Continuity, Differentiability and Differentiation ISC Class 12: Notes and Solved Questions

What Continuity, Differentiability and Differentiation is about

Weightage: Continuity, Differentiability and Differentiation is part of the unit Calculus, which carries 35 of the 80 marks in the ISC Mathematics, shared with 4 other chapters. Questions on it can appear in any section of the paper.Source: CISCE ISC Mathematics syllabus, examination year 2027, distribution of marks

Key concepts

Continuity at a point

f is continuous at x = a if (i) f(a) is defined, (ii) ${\lim }_{x\to a}f(x)$ exists, i.e. the left-hand limit (LHL) equals the right-hand limit (RHL), and (iii) this limit equals f(a). f is continuous on an interval if it is continuous at every point of it. Polynomials, sin x, cos x, $e^x$ and log x (x > 0) are continuous on their domains; sums, differences, products and quotients (where the denominator is non-zero) of continuous functions are continuous, and so are compositions. To find a constant that makes a function continuous, equate the limit to the function value.

Worked example

Find k so that $f(x)=\frac{x^2-9}{x-3}$ for x ≠ 3, f(3) = k, is continuous at x = 3.

Try it first, then show the solutionSolution

1. ${\lim }_{x\to 3}\frac{(x-3)(x+3)}{x-3}={\lim }_{x\to 3}(x+3)=6$
2. Continuity needs f(3) = 6, so k = 6.

Answer: k = 6

Piecewise functions and types of discontinuity

For a piecewise function, find the LHL using the piece valid just to the left of the joining point and the RHL using the piece just to the right; compare both with the value at the point (from whichever piece includes it). A removable discontinuity occurs when the limit exists but f(a) is undefined or different; redefining f(a) removes it. A jump (non-removable) occurs when LHL ≠ RHL. The greatest integer function [x] jumps at every integer; |x| has no breaks.

Worked example

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

Try it first, then show the solutionSolution

1. LHL = f(2) = 2k + 1 (from the first piece); RHL = 3(2) − 1 = 5.
2. Continuity: 2k + 1 = 5, so k = 2.

Answer: k = 2

Differentiability and its link with continuity

f is differentiable at a if $f^{\prime }(a)={\lim }_{h\to 0}\frac{f(a+h)-f(a)}{h}$ exists, i.e. the left-hand derivative $L{f}^{\prime }(a)$ equals the right-hand derivative $R{f}^{\prime }(a)$. Every differentiable function is continuous, but a continuous function need not be differentiable: f(x) = |x| has $L{f}^{\prime }(0)=-1$ and $R{f}^{\prime }(0)=1$. Graphically, f fails to be differentiable at corners (sharp turns), cusps, vertical tangents and breaks. For a piecewise function, check continuity first, then compare the derivatives of the two pieces at the join.

Worked example

Find a and b so that $f(x)=\{\begin{array}{ll}{x}^2,& x\le 1\\ ax+b,& x>1\end{array}$ is differentiable at x = 1.

Try it first, then show the solutionSolution

1. Continuity at 1: 1 = a + b.
2. Equal derivatives at 1: left 2x = 2, right a; so a = 2.
3. Then b = −1.

Answer: a = 2, b = −1

Chain rule and standard derivatives

Chain rule: if y = f(u) and u = g(x), $\frac{dy}{dx}=\frac{dy}{du}\cdot \frac{du}{dx}$. Standard results: $\frac{d}{dx}{e}^x=e^x$, $\frac{d}{dx}{a}^x=a^x\log a$, $\frac{d}{dx}\log x=\frac{1}{x}$; $\frac{d}{dx}{\sin }^{-1}x=\frac{1}{\sqrt{1-x^2}}$, $\frac{d}{dx}{\cos }^{-1}x=-\frac{1}{\sqrt{1-x^2}}$, $\frac{d}{dx}{\tan }^{-1}x=\frac{1}{1+x^2}$, $\frac{d}{dx}{\cot }^{-1}x=-\frac{1}{1+x^2}$, $\frac{d}{dx}{\sec }^{-1}x=\frac{1}{|x|\sqrt{x^2-1}}$. Work from the outside in, one layer at a time.

