as well, so it is differentiable. Every differentiable function is continuous, but $\left|x\right|$ shows the reverse is false. The chain rule then reaches every function you met here. You use it for inverse trig functions and for curves given by an equation. You use it with logs for powers like $x^x$, and with a parameter for paths. Differentiate twice for the second derivative. Rolle’s theorem and the mean value theorem are enrichment. They link the slope inside an interval to the values at its ends.
Check yourself
Which relationship between continuity and differentiability holds in general?
differentiable implies continuous, but not the reverse
continuous implies differentiable, but not the reverse
continuous implies differentiable, and differentiable implies continuous: the two are equivalent
neither property implies the other
Check your answer
✓ differentiable implies continuous, but not the reverse — (A) Differentiable always implies continuous; continuous never guarantees differentiable, $|x|$ at $x=0$ is the standing counterexample for the whole chapter.
continuous implies differentiable, but not the reverse — This states the false direction, $|x|$ at $x=0$ is the standing counterexample, continuous there, but not differentiable there. The true direction runs the other way.
continuous implies differentiable, and differentiable implies continuous: the two are equivalent — The two properties are not equivalent, $|x|$ at $x=0$ is continuous but not differentiable, so continuity does not force differentiability.
neither property implies the other — One direction genuinely always holds, differentiable always implies continuous, so the two properties are not independent of each other.
The chain rule, implicit differentiation, logarithmic differentiation, and parametric derivatives all developed later in this chapter. What are they, relative to the continuity/differentiability foundation?
replacements for the continuity test, used once differentiability is confirmed
methods that only apply to functions which fail the differentiability test
faster ways to compute a derivative the limit definition answers too slowly
separate new ideas about what it means for a function to be smooth
Check your answer
replacements for the continuity test, used once differentiability is confirmed — Continuity is a separate question, checked earlier and not superseded by any differentiation technique, these methods compute a derivative, they do not test whether one exists.
methods that only apply to functions which fail the differentiability test — These techniques are used precisely when a derivative exists and needs computing efficiently, they are not tools reserved for functions that already failed a differentiability test.
✓ faster ways to compute a derivative the limit definition answers too slowly — (C) Each of these techniques is a faster way to compute a derivative the limit definition would answer too slowly, not a new idea about smoothness.
separate new ideas about what it means for a function to be smooth — None of these techniques change what smoothness means, that question was already settled by the LHD/RHD definition earlier. They are computational shortcuts, not new definitions.
How do Rolle’s theorem and the Mean Value Theorem relate to the derivative-computing techniques covered earlier?
they replace the need to compute a derivative at all
they only apply to functions that are continuous but not differentiable
they turn a derivative into a claim about behaviour across an interval
they re-test whether a function is differentiable at a single point
Check your answer
they replace the need to compute a derivative at all — Both theorems still need $f'(x)$ to exist and be well-behaved on the interval, they use a computed derivative to conclude something further, they do not remove the need to compute one.
they only apply to functions that are continuous but not differentiable — Both theorems specifically require differentiability on the open interval as a hypothesis, a function that fails to be differentiable does not qualify.
✓ they turn a derivative into a claim about behaviour across an interval — (C) Rolle’s theorem and the MVT take a derivative that is already known to exist and turn it into a statement about the function’s behaviour across the whole interval $[a,b]$, not just its value at one point.
they re-test whether a function is differentiable at a single point — Testing differentiability at one point is the earlier LHD/RHD work, Rolle’s and the MVT instead use the derivative to make a claim about behaviour spread across the whole interval.
Check yourself: the whole chapter
A graph of $f$ has no break at $x=a$: it can be drawn without lifting the pen. What follows about whether $f$ has a single well-defined tangent line at $a$?
It must have one: an unbroken graph is always smooth.
Nothing follows for certain: a corner may leave no tangent.
It cannot have one: unbroken and smooth exclude each other.
It has one only if $f$ is a polynomial.
Check your answer
It must have one: an unbroken graph is always smooth. — Being unbroken never forces a single tangent line. $f(x) = |x|$ has no break at $x=0$, yet its graph has a corner there.
