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Class 12 Maths · Chapter 5 · Bihar Board

Continuity & Differentiability

Master the connection between continuity and differentiability with visual graphs, chain rule, logarithmic differentiation, step-marked solutions, MCQs, A-R, and PYQs.

8 Visual Graphs Chain Rule Step Marking
5Core Concepts
20MCQs
10Assertion-Reason
8PYQs
Section 01

Basic Concepts

Understanding continuity and differentiability

5.1 Introduction

In this chapter, we introduce two very important concepts — continuity and differentiability — and study the relation between them. We also learn to differentiate inverse trigonometric functions, exponential functions, and logarithmic functions.

Intuitive Idea

A function is continuous at a point if you can draw its graph around that point without lifting the pen from the paper.

5.2 Continuity at a Point

Definition — Continuity

A real function $f$ is continuous at $x = c$ if:

$$\lim_{x \to c} f(x) = f(c)$$

More elaborately: LHL = RHL = f(c)

If any one of these three values doesn't match, the function is discontinuous at that point.

Conditions in Detail

  • LHL: $\lim_{x \to c^-} f(x)$ must exist
  • RHL: $\lim_{x \to c^+} f(x)$ must exist
  • Value: $f(c)$ must be defined
  • All equal: LHL = RHL = f(c)

5.3 Differentiability at a Point

Definition — Derivative

The derivative of $f$ at $c$ is:

$$f'(c) = \lim_{h \to 0} \frac{f(c+h) - f(c)}{h}$$

If this limit exists, $f$ is differentiable at $c$.

Left and Right Hand Derivatives

$$Lf'(c) = \lim_{h \to 0^-} \frac{f(c+h) - f(c)}{h}$$
$$Rf'(c) = \lim_{h \to 0^+} \frac{f(c+h) - f(c)}{h}$$

$f$ is differentiable at $c$ if and only if Lf'(c) = Rf'(c) (both finite and equal).

5.4 Relationship Between Continuity and Differentiability

Key Theorem

Every differentiable function is continuous, but the converse is NOT true.

Differentiable $\Rightarrow$ Continuous
Continuous $\nRightarrow$ Differentiable

Example: $f(x) = |x|$ is continuous at $x=0$ but not differentiable there.

5.5 Algebra of Continuous Functions

If $f$ and $g$ are continuous at $c$, then:

  • $f \pm g$ is continuous at $c$
  • $f \cdot g$ is continuous at $c$
  • $\dfrac{f}{g}$ is continuous at $c$ (provided $g(c) \neq 0$)
  • Composite $f \circ g$ is continuous (if conditions hold)
Section 02

🎨 Visual Graphs

See continuity and differentiability in action

V1 Jump Discontinuity — LHL ≠ RHL

LHL = 1, RHL = 2 — Not Continuous X Y y = 1 (LHL) y = 2 (RHL) Jump of 1 unit O At x = 0: Left limit = 1, Right limit = 2 — Different → Discontinuous
When the left-hand and right-hand limits do not match, there's a "jump" in the graph — the function is discontinuous.

V2 Removable Discontinuity — LHL = RHL ≠ f(c)

Both limits = 1, but f(0) = 2 X Y y = 1 (limit) f(0) = 2 Both limits exist and equal 1, but f(0) = 2 → Discontinuous at x = 0 (Can be fixed by redefining f(0) = 1)
When both limits exist but don't match the function's value, it's a "removable" discontinuity — you can fix it by changing just one value.

V3 $y = |x|$ — Continuous at 0

LHL = RHL = f(0) = 0 X Y f(0) = 0 y = −x y = x LHL = 0, RHL = 0, f(0) = 0 → Continuous ✓
The absolute value function is continuous at $x = 0$ — the graph has no jumps or holes.

V4 $y = |x|$ — NOT Differentiable at 0

Sharp corner — left derivative ≠ right derivative X Y Sharp Corner Slope = −1 (left) Slope = +1 (right) Left derivative = −1, Right derivative = +1 → Not differentiable
At $x = 0$, the function is continuous, but the "sharp corner" makes left and right derivatives different. So it's not differentiable.

V5 Chain Rule — Function Inside Function

Differentiate outer, multiply by inner derivative Outer function h(t) = t³ h'(t) = 3t² Inner function g(x) = 2x+1 g'(x) = 2 × d/dx [h(g(x))] = h'(g(x)) · g'(x) = 3(2x+1)² · 2 = 6(2x+1)²
Chain Rule: differentiate the outer function (keeping inner as-is), then multiply by the derivative of the inner function.

V6 Exponential Growth vs Polynomial Growth

$e^x$ eventually beats any polynomial X Y y = x y = x² y = x³ y = eˣ For large x, eˣ grows faster than any polynomial
Exponential growth always outpaces polynomial growth for sufficiently large $x$. That's why $e^x$ is fundamental in calculus.

V7 $e^x$ and $\ln x$ — Mirror Images Across $y = x$

Inverse functions X Y y = x y = eˣ y = ln x O ln x and eˣ are inverse functions — reflected across y = x
Since $e^x$ and $\ln x$ are inverses, their graphs are mirror images across the line $y = x$.

