Continuity and Differentiability Class 12: Notes, Methods and Solved Questions

What Continuity and Differentiability is about

Weightage: This chapter is part of Unit III, Calculus, which carries 35 of the 80 theory marks together with Application of Derivatives, Integrals, Application of Integrals and Differential Equations.Source: CBSE Mathematics syllabus 2026-27, Unit III Calculus

In real board papers: asked in 8 of the 8 board papers we hold (2023 to 2026), on average 8.1 marks per paper. See the questions

Key concepts

Continuity at a point

NCERT section 5.2

A function f is continuous at x = a if f(a) is defined and

${\lim }_{x\to a^{-}}f(x)={\lim }_{x\to a^{+}}f(x)=f(a)$

(left-hand limit = right-hand limit = value). If any of these fails, f is discontinuous at a. A function is continuous on an interval if it is continuous at every point of it, and a continuous function is one continuous at every point of its domain.

For a piecewise function, check the points where the formula changes; elsewhere each piece (a polynomial, sine and so on) is continuous by itself. To find a constant that makes the function continuous, equate the left-hand limit, right-hand limit and value.

Worked example

Find k so that $f(x)=\{\begin{array}{ll}kx+1,& x\le 2\\ 3x-1,& x>2\end{array}$ is continuous at x = 2. [2 marks]

Try it first, then show the solutionSolution

1. LHL = f(2) = 2k + 1; RHL = ${\lim }_{x\to 2^{+}}(3x-1)=5$
2. Continuity: $2k+1=5\Rightarrow k=2$

Answer: k = 2

Standard continuous functions and algebra of continuity

NCERT section 5.2.1

Continuous everywhere on their domains: polynomials, sin x, cos x, |x|, $e^x$, log x (for x > 0), rational functions (except where the denominator is 0), and the inverse trigonometric functions on their domains.

If f and g are continuous at a, then so are $f+g$, $f-g$, $fg$, and $\frac{f}{g}$ (when $g(a)\ne 0$). A composition of continuous functions is continuous: for example $|\sin x|$ and $\sin (x^2)$ are continuous everywhere.

The greatest integer function [x] is discontinuous at every integer (it jumps by 1), and continuous at all other real numbers.

Worked example

Find all points of discontinuity of $f(x)=\frac{x^2+1}{x^2-5x+6}$. [1 mark]

Try it first, then show the solutionSolution

1. f is a rational function, continuous wherever the denominator is non-zero.
2. $x^2-5x+6=(x-2)(x-3)=0$ at x = 2 and x = 3.

Answer: x = 2 and x = 3 (f is not defined there)

Differentiability and its relation with continuity

NCERT section 5.3

f is differentiable at x = a if the derivative $f^{\prime }(a)={\lim }_{h\to 0}\frac{f(a+h)-f(a)}{h}$ exists, i.e. the left-hand derivative ${\lim }_{h\to 0^{-}}\frac{f(a+h)-f(a)}{h}$ equals the right-hand derivative ${\lim }_{h\to 0^{+}}\frac{f(a+h)-f(a)}{h}$.

Theorem: if f is differentiable at a, it is continuous at a. The converse is false: $f(x)=|x|$ is continuous at 0 but its LHD is −1 and RHD is +1, so it is not differentiable there (the graph has a corner).

Graphically, a function is not differentiable at a corner (sharp turn), a jump (break) or a vertical tangent.

Worked example

Show that $f(x)=|x-2|$ is continuous but not differentiable at x = 2. [2 marks]

Try it first, then show the solutionSolution

1. Continuity: LHL = RHL = 0 = f(2).
2. RHD: ${\lim }_{h\to 0^{+}}\frac{|h|-0}{h}=1$; LHD: ${\lim }_{h\to 0^{-}}\frac{|h|}{h}=-1$.
3. LHD ≠ RHD, so f is not differentiable at x = 2.

Answer: Continuous at 2, not differentiable (LHD = −1, RHD = 1).

Chain rule (derivatives of composite functions)

NCERT section 5.3.1

If y = f(u) and u = g(x), then

$\frac{dy}{dx}=\frac{dy}{du}\cdot \frac{du}{dx}$

Work from the outside in: differentiate the outer function, keeping the inside unchanged, then multiply by the derivative of the inside. For longer chains, keep multiplying: $\frac{d}{dx}\sin (\cos (x^2))=\cos (\cos x^2)\cdot (-\sin x^2)\cdot 2x$.