Worked example

Differentiate (a) $\sin (x^3)$, (b) ${\sin }^{-1}(x^2)$, (c) $e^{\tan x}$.

Try it first, then show the solutionSolution

1. (a) $\cos (x^3)\cdot 3x^2$
2. (b) $\frac{1}{\sqrt{1-x^4}}\cdot 2x$
3. (c) $e^{\tan x}{\sec }^{2}x$

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

Simplifying by substitution before differentiating

Inverse-trigonometric expressions often collapse with a substitution. With x = tan θ: ${\sin }^{-1}\frac{2x}{1+x^2}$, ${\cos }^{-1}\frac{1-x^2}{1+x^2}$ and ${\tan }^{-1}\frac{2x}{1-x^2}$ all equal 2θ = $2{\tan }^{-1}x$ (for suitable x). With x = sin θ: ${\sin }^{-1}(2x\sqrt{1-x^2})=2{\sin }^{-1}x$. Simplify first, then differentiate; this is far shorter and safer than the chain rule on the raw expression. The derivative of u with respect to v is $\frac{du/dx}{dv/dx}$.

Worked example

Differentiate ${\sin }^{-1}\frac{2x}{1+x^2}$ with respect to x, for |x| < 1.

Try it first, then show the solutionSolution

1. Put x = tan θ: $\frac{2\tan \theta }{1+{\tan }^2\theta }=\sin 2\theta$, and 2θ lies in $(-\frac{\pi }{2},\frac{\pi }{2})$.
2. So y = 2θ = $2{\tan }^{-1}x$ and $\frac{dy}{dx}=\frac{2}{1+x^2}$.

Answer: $\frac{2}{1+x^2}$

Implicit differentiation

When y is not given explicitly, differentiate both sides of the equation with respect to x, remembering that y is a function of x: $\frac{d}{dx}{y}^2=2y\frac{dy}{dx}$ and $\frac{d}{dx}(xy)=y+x\frac{dy}{dx}$. Collect the $\frac{dy}{dx}$ terms on one side and solve. Substitute a point only after finding the general expression.

Worked example

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

Try it first, then show the solutionSolution

1. $2x+y+x\frac{dy}{dx}+2y\frac{dy}{dx}=0$
2. $\frac{dy}{dx}=-\frac{2x+y}{x+2y}$; at (1, 2): $-\frac{4}{5}$

Answer: $-\frac{2x+y}{x+2y}$; −0.8 at (1, 2)

Logarithmic and parametric differentiation

Logarithmic differentiation: for $y=u^v$ (both u and v depending on x) or long products and quotients, take logs: log y = v log u; differentiate: $\frac{1}{y}\frac{dy}{dx}=v^{\prime }\log u+v\frac{u^{\prime }}{u}$. For a sum such as $y=u^v+w^z$, differentiate each term separately. Parametric differentiation: if x = f(t), y = g(t), then $\frac{dy}{dx}=\frac{dy/dt}{dx/dt}$. For the second derivative, $\frac{d^{2}y}{d{x}^2}=\frac{d}{dt}\left(\frac{dy}{dx}\right)÷\frac{dx}{dt}$.

Worked example

Find $\frac{dy}{dx}$ for (a) $y=x^x$ and (b) x = a(θ − sin θ), y = a(1 − cos θ).

Try it first, then show the solutionSolution

1. (a) log y = x log x ⇒ $\frac{1}{y}{y}^{\prime }=\log x+1$ ⇒ $y^{\prime }=x^x(1+\log x)$.
2. (b) $\frac{dx}{d\theta }=a(1-\cos \theta )$, $\frac{dy}{d\theta }=a\sin \theta$; $\frac{dy}{dx}=\frac{\sin \theta }{1-\cos \theta }=\frac{2\sin \frac{\theta }{2}\cos \frac{\theta }{2}}{2{\sin }^2\frac{\theta }{2}}=\cot \frac{\theta }{2}$.

Answer: (a) $x^x(1+\log x)$ (b) $\cot \frac{\theta }{2}$

Second-order derivatives

$\frac{d^{2}y}{d{x}^2}$ is the derivative of $\frac{dy}{dx}$. Typical ISC questions ask you to prove a relation between y, y′ and y″. Tips: differentiate once, rearrange to remove awkward roots or fractions (e.g. multiply by $\sqrt{1-x^2}$ or by x), square if needed, then differentiate again. The syllabus limits successive differentiation to the second order.