✓ Nothing follows for certain: a corner may leave no tangent. — (B) An unbroken graph need not be smooth. $f(x) = |x|$ has no break at $x=0$, but its two sides meet at a corner, so no single tangent line exists there.
It cannot have one: unbroken and smooth exclude each other. — The two are not opposites either. A great many functions, such as any polynomial, are both unbroken and smooth everywhere.
It has one only if $f$ is a polynomial. — Polynomials are one sufficient case, not the rule. Functions such as $\sin x$ or $e^x$ are smooth too, without being polynomials.
For $f$ at $x=a$, suppose $\lim_{x \to a^-} f(x) = \lim_{x \to a^+} f(x) = 5$, but $f(a) = 3$. Is $f$ continuous at $x=a$?
Yes, the left and right limits agree, so $f$ is continuous at $a$.
Yes, since $f(a)$ exists and the limit also exists, that is already enough.
No, $f$ is discontinuous because the two one-sided limits disagree.
No, $f$ is discontinuous since the shared limit does not equal $f(a)$.
Check your answer
Yes, the left and right limits agree, so $f$ is continuous at $a$. — Agreement between the two one-sided limits is not enough by itself. That shared value, $5$, must also equal $f(a) = 3$, and it does not.
Yes, since $f(a)$ exists and the limit also exists, that is already enough. — $f(a)$ and the limit both existing is not the same as them matching. Here they exist at $3$ and $5$, two different numbers.
No, $f$ is discontinuous because the two one-sided limits disagree. — The one-sided limits do agree, both at $5$. The break sits elsewhere: that shared value does not match $f(a) = 3$.
✓ No, $f$ is discontinuous since the shared limit does not equal $f(a)$. — (D) $\lim_{x \to a^-} f(x) = \lim_{x \to a^+} f(x) = 5$, so the two-sided limit exists and equals $5$. Since $f(a) = 3$ and $5$ is not $3$, the third condition fails, so $f$ is discontinuous at $a$.
At $x=a$, the left-hand derivative and the right-hand derivative of $f$ both equal $4$. What follows about the derivative of $f$ at $a$?
The derivative of $f$ at $a$ equals $4$: both one-sided derivatives exist and agree.
Nothing follows: LHD and RHD describe the function value, not its slope.
It cannot be settled without also checking the left-hand and right-hand limits of $f$ itself.
$f$ is continuous at $a$, but the derivative there is still undefined.
Check your answer
✓ The derivative of $f$ at $a$ equals $4$: both one-sided derivatives exist and agree. — (A) Differentiability at $a$ means exactly this: the left-hand and right-hand derivatives both exist and agree. Here both equal $4$, so the derivative of $f$ at $a$ is $4$.
Nothing follows: LHD and RHD describe the function value, not its slope. — LHD and RHD track the slope, the one-sided limit of the difference quotient, not the function value. Tracking the function value is LHL and RHL’s job, defined earlier in this chapter.
It cannot be settled without also checking the left-hand and right-hand limits of $f$ itself. — No extra check is needed here. LHD and RHD existing and agreeing is exactly what it means for the derivative at $a$ to exist, nothing further has to be verified first.
$f$ is continuous at $a$, but the derivative there is still undefined. — LHD equal to RHD equal to $4$ is precisely the definition of differentiability at $a$ being satisfied. The derivative at $a$ equals that shared value, $4$, directly.
$f$ is differentiable at $x=a$, and its derivative there equals $3$. As $h \to 0$, the product $[(f(a+h)-f(a))/h] \cdot h$ tends to which value, and what does that show?
It tends to $3$: this shows nothing new about continuity.
It is undefined, since dividing by $h$ as $h \to 0$ makes the fraction blow up.
It tends to $0$, forcing $f(a+h)$ toward $f(a)$: exactly continuity at $a$.
It tends to $3 h$, which shows continuity only in the special case $h = 0$.
Check your answer
It tends to $3$: this shows nothing new about continuity. — The factor of $h$ still matters, it is not dropped. A finite number, $3$, multiplied by something tending to $0$ gives $0$, not $3$ itself.
