Chapter 5 · Continuity and Differentiability · Go for it

Continuity and differentiability, the quick way

Continuity means no break in the graph. The rest of the chapter is finding derivatives, and nine moves cover most questions.

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The chapter in one minute

What continuity and derivatives are, and what the exam asks.

A function is continuous at a point when its graph has no break there. To test it, you compare three numbers: the left-hand limit, the right-hand limit and the value.

A function is differentiable at a point when it has a derivative (a slope) there. A break or a sharp corner in the graph stops this.

Most of the chapter is about finding derivatives of harder functions: a function inside a function, an equation that mixes $x$ and $y$ ⁠, $x$ in a power, and curves given by a parameter.

Most exam questions on this chapter are one of nine kinds, and each kind has a fixed recipe. You will learn all nine below.

9question types

13formula cards

3numbers that must match for continuity: LHL, RHL and the value

1rule behind most of the chapter: the chain rule

Formulas

The formulas you must know

Learn these cards. Almost everything in the chapter is built from them.

Continuity at

c

\begin{aligned} &\lim_{x\to c^-} f(x) \\ &= \lim_{x\to c^+} f(x) \\ &= f(c) \end{aligned}

Left-hand limit, right-hand limit and value: all three equal means continuous.

Derivative at

c

\begin{aligned} &f'(c) \\ &= \lim_{h\to0}\frac{f(c + h) - f(c)}{h} \end{aligned}

With $h$ negative this is the LHD, with $h$ positive the RHD. Equal means differentiable.

Chain rule

\frac{dy}{dx} = \frac{dy}{dt}\cdot\frac{dt}{dx}

Differentiate the outside, keep the inside, then multiply by the derivative of the inside.

Product rule

(uv)' = u'v + uv'

You need it for terms like $xy$ and $x\log x$ ⁠.

Quotient rule

\left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2}

Derivative of the top comes first on top.

Powers and trig

\frac{d}{dx}x^n = nx^{n - 1}

Also $(\sin x)' = \cos x$ ⁠, $(\cos x)' = -\sin x$ and $(\tan x)' = \sec^2 x$ ⁠.

Exponential and log

\frac{d}{dx}e^x = e^x

Also $\frac{d}{dx}\log x = \frac{1}{x}$ for $x > 0$ ⁠, and $\frac{d}{dx}a^x = a^x\log a$ for $a > 0$ ⁠.

Inverse sine

\frac{d}{dx}\sin^{-1}x = \frac{1}{\sqrt{1 - x^2}}

Inverse cosine is the same with a minus sign: $\frac{d}{dx}\cos^{-1}x = -\frac{1}{\sqrt{1 - x^2}}$ ⁠. Both hold for $-1 < x < 1$ ⁠.

Inverse tan

\frac{d}{dx}\tan^{-1}x = \frac{1}{1 + x^2}

The one used most, often after putting $x = \tan\theta$ ⁠.

y

term

\frac{d}{dx}(y^2) = 2y\frac{dy}{dx}

Treat $y$ as a function of $x$ ⁠: every $y$ term you differentiate gets a $\frac{dy}{dx}$ ⁠.

Log of

y

\frac{d}{dx}\log y = \frac{1}{y}\frac{dy}{dx}

The key step of logarithmic differentiation.

Parametric form

\frac{dy}{dx} = \frac{dy/dt}{dx/dt}

The derivative of $y$ goes on top. It needs $\frac{dx}{dt} \ne 0$ ⁠.

Second derivative

\frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{dy}{dx}\right)

Differentiate twice. Also written $y''$ or $y_2$ ⁠.

Question types

The question types and how to crack each one

For each type: how to spot it, what to do, and one real NCERT question worked the exam way.

1. Is $f$ continuous at the joint? Compare three numbers

Spot it: $f$ is given in pieces, and you are asked where it is continuous, or where it breaks.

Away from the joint, each piece is a polynomial (or another continuous function), so $f$ is continuous there.
At the joint $c$ ⁠, find three numbers: LHL from the piece for $x < c$ ⁠, RHL from the piece for $x > c$ ⁠, and $f(c)$ from the piece that includes $c$ ⁠.
All three equal: continuous at $c$ ⁠. Any two different: not continuous at $c$ ⁠.