Worked example

If y = a cos(log x) + b sin(log x), show that $x^2\frac{d^{2}y}{d{x}^2}+x\frac{dy}{dx}+y=0$.

Try it first, then show the solutionSolution

1. $y^{\prime }=\frac{-a\sin (\log x)+b\cos (\log x)}{x}$, so $x{y}^{\prime }=-a\sin (\log x)+b\cos (\log x)$.
2. Differentiate again: $x{y}^{\prime \prime }+y^{\prime }=\frac{-a\cos (\log x)-b\sin (\log x)}{x}=-\frac{y}{x}$.
3. Multiply by x: $x^2{y}^{\prime \prime }+x{y}^{\prime }+y=0$.

Answer: Proved

Maths in real life: Continuity, Differentiability and Differentiation

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

• Continuity and Differentiability of Piecewise Functions
• Implicit and Inverse Trigonometric Differentiation
• Second-Order and Parametric Derivatives

Continuity, Differentiability and Differentiation formula sheet

Continuity at a

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

LHL = RHL = value.

Derivative from first principles

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

Differentiable: LHD = RHD.

Chain rule

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

Exponential and log

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

log = natural logarithm.

Inverse trigonometric

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

cot⁻¹: $-\frac{1}{1+x^2}$; sec⁻¹: $\frac{1}{|x|\sqrt{x^2-1}}$.

Trigonometric

$\frac{d}{dx}\tan x={\sec }^{2}x,\frac{d}{dx}\sec x=\sec x\tan x,\frac{d}{dx}\text{cosec}x=-\text{cosec}x\cot x$

Parametric

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

dx/dt ≠ 0.

Logarithmic differentiation

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

Take logs, then differentiate.

Derivative of u with respect to v

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

Useful substitutions

${\sin }^{-1}\frac{2x}{1+x^2}={\cos }^{-1}\frac{1-x^2}{1+x^2}={\tan }^{-1}\frac{2x}{1-x^2}=2{\tan }^{-1}x$

For |x| < 1 (and x ≥ 0 for the cos⁻¹ form).

How the board asks Continuity, Differentiability and Differentiation

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

Tricks to remember Continuity, Differentiability and Differentiation

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

• Left piece, right piece

  LHL from the piece that applies just left of a; RHL from the piece just right of a.

  Use it when: Piecewise continuity (the error examiners report most).

• Smooth means differentiable

  A corner (|x|) is continuous but not smooth; a break is neither.

  Use it when: Graph-based items.

• Co- functions get a minus

  Derivatives of cos⁻¹, cot⁻¹, cosec⁻¹ carry a minus sign.

  Use it when: Inverse-trig derivatives.

• Log first when the power moves

  Variable in the base and the exponent? Take logs before differentiating.

  Use it when: $x^x$, $(\sin x{)}^x$.

• Simplify before you differentiate

  See $\frac{2x}{1+x^2}$ inside sin⁻¹? Put x = tan θ first.

  Use it when: Inverse-trig differentiation.

Continuity, Differentiability and Differentiation: 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 D style)

Read the situation, then answer every part. In the board paper this is a 5-mark question split into 1 + 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, Differentiability and Differentiation

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

• ✗Mistake: Using the wrong piece for a one-sided limit

  CISCE examiners' comments, 2025: in counting points of discontinuity of a piecewise function, candidates used the wrong pieces for the left- and right-hand limits.

  Wrong working

  At x = 1, uses −x² − 2 for the right-hand limit.

  Right working

  For x → 1⁺ use the piece valid for x ≥ 1; for x → 1⁻ the piece valid on [0, 1).

  ✓Do this: Write the interval next to each limit before substituting.

  Usually costs: 1 mark.

• ✗Mistake: Not being able to evaluate the limit

  CISCE examiners' comments, 2024: few candidates could evaluate the limit when finding k for continuity of a rational function at the point where it is 0/0.

  Wrong working

  Substitutes x = 3 and gets 0/0, then stops.