It is undefined, since dividing by $h$ as $h \to 0$ makes the fraction blow up. — The bracket is exactly the difference quotient that defines the derivative at $a$. Since $f$ is differentiable there, that bracket is known to tend to the finite number $3$, not to blow up.
✓ It tends to $0$, forcing $f(a+h)$ toward $f(a)$: exactly continuity at $a$. — (C) The bracket tends to the finite number $3$ while the factor $h$ tends to $0$, so their product tends to $3 \cdot 0 = 0$. That forces $f(a+h)$ to approach $f(a)$, which is exactly continuity at $a$.
It tends to $3 h$, which shows continuity only in the special case $h = 0$. — $h$ must actually be sent to its limit, $0$, rather than left symbolic. $3 h$ itself tends to $0$ as $h \to 0$, so the whole product tends to $0$.
Differentiate $y = (2x+1)^3$ using the chain rule.
$d y/d x = 3(2x+1)^2$: differentiate the outer power only.
$d y/d x = 6(2x+1)^2$: multiply the outer slope by the inner slope, $2$.
$d y/d x = 3(2x+1)^2 + 2$: add the inner slope instead of multiplying it in.
$d y/d x = 3(2x+1)^2 \cdot (2x+1)$: multiply by the inner function itself, not its slope.
Check your answer
$d y/d x = 3(2x+1)^2$: differentiate the outer power only. — The inner piece is not just $x$, it is $2x+1$, and its own slope, $2$, must be multiplied in. Dropping that factor is the single most common chain-rule slip.
✓ $d y/d x = 6(2x+1)^2$: multiply the outer slope by the inner slope, $2$. — (B) With $u = 2x+1$, $y = u^3$, so $d y/d u = 3 u^2$ and $d u/d x = 2$. The chain rule multiplies these: $d y/d x = 3(2x+1)^2 \cdot 2 = 6(2x+1)^2$.
$d y/d x = 3(2x+1)^2 + 2$: add the inner slope instead of multiplying it in. — The two slopes combine by multiplication, not addition, the outer slope at the inner value times the inner slope, never a sum of the two.
$d y/d x = 3(2x+1)^2 \cdot (2x+1)$: multiply by the inner function itself, not its slope. — The second factor needed is the inner function’s slope, $2$, not the inner function itself, $2x+1$.
You have already made a check like this, long before you met the word continuity.
A tax slab’s rate change. A tax rule charges one rate up to an income limit and a higher rate only above it. The maths: The tax total has no jump at the limit. Its slope still changes there, the same corner a left-hand and right-hand derivative test catches. Tax equals $0.10x$ up to $10$ lakh, and $1 + 0.20(x-10)$ above it. Both formulas give exactly $1$ lakh of tax at $x=10$, but their slopes differ.
A tank filling with water. A tank’s water level rises as a tap fills it. The maths: The water level never jumps from one height straight to a much higher one. It passes through every height in between, the same no-break test continuity itself uses.
A cyclist’s two linked turns. A cyclist’s distance depends on the wheel’s turns, and the wheel’s turns depend on the pedal’s turns. The maths: Distance depends on wheel turns, and wheel turns depend on pedal turns. Those two linked steps join exactly the way a chain rule links an outer step to an inner one. One pedal turn gives $2$ wheel turns, and one wheel turn gives $7$ metres. So one pedal turn gives $2 \times 7 = 14$ metres.
Two warm spots adding to one room’s reading. A room’s temperature equals the sun’s warming through the window plus the heater’s own warming. The maths: If each one changes without a jump through the day, their combined total changes without a jump too. That is exactly the sum rule an algebra of continuous functions relies on. At one moment the window adds $5$ degrees of warming and the heater adds $3$. The combined total is $5+3=8$ degrees.
A tailor’s seam between two panels. A tailor joins two curved panels of cloth along one seam. The maths: If the two edges arrive at different slopes, the seam shows as a visible fold. That same mismatch is what a left-hand and right-hand derivative test catches. One edge’s slope rises into the seam at $2$ and the other falls away at $-1$, so $2 \neq -1$.
Your turn. A courier charges Rs $50$ up to $2$ kg and Rs $90$ for anything heavier, even by a gram. Is the charge continuous at $2$ kg? If not, how large is the jump? Answer: No, it is not continuous. At $2$ kg the charge jumps from Rs $50$ to Rs $90$, a jump of Rs $40$.