To differentiate one function with respect to another, say u with respect to v, use $\frac{du}{dv}=\frac{du/dx}{dv/dx}$.

Worked example

Differentiate (a) $y=\sin (x^2+1)$, (b) $y=\sqrt{\cos 3x}$ with respect to x. [2 marks]

Try it first, then show the solutionSolution

1. (a) $\frac{dy}{dx}=\cos (x^2+1)\cdot 2x=2x\cos (x^2+1)$
2. (b) $\frac{dy}{dx}=\frac{1}{2\sqrt{\cos 3x}}\cdot (-\sin 3x)\cdot 3=-\frac{3\sin 3x}{2\sqrt{\cos 3x}}$

Answer: (a) $2x\cos (x^2+1)$ (b) $-\frac{3\sin 3x}{2\sqrt{\cos 3x}}$

Implicit differentiation

NCERT section 5.3.2

When x and y are related by an equation that is not solved for y (such as $x^2+xy+y^2=7$), differentiate both sides with respect to x, treating y as a function of x: every time you differentiate a term containing y, multiply by $\frac{dy}{dx}$ (chain rule). Then collect the $\frac{dy}{dx}$ terms and solve.

For products such as xy use the product rule: $\frac{d}{dx}(xy)=y+x\frac{dy}{dx}$.

Worked example

If $x^2+xy+y^2=7$, find $\frac{dy}{dx}$ and its value at (1, 2). [2 marks]

Try it first, then show the solutionSolution

1. $2x+\left(y+x\frac{dy}{dx}\right)+2y\frac{dy}{dx}=0$
2. $\frac{dy}{dx}(x+2y)=-(2x+y)\Rightarrow \frac{dy}{dx}=-\frac{2x+y}{x+2y}$
3. At (1, 2): $-\frac{2+2}{1+4}=-\frac{4}{5}$

Answer: $\frac{dy}{dx}=-\frac{2x+y}{x+2y}$; $-\frac{4}{5}$ at (1, 2)

Derivatives of sin⁻¹x, cos⁻¹x and tan⁻¹x

NCERT section 5.3.3

Using implicit differentiation on $\sin y=x$ (with y in the principal range, where $\cos y\ge 0$): $\cos y\frac{dy}{dx}=1$, so $\frac{dy}{dx}=\frac{1}{\cos y}=\frac{1}{\sqrt{1-x^2}}$. Similarly:

• $\frac{d}{dx}{\sin }^{-1}x=\frac{1}{\sqrt{1-x^2}}$, $-1<x<1$
• $\frac{d}{dx}{\cos }^{-1}x=-\frac{1}{\sqrt{1-x^2}}$, $-1<x<1$
• $\frac{d}{dx}{\tan }^{-1}x=\frac{1}{1+x^2}$

Combine with the chain rule: $\frac{d}{dx}{\tan }^{-1}(g(x))=\frac{g^{\prime }(x)}{1+g(x{)}^2}$. (The derivatives of ${\cot }^{-1}$, ${\sec }^{-1}$ and ${\csc }^{-1}$ are not required as standard results.)

Worked example

Differentiate (a) ${\sin }^{-1}(2x)$, (b) ${\tan }^{-1}(x^3)$, (c) $x{\tan }^{-1}x$ with respect to x. [2 marks]

Try it first, then show the solutionSolution

1. (a) $\frac{1}{\sqrt{1-4x^2}}\cdot 2=\frac{2}{\sqrt{1-4x^2}}$
2. (b) $\frac{3x^2}{1+x^6}$
3. (c) Product rule: ${\tan }^{-1}x+\frac{x}{1+x^2}$

Answer: (a) $\frac{2}{\sqrt{1-4x^2}}$ (b) $\frac{3x^2}{1+x^6}$ (c) ${\tan }^{-1}x+\frac{x}{1+x^2}$

Exponential and logarithmic functions and their derivatives

NCERT section 5.4

The exponential function $y=a^x$ (a > 0, a ≠ 1) has domain R and range $(0,\infty )$; its inverse is the logarithm $y={\log }_{a}x$, domain $(0,\infty )$. With base e we get $e^x$ and the natural logarithm log x (written ln x in some books).