Exercise 5.1 Q6. Find all points of discontinuity of $f$ ⁠, where

f(x) = \begin{cases} 2x + 3, & x \le 2 \\ 2x - 3, & x > 2 \end{cases}

Both pieces are polynomials, so $f$ is continuous at every $x \ne 2$ ⁠. At $x = 2$ ⁠,

\begin{aligned} \text{LHL} &= \lim_{h\to0}\big(2(2 - h) + 3\big) \\ &= 7 \\ \text{RHL} &= \lim_{h\to0}\big(2(2 + h) - 3\big) \\ &= 1 \end{aligned}

and $f(2) = 2(2) + 3 = 7$ ⁠. Since $\text{LHL} \ne \text{RHL}$ ⁠, $f$ is not continuous at $x = 2$ ⁠.

∴ $x = 2$ is the only point of discontinuity

See a full exam answer · Go deeper: see it visually · Basics: left-hand and right-hand limits

In past papers (4): click to see

Board 2026 Q21(a) · Board 2025 Q5 · Board 2024 Q3 · Sample 2024-25 Q17

2. Find $k$ so that $f$ is continuous

Spot it: a function in pieces with an unknown $k$ (or $a$ and $b$ ⁠), and the words "find the value of $k$ ".

Find $f(c)$ ⁠, LHL and RHL at the joint, each in terms of $k$ ⁠.
Write the condition for continuity: $\text{LHL} = \text{RHL} = f(c)$ ⁠.
Solve the equation for $k$ ⁠.

Exercise 5.1 Q27. Find $k$ so that $f$ is continuous at $x = 2$ ⁠, where

f(x) = \begin{cases} kx^2, & x \le 2 \\ 3, & x > 2 \end{cases}

\begin{aligned} f(2) &= k(2)^2 = 4k \\ \text{LHL} &= \lim_{h\to0}k(2 - h)^2 \\ &= 4k \\ \text{RHL} &= \lim_{h\to0}3 = 3 \end{aligned}

For continuity at $x = 2$ ⁠, $\text{LHL} = \text{RHL} = f(2)$ ⁠. So

4k = 3

∴ $k = \frac{3}{4}$

See a full exam answer · Go deeper: see it visually · Basics: piecewise functions

In past papers (7): click to see

Board 2025 Q20 · Board 2023 Q13 · Sample 2026-27 Q8 · Sample 2025-26 Q7 · Sample 2023-24 Q4 · Sample 2022-23 Q4 · Sample 2020-21 Q21

3. Is $f$ differentiable at a point? Compare LHD and RHD

Spot it: a modulus such as $|x - 1|$ ⁠, or a function in pieces, and the word "differentiable".

Work out $f(c)$ and $f(c + h)$ in their simplest form.
LHD: divide $f(c + h) - f(c)$ by $h$ and let $h$ go to 0 from below ( $h$ negative, so $|h|$ is $-h$ ⁠). RHD: the same from above ( $|h|$ is $h$ ⁠).
LHD and RHD equal: differentiable. Different: not differentiable (the graph has a corner).

Exercise 5.2 Q9. Prove that $f(x) = |x - 1|$ is not differentiable at $x = 1$ ⁠.

Here $f(1) = 0$ and $f(1 + h) = |h|$ ⁠.

\begin{aligned} \text{LHD} &= \lim_{h\to0^-}\frac{|h|}{h} \\ &= \lim_{h\to0^-}\frac{-h}{h} = -1 \\ \text{RHD} &= \lim_{h\to0^+}\frac{|h|}{h} \\ &= \lim_{h\to0^+}\frac{h}{h} = 1 \end{aligned}

∴ $\text{LHD} \ne \text{RHD}$ ⁠, so $f$ is not differentiable at $x = 1$

See a full exam answer · Go deeper: see it visually · Basics: slope at a point

In past papers (8): click to see

Board 2025 Q5 · Board 2024 Q21(a) · Sample 2026-27 Q8 · Sample 2024-25 Q17 · Sample 2024-25 Q19 · Sample 2024-25 Q34(a) · Sample 2023-24 Q17 · Sample 2020-21 Q31(a)

4. The chain rule: a function inside a function

Spot it: something other than a plain $x$ inside $\sin$ ⁠, $\cos$ ⁠, $\log$ ⁠, $e$ or a power, like $\sin(x^2 + 5)$ or $e^{\cos x}$ ⁠.

Name the inside: put $t$ equal to it.
Differentiate the outside with respect to $t$ ⁠, and the inside with respect to $x$ ⁠.
Multiply the two, then put the inside back in place of $t$ ⁠.