LATER
Where continuity and differentiability go next
University first-year real analysis makes continuity and differentiability precise, using two small numbers: epsilon, a target distance, and delta, a step size that keeps the function within epsilon of the limit. Engineering signal processing runs the same left and right slope test on a signal with a sudden change in direction, checking whether the rate of change still agrees from both sides.
Exercise 5.1 tests continuity at a point. You find the LHL, the RHL and the value, then check that all three agree. Some ask you to choose a constant so that two pieces join up.
Exercise 5.1: Continuity at a point
practice Show that $f(x) = 4 x - 1$ is continuous at $x = 2$.
practice Show that $f(x) = -3 x + 5$ is continuous at $x = -1$.
practice Examine the continuity of $f(x) = 2 x^2 + 3$ at $x = 1$.
practice Examine the continuity of $f(x) = x^3 - 4 x$ at $x = -1$.
practice A function is defined by $f(x) = 3$ for $x \neq 2$, and $f(2) = 7$. Examine the continuity of $f$ at $x = 2$.
practice Prove that the constant function $f(x) = 9$ is continuous at every real number $c$.
practice Prove that the identity function $f(x) = x$ is continuous at every real number $c$.
practice Is $f(x) = |x - 3|$ continuous at $x = 3$?
practice Examine the continuity of $f(x) = x + 1$ for $x \leq 2$, and $f(x) = 3$ for $x > 2$, at $x = 2$.
practice Examine the continuity of $f(x) = 2 x + 1$ for $x \leq 1$, and $f(x) = x + 4$ for $x > 1$, at $x = 1$.
practice A function is defined by $f(x) = x + 3$ for $x < 0$, $f(0) = 1$, and $f(x) = x - 3$ for $x > 0$. Examine the continuity of $f$ at $x = 0$.
practice Show that $f(x) = (x^2 - 25)/(x + 5)$, $x \neq -5$, is continuous on its domain.
practice Find the value of $k$ so that $f(x) = k x + 3$ for $x \leq 2$, and $f(x) = x - 1$ for $x > 2$, is continuous at $x = 2$.
practice Find the value of $k$ so that $f(x) = 5$ for $x \leq 3$, and $f(x) = k x - 4$ for $x > 3$, is continuous at $x = 3$.
practice Find the value of $k$ so that $f(x) = k x^2$ for $x \leq 1$, and $f(x) = 2 x + 1$ for $x > 1$, is continuous at $x = 1$.
practice Find $a$ and $b$ so that $f(x) = 2$ for $x \leq 0$, $f(x) = a x + b$ for $0 < x < 2$, and $f(x) = 6$ for $x \geq 2$, is a continuous function.
practice Prove that $f(x) = x^5$ is continuous at $x = 2$.
practice A function is defined by $f(x) = 2$ for $x \leq 0$, and $f(x) = -2$ for $x > 0$. Examine the continuity of $f$ at $x = 0$.
practice Examine the continuity of $f(x) = \sin x + \cos x$ at $x = 0$.
practice Examine the continuity of $f(x) = \sin x \cdot \cos x$ at $x = \pi/2$.
practice Examine the continuity of $f(x) = \tan x$ at $x = \pi/4$.
practice Examine the continuity of the greatest integer function $f(x) = \lfloor x\rfloor$ at $x = 2$.
practice Examine the continuity of the greatest integer function $f(x) = \lfloor x\rfloor$ at $x = 2.5$.
practice The fractional-part function is $g(x) = x - \lfloor x\rfloor$. Examine the continuity of $g$ at $x = 3$.
practice Examine the continuity of $f(x) = |x - 1| + |x + 1|$ at $x = -1$.
practice Examine the continuity of $f(x) = 1/x$ for $x < 0$, and $f(x) = x + 1$ for $x \geq 0$, at $x = 0$.
practice Where is $f(x) = 1/(x^2 - 4)$ not defined? Is $f$ continuous at every point of its domain?
practice Where is $f(x) = 1/(x^2 - 3 x + 2)$ not defined? Is $f$ continuous at every point of its domain?
practice For what value of $k$ is $f(x) = k x + 1$ for $x \leq 3$, and $f(x) = 3 x - 5$ for $x > 3$, continuous at $x = 3$?