Laws of logarithms: $\log (mn)=\log m+\log n$, $\log \frac{m}{n}=\log m-\log n$, $\log m^n=n\log m$, and ${\log }_{a}x=\frac{\log x}{\log a}$.

Derivatives:

• $\frac{d}{dx}{e}^x=e^x$; $\frac{d}{dx}{a}^x=a^x\log a$
• $\frac{d}{dx}\log x=\frac{1}{x}$, x > 0; $\frac{d}{dx}{\log }_{a}x=\frac{1}{x\log a}$

Worked example

Differentiate (a) $e^{3x}\log (2x)$, (b) $5^{\sin x}$. [2 marks]

Try it first, then show the solutionSolution

1. (a) $3e^{3x}\log (2x)+e^{3x}\cdot \frac{1}{2x}\cdot 2=e^{3x}\left(3\log 2x+\frac{1}{x}\right)$
2. (b) $5^{\sin x}\log 5\cdot \cos x$

Answer: (a) $e^{3x}\left(3\log 2x+\frac{1}{x}\right)$ (b) $5^{\sin x}\cos x\log 5$

Logarithmic differentiation

NCERT section 5.5

Take logarithms first when (1) both base and power vary, as in $x^{\sin x}$ or $(\cos x{)}^x$, or (2) the function is a long product or quotient.

Method for $y=[u(x){]}^{v(x)}$: $\log y=v\log u$. Differentiate: $\frac{1}{y}\frac{dy}{dx}=v^{\prime }\log u+v\frac{u^{\prime }}{u}$, so $\frac{dy}{dx}=u^v\left(v^{\prime }\log u+\frac{v{u}^{\prime }}{u}\right)$.

For a sum such as $y=x^x+(\sin x{)}^x$, put $u=x^x$ and $v=(\sin x{)}^x$, differentiate each separately by logs, then add: $\frac{dy}{dx}=\frac{du}{dx}+\frac{dv}{dx}$. (Taking the log of a sum directly does not help.)

Worked example

Differentiate $y=x^{\cos x}$, x > 0. [3 marks]

Try it first, then show the solutionSolution

1. $\log y=\cos x\log x$
2. $\frac{1}{y}\frac{dy}{dx}=-\sin x\log x+\frac{\cos x}{x}$
3. $\frac{dy}{dx}=x^{\cos x}\left(\frac{\cos x}{x}-\sin x\log x\right)$

Answer: $x^{\cos x}\left(\frac{\cos x}{x}-\sin x\log x\right)$

Derivatives of parametric functions

NCERT section 5.6

If x = f(t) and y = g(t), then

$\frac{dy}{dx}=\frac{dy/dt}{dx/dt}$, provided $\frac{dx}{dt}\ne 0$

For the second derivative, differentiate $\frac{dy}{dx}$ (a function of t) with respect to t and divide by $\frac{dx}{dt}$ again:

$\frac{d^{2}y}{d{x}^2}=\frac{d}{dt}\left(\frac{dy}{dx}\right)\cdot \frac{dt}{dx}$

A common error is to write $\frac{d^{2}y}{d{x}^2}=\frac{d^{2}y/d{t}^2}{d^{2}x/d{t}^2}$, which is wrong.

Worked example

If $x=t^2$ and $y=t^3$, find $\frac{dy}{dx}$ and $\frac{d^{2}y}{d{x}^2}$. [3 marks]

Try it first, then show the solutionSolution

1. $\frac{dx}{dt}=2t$, $\frac{dy}{dt}=3t^2$, so $\frac{dy}{dx}=\frac{3t^2}{2t}=\frac{3t}{2}$
2. $\frac{d}{dt}\left(\frac{3t}{2}\right)=\frac{3}{2}$
3. $\frac{d^{2}y}{d{x}^2}=\frac{3}{2}\cdot \frac{1}{2t}=\frac{3}{4t}$

Answer: $\frac{dy}{dx}=\frac{3t}{2}$, $\frac{d^{2}y}{d{x}^2}=\frac{3}{4t}$

Second-order derivatives

NCERT section 5.7

The second derivative is the derivative of the derivative: $\frac{d^{2}y}{d{x}^2}=\frac{d}{dx}\left(\frac{dy}{dx}\right)$, also written y″ or $y_2$.