Exercise 5.2 Q1. Differentiate $\sin(x^2 + 5)$ ⁠.

Let $y = \sin(x^2 + 5)$ ⁠. Put $t = x^2 + 5$ ⁠, so $y = \sin t$ ⁠.

Then $\frac{dy}{dt} = \cos t$ and $\frac{dt}{dx} = 2x$ ⁠.

\begin{aligned} \frac{dy}{dx} &= \frac{dy}{dt}\cdot\frac{dt}{dx} \\ &= \cos t\cdot 2x \end{aligned}

∴ $\frac{dy}{dx} = 2x\cos(x^2 + 5)$

See a full exam answer · Go deeper: see it visually · Basics: composite functions

In past papers (4): click to see

Board 2024 Q13 · Board 2024 Q21(b) · Board 2023 Q24 · Sample 2025-26 Q22

5. Implicit: $x$ and $y$ mixed in one equation

Spot it: an equation such as $x^2 + xy + y^2 = 100$ that you cannot easily solve for $y$ ⁠.

Differentiate every term on both sides with respect to $x$ ⁠.
Each $y$ term gets a $\frac{dy}{dx}$ ⁠: $y^2$ gives $2y\frac{dy}{dx}$ ⁠, and $xy$ needs the product rule.
Bring the $\frac{dy}{dx}$ terms to one side, take $\frac{dy}{dx}$ out as a common factor, and divide.

Exercise 5.3 Q5. Find $\frac{dy}{dx}$ if $x^2 + xy + y^2 = 100$ ⁠.

Differentiate both sides with respect to $x$ ⁠:

\begin{aligned} &2x + y + x\frac{dy}{dx} \\ &\quad + 2y\frac{dy}{dx} = 0 \end{aligned}

(x + 2y)\frac{dy}{dx} = -(2x + y)

∴ $\frac{dy}{dx} = -\frac{2x + y}{x + 2y}$

See a full exam answer · Go deeper: see it visually · Basics: solving for dy/dx

In past papers (6): click to see

Board 2026 Q21(b) · Board 2025 Q21(b) · Board 2025 Q32(a) · Board 2024 Q27(a) · Sample 2026-27 Q22 · Sample 2022-23 Q24

6. Inverse trig of a fraction: put $x = \tan\theta$ first

Spot it: $\sin^{-1}$ ⁠, $\cos^{-1}$ or $\tan^{-1}$ of a fraction such as $\frac{2x}{1 + x^2}$ or $\frac{1 - x^2}{1 + x^2}$ ⁠.

Spot a formula in $\tan\theta$ hiding in the fraction, and put $x = \tan\theta$ ⁠.
Use the given interval to see where $2\theta$ lies, then simplify, for example $\cos^{-1}(\cos 2\theta)$ to $2\theta$ ⁠.
Write $y$ in $x$ (here $2\tan^{-1}x$ ⁠) and differentiate with the inverse-tan card.

Exercise 5.3 Q11. Find $\frac{dy}{dx}$ if $y = \cos^{-1}\frac{1 - x^2}{1 + x^2}$ ⁠, $0 < x < 1$ ⁠.

Put $x = \tan\theta$ ⁠. As $0 < x < 1$ ⁠, $0 < 2\theta < \frac{\pi}{2}$ ⁠.

\begin{aligned} y &= \cos^{-1}\frac{1 - \tan^2\theta}{1 + \tan^2\theta} \\ &= \cos^{-1}(\cos 2\theta) = 2\theta \\ &= 2\tan^{-1}x \end{aligned}

∴ $\frac{dy}{dx} = \frac{2}{1 + x^2}$

See a full exam answer · Go deeper: see it visually · Basics: trigonometric identities

In past papers (1): click to see

Sample 2026-27 Q7

7. Logarithmic differentiation: $x$ in the power

Spot it: a power whose exponent contains $x$ ⁠, like $x^x$ ⁠, $x^{\sin x}$ or $(\log x)^{\cos x}$ ⁠, or a long product.

For a sum of such powers, name the parts $u$ and $v$ and do each part on its own.
Take log of both sides, then differentiate: the left side gives $\frac{1}{u}\frac{du}{dx}$ ⁠.
Multiply by $u$ ⁠, and write $u$ back in terms of $x$ ⁠.

Exercise 5.5 Q4. Differentiate $x^x - 2^{\sin x}$ ⁠.

Let $u = x^x$ and $v = 2^{\sin x}$ ⁠, so $y = u - v$ (here $x > 0$ ⁠).