$1$
$3$
$-1$
$5$
practice At which of these points is $f(x) = \lfloor x\rfloor$ NOT continuous?
$x = 1.5$
$x = 2$
$x = -0.5$
$x = 0.25$
practice Examine the continuity of $f(x) = \sin x - \cos x$ at $x = 0$.
practice Examine the continuity of $f(x) = |\cos x|$ at $x = \pi/2$.
practice Examine the continuity of $f(x) = x + \lfloor x\rfloor$ at $x = 1$.
practice Show that $f(x) = \sin(x^2 + 1)$ is a continuous function.
Discontinuous at $x = 0$; $\text{LHL} = 3$, $\text{RHL} = -3$, whatever value $f(0)$ is given.
Continuous for every $x \neq -5$: away from $x = -5$ this is a quotient of continuous polynomials with a nonzero denominator, so it is continuous by the algebra of continuous functions.
Continuous everywhere: $x^2 + 1$ is a polynomial (continuous everywhere) and sin is continuous everywhere, so the composite is continuous everywhere by the composition rule.
In Exercise 5.2 you work out the LHD and the RHD from the limit. Then you decide whether the function is differentiable at the point. Watch for corners, where the function is continuous but the two slopes differ.
Exercise 5.2: Differentiability
practice Compute the left-hand and right-hand derivatives of $f(x) = |x - 3|$ at $x = 3$, and decide whether $f$ is differentiable there.
practice Compute the left-hand and right-hand derivatives of $f(x) = |x + 1|$ at $x = -1$, and decide whether $f$ is differentiable there.
practice Show that $f(x) = |x|$ is differentiable at $x = 2$, and find $f'(2)$.
practice Show that $f(x) = |x|$ is differentiable at $x = -3$, and find $f'(-3)$.
practice Use the difference-quotient definition to show that $f(x) = x^2$ is differentiable at $x = 0$.
practice A function is defined by $f(x) = |x| + |x - 1|$. Compute the left-hand and right-hand derivatives at $x = 0$, and decide whether $f$ is differentiable there.
practice For the same function $f(x) = |x| + |x - 1|$, compute the left-hand and right-hand derivatives at $x = 1$, and decide whether $f$ is differentiable there.
practice Decide whether the greatest integer function $f(x) = \lfloor x\rfloor$ is differentiable at $x = 2$.
practice If $f$ is differentiable at $x = a$, must $f$ be continuous at $a$? Justify your answer.
practice If $f$ is continuous at $x = a$, must $f$ be differentiable at $a$? Justify your answer with an example.
practice Show that $f(x) = x |x|$ is differentiable at $x = 0$, and find $f'(0)$.
practice A function is defined by $f(x) = 2 x + 3$ for $x < 1$, and $f(x) = 5$ for $x \geq 1$. Check continuity at $x = 1$, then decide whether $f$ is differentiable there.
practice A function is defined by $f(x) = x^2$ for $x \leq 1$, and $f(x) = 2 x - 1$ for $x > 1$. Check continuity at $x = 1$, then decide whether $f$ is differentiable there.
practice At how many points is $f(x) = |x - 1| + |x - 2|$ not differentiable?
$0$
$1$
$2$
$3$
Answers
Not differentiable at $x = 3$: $\text{LHD} = -1$, $\text{RHD} = 1$.
Not differentiable at $x = -1$: $\text{LHD} = -1$, $\text{RHD} = 1$.
$f'(2) = 1$; differentiable, since $\text{LHD} = \text{RHD} = 1$.
$f'(-3) = -1$; differentiable, since $\text{LHD} = \text{RHD} = -1$.
$f'(0) = 0$; differentiable.
Not differentiable at $x = 0$: $\text{LHD} = -2$, $\text{RHD} = 0$.
Not differentiable at $x = 1$: $\text{LHD} = 0$, $\text{RHD} = 2$.
Not differentiable at $x = 2$: $f$ is not even continuous there ($\text{LHL} = 1$, $\text{RHL} = 2$).