A frequent board question asks you to prove a relation between y, y′ and y″ (a differential equation). Strategy: find y′, then y″, and substitute; or, when y contains constants or roots, rearrange first (square, cross-multiply) so that differentiating again produces the required form.

Worked example

If $y=(a+bx)e^{2x}$, show that $\frac{d^{2}y}{d{x}^2}-4\frac{dy}{dx}+4y=0$. [3 marks]

Try it first, then show the solutionSolution

1. $y^{\prime }=b{e}^{2x}+2(a+bx)e^{2x}=b{e}^{2x}+2y$
2. $y^{\prime \prime }=2b{e}^{2x}+2y^{\prime }$
3. From the first line $b{e}^{2x}=y^{\prime }-2y$, so $y^{\prime \prime }=2(y^{\prime }-2y)+2y^{\prime }=4y^{\prime }-4y$
4. Hence $y^{\prime \prime }-4y^{\prime }+4y=0$.

Answer: Shown: $y^{\prime \prime }-4y^{\prime }+4y=0$

Maths in real life: Continuity and Differentiability

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Topics in this chapter

• Implicit Differentiation
• Logarithmic Differentiation
• Parametric and Second Order Derivatives

Continuity and Differentiability formula sheet

Continuity at a

${\lim }_{x\to a^{-}}f(x)={\lim }_{x\to a^{+}}f(x)=f(a)$

Derivative at a (first principles)

$f^{\prime }(a)={\lim }_{h\to 0}\frac{f(a+h)-f(a)}{h}$

Chain rule

$\frac{dy}{dx}=\frac{dy}{du}\cdot \frac{du}{dx}$

Inverse trigonometric derivatives

$\frac{d}{dx}{\sin }^{-1}x=\frac{1}{\sqrt{1-x^2}},\frac{d}{dx}{\cos }^{-1}x=-\frac{1}{\sqrt{1-x^2}},\frac{d}{dx}{\tan }^{-1}x=\frac{1}{1+x^2}$

Exponential

$\frac{d}{dx}{e}^x=e^x,\frac{d}{dx}{a}^x=a^x\log a$

Logarithm

$\frac{d}{dx}\log x=\frac{1}{x},\frac{d}{dx}{\log }_{a}x=\frac{1}{x\log a}$

Logarithmic differentiation

$y=u^v\Rightarrow \frac{dy}{dx}=u^v\left(v^{\prime }\log u+\frac{v{u}^{\prime }}{u}\right)$

Parametric derivative

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

Parametric second derivative

$\frac{d^{2}y}{d{x}^2}=\frac{d}{dt}\left(\frac{dy}{dx}\right)\cdot \frac{dt}{dx}$

Derivative of u with respect to v

$\frac{du}{dv}=\frac{du/dx}{dv/dx}$

Standard trigonometric derivatives

$(\tan x{)}^{\prime }={\sec }^{2}x,(\sec x{)}^{\prime }=\sec x\tan x,(\cot x{)}^{\prime }=-{\csc }^{2}x,(\csc x{)}^{\prime }=-\csc x\cot x$

How the board asks Continuity and Differentiability

The question patterns to expect, and the method that works for each.

Tricks to remember Continuity and Differentiability

Short memory aids. Say each one aloud once, then use it on a question the same day.

• Three equal for continuity

  LHL = RHL = value: 'left, right, here' must all agree.

  Use it when: Every continuity question.

• Corners kill derivatives

  A sharp corner (like |x| at 0) is continuous but not differentiable; a break is neither.

  Use it when: Assertion-reason on continuity versus differentiability.

• Outside in

  Chain rule: differentiate the outer layer, leave the inside alone, then multiply by the inside's derivative, layer by layer.

  Use it when: Composite functions.

• Variable up, variable down: log it

  If x appears in both base and power, take logs. For a sum of such terms, split first.

  Use it when: Logarithmic differentiation.