Take log: $\log u = x\log x$ ⁠. Differentiate:

\begin{aligned} \frac{1}{u}\frac{du}{dx} &= \log x + 1 \\ \frac{du}{dx} &= x^x(1 + \log x) \end{aligned}

In the same way, $\log v = \sin x\log 2$ ⁠, so

\begin{aligned} \frac{1}{v}\frac{dv}{dx} &= \cos x\log 2 \\ \frac{dv}{dx} &= 2^{\sin x}\cos x\log 2 \end{aligned}

∴ $\frac{dy}{dx} = x^x(1 + \log x)$ $-\ 2^{\sin x}\cos x\log 2$

See a full exam answer · Go deeper: see it visually · Basics: exponentials and logarithms

In past papers (5): click to see

Board 2024 Q27(b) · Sample 2026-27 Q28(b) · Sample 2025-26 Q26(a) · Sample 2024-25 Q23(b) · Sample 2020-21 Q30

8. Parametric form: $x$ and $y$ given through $t$ or $\theta$

Spot it: two equations, one for $x$ and one for $y$ ⁠, both in the same third letter.

Differentiate $x$ and $y$ separately with respect to the parameter.
Divide, with $y$ on top: $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$ ⁠.
Simplify, and leave the answer in the parameter.

Exercise 5.6 Q9. Find $\frac{dy}{dx}$ if $x = a\sec\theta$ ⁠, $y = b\tan\theta$ ⁠.

\begin{aligned} \frac{dx}{d\theta} &= a\sec\theta\tan\theta \\ \frac{dy}{d\theta} &= b\sec^2\theta \end{aligned}

\begin{aligned} \frac{dy}{dx} &= \frac{dy/d\theta}{dx/d\theta} \\ &= \frac{b\sec^2\theta}{a\sec\theta\tan\theta} \\ &= \frac{b}{a}\cdot\frac{\sec\theta}{\tan\theta} \end{aligned}

As $\frac{\sec\theta}{\tan\theta} = \frac{1}{\sin\theta}$ ⁠,

∴ $\frac{dy}{dx} = \frac{b}{a}\operatorname{cosec}\theta$

See a full exam answer · Go deeper: see it visually

In past papers (8): click to see

Board 2026 Q34 · Board 2025 Q21(a) · Board 2025 Q32(b) · Sample 2026-27 Q28(a) · Sample 2025-26 Q26(b) · Sample 2024-25 Q23(a) · Sample 2024-25 Q34(b) · Sample 2020-21 Q31(b)

9. Second derivative: differentiate twice, then prove

Spot it: $\frac{d^2y}{dx^2}$ or $y''$ in the question, often "prove that $\frac{d^2y}{dx^2} + y = 0$ ".

Find $\frac{dy}{dx}$ ⁠.
Differentiate again to get $\frac{d^2y}{dx^2}$ ⁠.
Put them into the left side of what you must prove, simplify, and end with "Hence proved".

Exercise 5.7 Q11. If $y = 5\cos x - 3\sin x$ ⁠, prove that $\frac{d^2y}{dx^2} + y = 0$ ⁠.

\begin{aligned} \frac{dy}{dx} &= -5\sin x - 3\cos x \\ \frac{d^2y}{dx^2} &= -5\cos x + 3\sin x \\ &= -(5\cos x - 3\sin x) \\ &= -y \end{aligned}

∴ $\frac{d^2y}{dx^2} + y = -y + y = 0$ ⁠. Hence proved.

See a full exam answer · Go deeper: see it visually · Basics: derivative rules

In past papers (9): click to see

Board 2026 Q34 · Board 2025 Q32(b) · Board 2023 Q38(ii) · Sample 2026-27 Q28(a) · Sample 2025-26 Q26(b) · Sample 2024-25 Q34(b) · Sample 2023-24 Q31 · Sample 2022-23 Q16 · Sample 2020-21 Q31(b)

Shortcuts

Shortcuts and quick tips

Small habits that save time and marks.