Yes, differentiability always forces continuity at the same point (differentiable implies continuous).
No, continuity does not force differentiability. $f(x) = |x|$ is continuous at $x = 0$, yet $\text{LHD} = -1$ and $\text{RHD} = 1$ there disagree, so $f$ is not differentiable at 0.
$f'(0) = 0$; differentiable.
Not differentiable at $x = 1$: $\text{LHD} = 2$, $\text{RHD} = 0$ (though $f$ is continuous there).
$f'(1) = 2$; differentiable, since $\text{LHD} = \text{RHD} = 2$: the two branches join smoothly.
Exercise 5.5 uses the slopes of $e^x$ and $\ln x$, with the chain rule. Functions such as $e^{\sin x}$ and $\ln(x^2 + 1)$ come up often. They prepare you for logarithmic differentiation.
Exercise 5.5: Derivatives of exponential and logarithmic functions
practice Differentiate $y = e^{3x}$ with respect to $x$.
practice Differentiate $y = e^{-x}$ with respect to $x$.
practice Differentiate $y = \ln(2x)$ with respect to $x$, for $x > 0$.
practice Differentiate $y = \ln(x^2)$ with respect to $x$, for $x \neq 0$.
practice Differentiate $y = 5^x$ with respect to $x$.
practice Differentiate $y = e^{x^2}$ with respect to $x$.
practice Find $d y/d x$ at $x = 0$ for $y = e^{2x}$.
practice Find $d y/d x$ at $x = 1$ for $y = \ln(3x)$.
practice Which of these is $d/d x (3^x)$?
$3^x$
$x \cdot 3^{x-1}$
$3^x \cdot \ln(3)$
$3^x \cdot \ln(x)$
practice Differentiate $y = e^{\sin x}$ with respect to $x$.
practice Differentiate $y = \ln(x^2+1)$ with respect to $x$.
practice Differentiate $y = e^x \cdot \ln x$ with respect to $x$, for $x > 0$.
practice Differentiate $y = \ln(\sin x)$ with respect to $x$, for $0 < x < \pi$.
practice Find $d y/d x$ at $x = 0$ for $y = e^{\cos x}$.
practice Differentiate $y = \ln((x+1)/(x-1))$ with respect to $x$, for $x > 1$.
practice Differentiate $y = e^x \sin x + \ln(x^2+1)$ with respect to $x$.
practice Find $d y/d x$ at $x = 0$ for $y = \ln(e^x+1)$.
In Exercise 5.6 you take logs before you differentiate. Some functions, such as $x^x$, have $x$ in both the base and the power. Others are long products or quotients. The skill is to spot when logs make the work shorter. For a sum such as $y = x^x + (\sin x)^{\cos x}$, take logs of each term on its own, because $\ln(u + v)$ is not $\ln u + \ln v$.
Exercise 5.6: Logarithmic differentiation
practice Differentiate $y = x^{2x}$ with respect to $x$, for $x > 0$.
practice Differentiate $y = (2x)^x$ with respect to $x$, for $x > 0$.
practice Differentiate $y = x^{x+1}$ with respect to $x$, for $x > 0$.
practice Differentiate $y = (3x)^{3x}$ with respect to $x$, for $x > 0$.
practice Find $d y/d x$ at $x = 1$ for $y = x^x$.
practice Find $d y/d x$ at $x = 1$ for $y = x^{2x}$.
practice Differentiate $y = (\sin x)^x$ with respect to $x$, for $0 < x < \pi$.
practice Differentiate $y = x^{\sin x}$ with respect to $x$, for $x > 0$.
practice Differentiate $y = (x+1)(x+2) / ((x+3)(x+4))$ with respect to $x$, for $x > -1$.
practice Find $d y/d x$ at $x = 0$ for $y = (x+1)^x$, for $x > -1$.
practice Differentiate $y = x^x \cdot (\sin x)^x$ with respect to $x$, for $0 < x < \pi$.
practice Find $d y/d x$ at $x = 1$ for $y = x^x \cdot (x+1)^x$.
practice Differentiate $y = x^x + (\sin x)^{\cos x}$ with respect to $x$, for $0 < x < \pi$.