• Divide by dx/dt twice

  Parametric second derivative: differentiate dy/dx with respect to t, then divide by dx/dt again.

  Use it when: Avoiding the d²y/dt² ÷ d²x/dt² error.

Continuity and Differentiability: board-level solved questions

Ordered from board level to HOTS. Cover the solution, write your own, then compare step by step.

Case-study practice (Section E style)

Read the situation, then answer every part. In the board paper this is a 4-mark question split into 1 + 1 + 2 marks.

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

Where students lose marks in Continuity and Differentiability

The slips examiners see most often, and what to write instead.

• ✗Mistake: Checking only one side in continuity

  Taking f(a) from one formula and stopping.

  ✓Do this: Compute both one-sided limits and the value separately, then compare.

  Usually costs: 1 mark.

• ✗Mistake: Forgetting dy/dx on y-terms in implicit differentiation

  Differentiating y² as 2y.

  Wrong working

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

  Right working

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

  ✓Do this: Every y-term gets a dy/dx factor.

  Usually costs: Most of the marks.

• ✗Mistake: Taking log of a sum

  Applying logarithmic differentiation to $x^x+(\sin x{)}^x$ directly.

  ✓Do this: Write y = u + v, differentiate u and v separately by logs, then add.

  Usually costs: 2-3 marks.

• ✗Mistake: Wrong parametric second derivative

  Dividing second derivatives.

  Wrong working

  $\frac{d^{2}y}{d{x}^2}=\frac{d^{2}y/d{t}^2}{d^{2}x/d{t}^2}$

  Right working

  $\frac{d^{2}y}{d{x}^2}=\frac{d}{dt}\left(\frac{dy}{dx}\right)÷\frac{dx}{dt}$

  ✓Do this: Differentiate dy/dx w.r.t. t, then divide by dx/dt.

  Usually costs: 2 marks.

• ✗Mistake: Using inverse-trig identities that are no longer in the syllabus

  Old methods like 2tan⁻¹x = tan⁻¹(2x/(1 − x²)).

  ✓Do this: Differentiate directly with the chain rule unless a simple principal-value step is enough.

  Usually costs: Time, and risk of domain errors.

NCERT exercise guide: Continuity and Differentiability

What each exercise in the NCERT chapter tests, how to approach it, and a question of the same kind solved step by step. Questions here are written in our own words.

Continuity and Differentiability: previous year questions

• Asked in 8 of the 8 board papers we hold (2023 to 2026), on average 8.1 marks per paper; 43 questions in all, counting OR alternatives. In 4 of 4 sample papers.

Counted from the real papers on our previous year papers page, not estimated. Average marks are over every board paper we hold, with an OR pair counted once. Chapter tags are ours.

Solved board questions from Continuity and Differentiability

Real questions from recent papers. Try each one first, then open our step-by-step solution with marks per step.

Every Continuity and Differentiability question, by year

Practice questions on Continuity and Differentiability

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

1. Q1 · 1 mark · MCQBoard level

   The function f(x) = [x] (greatest integer) is discontinuous at:
   (a) every real number
   (b) every integer
   (c) x = 0 only
   (d) no point

   Hint

   Where does it jump?

   Answer

   (b) every integer

2. Q2 · 1 mark · MCQBoard level

   The derivative of $\log (\sin x)$ is:
   (a) $\tan x$
   (b) $\cot x$
   (c) $\frac{1}{\sin x}$
   (d) $-\cot x$

   Hint

   Chain rule.

   Answer

   (b) cot x

3. Q3 · 1 mark · Very short answerBoard level

   Differentiate $e^{\sqrt{x}}$ with respect to x.

   Hint

   Inside derivative is $\frac{1}{2\sqrt{x}}$.

   Answer

   $\frac{e^{\sqrt{x}}}{2\sqrt{x}}$

4. Q4 · 1 mark · Assertion-reasonHarder board level

   Assertion (A): $f(x)=\{\begin{array}{ll}x\sin \frac{1}{x},& x\ne 0\\ 0,& x=0\end{array}$ is continuous at x = 0.
   Reason (R): $-|x|\le x\sin \frac{1}{x}\le |x|$ for $x\ne 0$.
   (a) Both true, R explains A. (b) Both true, R does not explain A. (c) A true, R false. (d) A false, R true.