Polynomials, $\sin x$ ⁠, $\cos x$ ⁠, $e^x$ and $|x|$ are continuous everywhere, and so are their sums, products and composites, and their quotients wherever the bottom is not 0. When every piece is built from these, only the joints need checking.
The value $f(c)$ comes from the piece whose condition includes $c$ ⁠, such as $x \le 2$ when $c$ is 2.
Differentiable at a point means continuous there. The reverse is not true: $|x - 1|$ is continuous at 1 but not differentiable there.
Chain rule from the outside in, multiplying as you go: $\frac{d}{dx}e^{\cos x} = -e^{\cos x}\sin x$ ⁠.
In implicit questions, $xy$ needs the product rule: $\frac{d}{dx}(xy) = y + x\frac{dy}{dx}$ ⁠.
A number on its own differentiates to 0, so the 100 in $x^2 + xy + y^2 = 100$ gives 0.
$x$ in the power? Take log of both sides first. For a sum of two such powers, never take log of the sum: do each part on its own.
For $\frac{2x}{1 + x^2}$ ⁠, $\frac{1 - x^2}{1 + x^2}$ or $\frac{2x}{1 - x^2}$ inside an inverse trig function, put $x = \tan\theta$ and simplify before you differentiate. Check the given interval first: it decides whether the simplification is allowed.
In parametric questions the answer stays in $t$ or $\theta$ ⁠. There is no need to remove the parameter.
An implicit answer may contain both $x$ and $y$ ⁠. That is fine.

Watch out

Mistakes that cost marks

Each one is common, and each has an easy fix.

Writing $\frac{d}{dx}\sin(x^2) = \cos(x^2)$ → multiply by the derivative of the inside: the answer is $2x\cos(x^2)$ ⁠.
Taking $f(c)$ from the wrong piece → use the piece whose condition includes $c$ ⁠.
Forgetting $\frac{dy}{dx}$ on a $y$ term → $\frac{d}{dx}(y^2) = 2y\frac{dy}{dx}$ ⁠, not just $2y$ ⁠.
Treating $x^x$ as a power and writing $x\cdot x^{x - 1}$ → take logs first: $\frac{d}{dx}x^x = x^x(1 + \log x)$ ⁠.
Putting the derivative of $x$ on top → $y$ goes on top: $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$ ⁠.
"Continuous, so differentiable" → not always: $|x - 1|$ is continuous at 1 but has a corner there, so work out LHD and RHD.
Writing $\cos^{-1}(\cos 2\theta) = 2\theta$ with no interval → write the interval line first; it is what makes this step true.
Stopping at $\frac{d^2y}{dx^2}$ in a "prove that" question → finish with the target line and "Hence proved".

Marks

How the examiner gives marks

From CBSE's own marking schemes. Write these lines and the marks follow.

Continuity: write LHL, RHL and $f(c)$ each on its own line, then the conclusion in words ("so $f$ is continuous at $c$ "). Conclusions in words carry marks.
Differentiability: the function written in pieces, the LHD as a limit, the RHD, and the line "LHD = RHD, so $f$ is differentiable" are each worth half a mark.
Chain rule "show that" (2 marks): the derivative by the chain rule is half a mark, the simplification 1 mark, and the final target line half a mark.
Logarithmic differentiation: in the 2025-26 sample paper each stage of the working earned 1 mark, so show the log line, the differentiated line and the final $\frac{dy}{dx}$ ⁠.
"Prove that" questions: end with the target statement and "Hence proved". That line carries half a mark.
A slip is penalised only once, so keep going. An MCQ earns 1 mark for the right option only.

Practice

Practise these first

Easiest first. Together they cover every type above.

Exercise 5.1 Q10 continuity at the joint: three numbers that match
Exercise 5.1 Q29 find $k$ with two straight-line pieces
Exercise 5.2 Q2 the chain rule: cos of sin
Exercise 5.7 Q3 a second derivative with the product rule
Exercise 5.6 Q1 parametric form with $t$
Exercise 5.3 Q4 implicit, with the product rule for $xy$
Exercise 5.4 Q5 the chain rule three layers deep, with log and $e^x$
Exercise 5.7 Q15 show that $y'' = 49y$
Exercise 5.3 Q10 put $x = \tan\theta$ ⁠, then $\tan 3\theta$
Exercise 5.5 Q3 logarithmic differentiation of $(\log x)^{\cos x}$
Miscellaneous Q20 LHD and RHD with two modulus terms
Exercise 5.1 Q26 find $k$ when the limit needs $\frac{\sin h}{h}$

You can do this

Nine moves, one at a time. Work through the practice list, write every line the examiner looks for, and you are ready for the exam.

Ready to go deeper? Open the interactive lesson

Past papers

35 CBSE board and sample-paper questions on Continuity and Differentiability, each with a full-marks exam answer. Practise them →