The curves in Exercise 5.7 are given by a parameter $t$. Find $d x/d t$ and $d y/d t$, then divide. Many use $\cos t$ and $\sin t$, as the worked circle does.
Exercise 5.7: Derivatives of parametric functions
practice If $x = t^2$ and $y = t^3$, find $d y/d x$.
practice If $x = 4t$ and $y = t^2$, find $d y/d x$.
practice If $x = 2 \cos \theta$ and $y = 2 \sin \theta$, find $d y/d x$.
practice If $x = e^t$ and $y = e^{2t}$, find $d y/d x$.
practice If $x = a t$ and $y = a/t$, for a positive constant $a$, find $d y/d x$.
practice If $x = t - 1/t$ and $y = t + 1/t$, find $d y/d x$.
practice If $x = t^3 - 3t$ and $y = t^2 - 2$, find $d y/d x$.
practice If $x = \ln t$ and $y = \sin t$, for $t > 0$, find $d y/d x$.
practice If $x = a (\theta + \sin \theta)$ and $y = a (1 - \cos \theta)$, find $d y/d x$.
practice If $x = 3 \sec \theta$ and $y = 4 \tan \theta$, find $d y/d x$.
practice If $x = a (\sin \theta - \theta \cos \theta)$ and $y = a (\cos \theta + \theta \sin \theta)$, find $d y/d x$.
practice If $x = t + \sin t \cos t$ and $y = \sin^2 t$, find $d y/d x$.
In Exercise 5.8 you differentiate twice. The functions come as formulas, as equations, or with a parameter. With a parameter $t$, find $d y/d x$ first, then differentiate it with respect to $t$ and divide by $d x/d t$. It is never $d^2 y / d t^2$ divided by $d^2 x / d t^2$. Some ask you to show that a function fits an equation with its second derivative in it.
Exercise 5.8: Second-order derivatives
practice Find $d^2 y / d x^2$ if $y = x^4 - 6x^2 + 5$.
practice Find $d^2 y / d x^2$ if $y = \sin 3x$.
practice Find $d^2 y / d x^2$ if $y = e^{2x}$.
practice Find $d^2 y / d x^2$ if $y = \ln x$, for $x > 0$.
practice Find $d^2 y / d x^2$ if $y = x \cos x$.
practice Find $d^2 y / d x^2$ if $y = \tan^{-1}(x)$.
practice If $x^2 + y^2 = 25$, find $d^2 y / d x^2$ in terms of $x$ and $y$.
practice If $x = t^2$ and $y = t^3$, find $d^2 y / d x^2$ in terms of $t$.
practice Find $d^2 y / d x^2$ if $y = x^2 \ln x$.
practice If $y = e^x \sin x$, show that $d^2 y / d x^2 - 2 d y/d x + 2y = 0$.
practice If $y = a \cos(\ln x) + b \sin(\ln x)$, show that $x^2 \: d^2 y / d x^2 + x \: d y/d x + y = 0$.
practice If $x = a (\theta - \sin \theta)$ and $y = a (1 - \cos \theta)$, find $d^2 y / d x^2$ in terms of $\theta$.
Answers
$12x^2-12$
$-9 \sin 3x$
$4 e^{2x}$
$-1/x^2$
$-2 \sin x - x \cos x$
$-2x/(1+x^2)^2$
$-25/y^3$
$3/(4t)$
$2 \ln x + 3$
Verified: $d^2 y / d x^2 = 2 e^x \cos x$, and $2 d y/d x - 2y$ also equals $2 e^x \cos x$, so the difference is $0$.
Verified: substituting $d y/d x$ and $d^2 y / d x^2$ into the left side gives $0$ for every $x > 0$.
The Miscellaneous Exercise mixes all the methods you have met. Questions 5 to 8, 11 to 13 and 16 check Rolle’s theorem and the mean value theorem, which are enrichment this year. Test both conditions before you solve for $c$.