  Right working

  Factorise and cancel (x − 3) first, then substitute.

  ✓Do this: A 0/0 form means there is a common factor to cancel.

  Usually costs: 1-2 marks.

• ✗Mistake: Wrong logarithm laws in implicit differentiation

  CISCE examiners' comments, 2025: proving a derivative from $x^y=e^{x-y}$, candidates applied log properties incorrectly, differentiated implicitly wrongly and substituted incorrectly.

  Wrong working

  log(e^{x − y}) = (x − y) log e^x … or log(u + v) = log u + log v.

  Right working

  log of $x^y$ is y log x; log of $e^{x-y}$ is x − y. So y log x = x − y, $y=\frac{x}{1+\log x}$, then differentiate.

  ✓Do this: Write the log step on its own line before differentiating.

  Usually costs: 1-2 marks.

• ✗Mistake: Differentiating an inverse-trig expression without simplifying

  CISCE examiners' comments, 2025: differentiating ${\sin }^{-1}\frac{2^{x+1}{3}^x}{1+36^x}$, candidates did not simplify using inverse-trigonometric properties and misapplied the chain rule.

  Wrong working

  Applies the chain rule to the whole fraction, with long algebra and errors.

  Right working

  Write it as ${\sin }^{-1}\frac{2\cdot 6^x}{1+(6^x{)}^2}=2{\tan }^{-1}{6}^x$ (for suitable x), then differentiate: $\frac{2\cdot 6^x\log 6}{1+36^x}$.

  ✓Do this: Look for the $\frac{2t}{1+t^2}$ pattern first.

  Usually costs: 1-2 marks.

• ✗Mistake: Reading differentiability from a graph

  CISCE examiners' comments, 2024: candidates did not understand how to tell from a graph where a function is not differentiable.

  Wrong working

  Chooses points where the graph is highest or lowest.

  Right working

  Non-differentiable at sharp corners, cusps, vertical tangents and breaks; smooth turning points are differentiable.

  ✓Do this: Ask: does the curve have a unique tangent here?

  Usually costs: 1 mark.

Practice questions on Continuity, Differentiability and Differentiation

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

   $\frac{d}{dx}\log (\sin x)$ equals:
   (a) tan x
   (b) cot x
   (c) $\frac{1}{\sin x}$
   (d) cos x

   Hint

   Chain rule.

   Answer

   (b) cot x

2. Q2 · 1 mark · MCQBoard level

   The greatest integer function f(x) = [x] is discontinuous at:
   (a) x = 0 only
   (b) all integers
   (c) no point
   (d) all non-integers

   Hint

   Where does it jump?

   Answer

   (b)

3. Q3 · 1 mark · MCQBoard level

   If f is differentiable at x = a, then f is:
   (a) not necessarily continuous at a
   (b) continuous at a
   (c) discontinuous at a
   (d) equal to zero at a

   Hint

   The converse fails.

   Answer

   (b)

4. Q4 · 1 mark · Very short answerBoard level

   Differentiate $e^{x^2}$ with respect to x.

   Hint

   Chain rule.

   Answer

   $2x{e}^{x^2}$

5. Q5 · 2 marks · Short answerBoard level

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

   Hint

   Equate the two sides at 2.

   Answer

   2k + 1 = 5, k = 2

6. Q6 · 2 marks · Short answerHarder board level

   If $x^2+xy+y^2=7$, find $\frac{dy}{dx}$ at (1, 2).

   Hint

   Implicit differentiation.

   Answer

   $-\frac{2x+y}{x+2y}=-\frac{4}{5}$

7. Q7 · 2 marks · Short answerHarder board level

   Differentiate $y=x^x$ with respect to x.

   Hint

   Take logs.

   Answer

   $x^x(1+\log x)$

8. Q8 · 2 marks · Short answerHOTS / competency

   How many points of discontinuity does $f(x)=\{\begin{array}{ll}2x,& x<1\\ x^2+1,& 1\le x<2\\ 3x-1,& x\ge 2\end{array}$ have? Justify.

   Hint

   Check x = 1 and x = 2.

   Answer

   At 1: LHL 2, RHL = f(1) = 2. At 2: LHL 5, RHL = f(2) = 5. Continuous at both joins (and each piece is a polynomial), so there are no points of discontinuity.