   Hint

   Squeeze the function between ±|x|.

   Answer

   (a). The bounds tend to 0, so the limit is 0 = f(0).

5. Q5 · 2 marks · Very short answerHarder board level

   Find dy/dx if $y={\tan }^{-1}\left(\frac{x}{2}\right)+{\sin }^{-1}(x^2)$.

   Hint

   Chain rule on each.

   Answer

   $\frac{1/2}{1+x^2/4}+\frac{2x}{\sqrt{1-x^4}}=\frac{2}{4+x^2}+\frac{2x}{\sqrt{1-x^4}}$

6. Q6 · 2 marks · Very short answerHarder board level

   If $e^y(x+2)=1$, show that $\frac{dy}{dx}=-e^y$.

   Hint

   Differentiate the product.

   Answer

   $e^y{y}^{\prime }(x+2)+e^y=0\Rightarrow y^{\prime }=-\frac{1}{x+2}$, and from the given relation $\frac{1}{x+2}=e^y$, so $y^{\prime }=-e^y$.

7. Q7 · 3 marks · Short answerHarder board level

   Differentiate $\frac{(x+1{)}^2\sqrt{x-1}}{(x+4{)}^3{e}^x}$ using logarithms.

   Hint

   Expand the log first.

   Answer

   $\log y=2\log (x+1)+\frac{1}{2}\log (x-1)-3\log (x+4)-x$. So $y^{\prime }=y\left(\frac{2}{x+1}+\frac{1}{2(x-1)}-\frac{3}{x+4}-1\right)$.

8. Q8 · 2 marks · Very short answerHarder board level

   If $x=a{\cos }^3\theta$ and $y=a{\sin }^3\theta$, find $\frac{dy}{dx}$ at $\theta =\frac{\pi }{6}$.

   Hint

   Many factors cancel.

   Answer

   $\frac{dy}{dx}=\frac{3a{\sin }^2\theta \cos \theta }{-3a{\cos }^2\theta \sin \theta }=-\tan \theta$; at π/6: $-\frac{1}{\sqrt{3}}$.

9. Q9 · 3 marks · Short answerHarder board level

   If $y=A{e}^{mx}+B{e}^{nx}$, show that $y^{\prime \prime }-(m+n)y^{\prime }+mny=0$.

   Hint

   Find y′ and y″ and substitute.

   Answer

   $y^{\prime }=mA{e}^{mx}+nB{e}^{nx}$, $y^{\prime \prime }=m^{2}A{e}^{mx}+n^{2}B{e}^{nx}$. Then $y^{\prime \prime }-(m+n)y^{\prime }+mny=A{e}^{mx}(m^2-m^2-mn+mn)+B{e}^{nx}(n^2-mn-n^2+mn)=0$.

10. Q10 · 3 marks · Short answerHOTS / competency

    Find the values of a and b such that $f(x)=\{\begin{array}{ll}{x}^2+ax+b,& x<2\\ 3x+1,& x\ge 2\end{array}$ is differentiable at x = 2.

    Hint

    Continuity and equal derivatives.

    Answer

    Continuity: 4 + 2a + b = 7. Derivatives: 2(2) + a = 3 ⇒ a = −1. Then b = 7 − 4 + 2 = 5. So a = −1, b = 5.

11. Q11 · 2 marks · Very short answerHOTS / competency

    If $f(x+y)=f(x)f(y)$ for all x, y, with f(0) = 1, f′(0) = 2 and f(3) = 5, use the definition of derivative to find f′(3).

    Hint

    $f(3+h)=f(3)f(h)$.

    Answer

    $f^{\prime }(3)={\lim }_{h\to 0}\frac{f(3)f(h)-f(3)}{h}=f(3)\lim \frac{f(h)-f(0)}{h}=f(3)f^{\prime }(0)=5\times 2=10$.

12. Q12 · 2 marks · Very short answerHarder board level

    Find the second derivative of $y=x^3\log x$, x > 0.

    Hint

    Product rule twice.

    Answer

    $y^{\prime }=3x^2\log x+x^2$; $y^{\prime \prime }=6x\log x+3x+2x=6x\log x+5x$.