Miscellaneous: Rolle’s theorem, the Mean Value Theorem, and mixed differentiation
practice Differentiate $y = \sin(x^2+1)$ with respect to $x$.
practice Differentiate $y = e^{3x}$ with respect to $x$.
practice Differentiate $y = \tan^{-1}(3x)$ with respect to $x$.
practice Differentiate $y = \ln(5x+2)$ with respect to $x$.
practice Verify Rolle’s theorem for $f(x) = x^2 - 4x + 3$ on $[1,3]$, and find the value of $c$.
practice Verify Rolle’s theorem for $f(x) = x^2 - 6x + 8$ on $[2,4]$, and find the value of $c$.
practice Verify the Mean Value Theorem for $f(x) = x^2 + 2x - 1$ on $[0,2]$, and find the value of $c$.
practice Verify the Mean Value Theorem for $f(x) = x^2 - 2x + 3$ on $[0,3]$, and find the value of $c$.
practice Using logarithmic differentiation, find $d y/d x$ if $y = x^{\sin x}$, for $x > 0$.
practice If $\sin(x+y) = y \cos x$, find $d y/d x$.
practice Verify Rolle’s theorem for $f(x) = \sin x$ on $[0, \pi]$, and find the value of $c$.
practice Verify the Mean Value Theorem for $f(x) = 2x^2 - 3x + 1$ on $[1,3]$, and find the value of $c$.
practice Does Rolle’s theorem apply to $f(x) = |x|$ on $[-1,1]$? Justify your answer.
practice At how many points in $\mathbb{R}$ is $f(x) = |x-2| + |x+2|$ not differentiable?
$0$
$1$
$2$
infinitely many
practice Find $f'(0)$ for $f(x) = x |x|$, using the definition of the derivative as a limit.
practice Verify the Mean Value Theorem for $f(x) = x^3$ on $[-1,2]$, and find the value of $c$.
Answers
$2x \cos(x^2+1)$
$3 e^{3x}$
$3/(1+9x^2)$
$5/(5x+2)$
$2$
$3$
$1$
$3/2$
$x^{\sin x} (\cos x \ln x + \sin x / x)$
$(y \sin x + \cos(x+y))/(\cos x - \cos(x+y))$
$\pi/2$
$2$
No, $f(-1) = f(1) = 1$, but $f$ is not differentiable at $x=0$, a point inside $(-1,1)$, so the differentiability hypothesis fails.
(C) $2$.
$0$
$1$
Board paper practice
practice At which value of $x$ is the function $f(x) = \left|x-5\right|$ not differentiable? (1 mark)
$-5$
$0$
$5$
$f$ is differentiable everywhere
practice Assertion (A). The function $f(x) = x^3$ is differentiable at every real $x$. Reason (R). Every polynomial function is differentiable at every point of its domain. (1 mark)
Both A and R are true and R is the correct explanation of A.
Both A and R are true but R is not the correct explanation of A.
A is true but R is false.
A is false but R is true.
practice Differentiate $y = \sec(2x+5)$ with respect to $x$. (2 marks)
practice Find the value of $k$ for which $f(x) = \sin 7x / x$ for $x \neq 0$, and $f(x) = k$ for $x = 0$, is continuous at $x = 0$. (2 marks)
practice If $x^3 + y^3 = 3x y$, find $d y/d x$. (5 marks)
practice A curve is given in parametric form by $x = \sin^3 \theta$ and $y = \cos^3 \theta$. (i) Find $d x / d \theta$ and $d y / d \theta$. (1 mark) (ii) Find $d y/d x$. (1 mark) (iii) Find $d^2 y / d x^2$. (2 marks)
practice Find $d^2 y / d x^2$ if $y = (2x+1)^5$. (2 marks)
Answers
(C) $5$.
(A) Both A and R are true and R is the correct explanation of A.
$d y/d x = 2 \sec(2x+5) \tan(2x+5)$.
$k = 7$.
$d y/d x = (y - x^2)/(y^2 - x)$.
(i) $d x / d \theta = 3 \sin^2 \theta \cos \theta$, $d y / d \theta = -3 \cos^2 \theta \sin \theta$. (ii) $d y/d x = -\cos \theta / \sin \theta$. (iii) $d^2 y / d x^2 = 1/(3 \sin^4 \theta \cos \theta)$.