9. Q9 · 3 marks · Short answerHarder board level

   If x = a cos³θ and y = a sin³θ, find $\frac{dy}{dx}$.

   Hint

   Parametric.

   Answer

   $\frac{dx}{d\theta }=-3a{\cos }^2\theta \sin \theta$, $\frac{dy}{d\theta }=3a{\sin }^2\theta \cos \theta$, so $\frac{dy}{dx}=-\tan \theta$.

10. Q10 · 3 marks · Short answerHarder board level

    If $y=({\sin }^{-1}x{)}^2$, show that $(1-x^2)\frac{d^{2}y}{d{x}^2}-x\frac{dy}{dx}-2=0$.

    Hint

    Multiply y′ by √(1 − x²) and square.

    Answer

    $y^{\prime }=\frac{2{\sin }^{-1}x}{\sqrt{1-x^2}}$, so $(1-x^2)y^{\prime 2}=4y$. Differentiate: $(1-x^2)2y^{\prime }{y}^{\prime \prime }-2x{y}^{\prime 2}=4y^{\prime }$; divide by 2y′: $(1-x^2)y^{\prime \prime }-x{y}^{\prime }-2=0$.

11. Q11 · 3 marks · Short answerHarder board level

    Differentiate sin(x³) with respect to x³.

    Hint

    u = sin(x³), v = x³.

    Answer

    $\frac{du}{dx}=3x^2\cos x^3$, $\frac{dv}{dx}=3x^2$, so $\frac{du}{dv}=\cos (x^3)$.

12. Q12 · 5 marks · Long answerHarder board level

    (a) Differentiate $y=x^{\sin x}+(\sin x{)}^x$. [3]
    (b) Find a, b for which $f(x)=\{\begin{array}{ll}{x}^2,& x\le 1\\ ax+b,& x>1\end{array}$ is differentiable at x = 1. [2]

    Hint

    Differentiate each term with logs.

    Answer

    (a) $x^{\sin x}\left(\cos x\log x+\frac{\sin x}{x}\right)+(\sin x{)}^x(\log \sin x+x\cot x)$. (b) Continuity a + b = 1; derivatives 2 = a; so a = 2, b = −1.

13. Q13 · 2 marks · Short answerHarder board level

    The function $f(x)=\frac{x^2-4}{x-2}$ is not defined at x = 2. What type of discontinuity is this, and how can it be removed?

    Hint

    Does the limit exist?

    Answer

    The limit is 4 (f(x) = x + 2 for x ≠ 2), so the discontinuity is removable: define f(2) = 4.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1. 1.Condition for continuity at x = a?Show answer

   LHL = RHL = f(a).

2. 2.Condition for differentiability at a?Show answer

   LHD = RHD (finite).

3. 3.Is every continuous function differentiable?Show answer

   No; |x| is continuous but not differentiable at 0.

4. 4.$\frac{d}{dx}{\tan }^{-1}x$ = ?Show answer

   $\frac{1}{1+x^2}$.

5. 5.$\frac{d}{dx}{a}^x$ = ?Show answer

   $a^x\log a$.

6. 6.Parametric derivative?Show answer

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

7. 7.When do you use logarithmic differentiation?Show answer

   When the variable is in both base and exponent, or for long products and quotients.

8. 8.What is a removable discontinuity?Show answer

   A point where the limit exists but differs from (or the function lacks) the value; redefining f(a) removes it.

Quick revision checklist

The night before the exam: can you do each of these without looking?

• ☐Can you find k for continuity of a 0/0 rational function?
• ☐Can you test a piecewise function at its joins with the right pieces?
• ☐Can you find a and b for differentiability?
• ☐Can you write the derivatives of all inverse-trig functions?
• ☐Can you simplify ${\sin }^{-1}\frac{2x}{1+x^2}$ before differentiating?
• ☐Can you differentiate $x^y=y^x$ implicitly?
• ☐Can you do a parametric derivative and simplify with half-angles?
• ☐Can you prove a second-order relation like $x^2{y}^{\prime \prime }+x{y}^{\prime }+y=0$?

Continuity, Differentiability and Differentiation: frequently asked questions