V8 Logarithmic Differentiation — When to Use

For $f(x)^{g(x)}$ — take log first! Hard to differentiate y = xˢⁱⁿˣ Base & exponent both vary → Take log Easy to differentiate log y = sin x · log x Now product rule applies → Result dy/dx = y · d/dx(log y) Substitute y back Use when function is of form [f(x)]^g(x) or has many multiplied terms
Logarithmic differentiation converts tricky exponential/power forms into simpler products that can be differentiated easily.
Section 03

Continuity — Deep Dive

Checking continuity at a point

C1 Test for Continuity at a Point $x = c$

Step-by-step procedure:

  1. Find $f(c)$ — the value of the function at $c$
  2. Find LHL: $\lim_{x \to c^-} f(x)$
  3. Find RHL: $\lim_{x \to c^+} f(x)$
  4. Check if LHL = RHL = f(c)
  5. If yes → Continuous; if not → Discontinuous
Point to Remember
  • Polynomial functions are continuous everywhere
  • Rational functions are continuous except where denominator = 0
  • $\sin x$, $\cos x$ are continuous everywhere
  • $\tan x$, $\sec x$, $\cot x$, $\csc x$ are continuous in their domains
  • Greatest integer function $[x]$ is discontinuous at every integer

C2 Points of Discontinuity

Common functions and their discontinuity points:

FunctionPoints of Discontinuity
$f(x) = \frac{1}{x}$$x = 0$
$f(x) = [x]$ (Greatest integer)All integers $x \in \mathbb{Z}$
$f(x) = \tan x$$x = (2n+1)\pi/2$
$f(x) = \frac{p(x)}{q(x)}$Where $q(x) = 0$
$f(x) = \sin\frac{1}{x}$$x = 0$ (oscillating)
Section 04

Differentiability — Deep Dive

Types, rules, and techniques

D1 Chain Rule

If $f = v \circ u$, i.e., $f(x) = v(u(x))$, then:

$$\frac{df}{dx} = \frac{dv}{dt} \cdot \frac{dt}{dx}, \quad \text{where } t = u(x)$$

Example: $\frac{d}{dx}[\sin(x^2)] = \cos(x^2) \cdot 2x = 2x\cos(x^2)$

D2 Derivatives of Inverse Trigonometric Functions

$$\frac{d}{dx}(\sin^{-1}x) = \frac{1}{\sqrt{1-x^2}}, \quad x \in (-1, 1)$$
$$\frac{d}{dx}(\cos^{-1}x) = -\frac{1}{\sqrt{1-x^2}}, \quad x \in (-1, 1)$$
$$\frac{d}{dx}(\tan^{-1}x) = \frac{1}{1+x^2}, \quad x \in \mathbb{R}$$
$$\frac{d}{dx}(\cot^{-1}x) = -\frac{1}{1+x^2}, \quad x \in \mathbb{R}$$
$$\frac{d}{dx}(\sec^{-1}x) = \frac{1}{|x|\sqrt{x^2-1}}, \quad |x| > 1$$
$$\frac{d}{dx}(\csc^{-1}x) = -\frac{1}{|x|\sqrt{x^2-1}}, \quad |x| > 1$$

D3 Derivatives of Exponential & Logarithmic Functions

$$\frac{d}{dx}(e^x) = e^x$$
$$\frac{d}{dx}(a^x) = a^x \log a, \quad a > 0$$
$$\frac{d}{dx}(\log x) = \frac{1}{x}, \quad x > 0$$
$$\frac{d}{dx}(\log_a x) = \frac{1}{x \log a}$$

D4 Implicit Differentiation

When $y$ is not expressed explicitly in terms of $x$, differentiate both sides with respect to $x$ and then solve for $\frac{dy}{dx}$.

Example: $x^2 + y^2 = 25$

Differentiating: $2x + 2y \frac{dy}{dx} = 0 \Rightarrow \frac{dy}{dx} = -\frac{x}{y}$

D5 Logarithmic Differentiation

Used for functions of the form $y = [f(x)]^{g(x)}$ or products of many factors.

Method: Take $\log$ on both sides → differentiate → solve for $\frac{dy}{dx}$

Example: $y = x^{\sin x}$

$\log y = \sin x \cdot \log x$

$\frac{1}{y}\frac{dy}{dx} = \cos x \log x + \frac{\sin x}{x}$

$\frac{dy}{dx} = x^{\sin x}\left(\cos x \log x + \frac{\sin x}{x}\right)$

D6 Parametric Differentiation

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

$$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{g'(t)}{f'(t)}, \quad f'(t) \neq 0$$

D7 Second Order Derivative

The derivative of $\frac{dy}{dx}$ with respect to $x$ is called second order derivative:

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

📐 All Important Formulas

Memorize these for board exams

F1 Continuity Conditions

$$\lim_{x \to c^-} f(x) = \lim_{x \to c^+} f(x) = f(c)$$

F2 Derivative Definition

$$f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$$

F3 Standard Derivatives

$\frac{d}{dx}(x^n) = nx^{n-1}$
$\frac{d}{dx}(\sin x) = \cos x$,   $\frac{d}{dx}(\cos x) = -\sin x$
$\frac{d}{dx}(\tan x) = \sec^2 x$,   $\frac{d}{dx}(\cot x) = -\csc^2 x$
$\frac{d}{dx}(\sec x) = \sec x \tan x$,   $\frac{d}{dx}(\csc x) = -\csc x \cot x$
$\frac{d}{dx}(e^x) = e^x$,   $\frac{d}{dx}(\log x) = \frac{1}{x}$

F4 Inverse Trig Derivatives

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

F5 Differentiation Rules

Product: $(uv)' = u'v + uv'$
Quotient: $\left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2}$
Chain: $\frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x)$

F6 Chain Rule Extended

$$\frac{df}{dx} = \frac{dv}{dt} \cdot \frac{dt}{dx}$$

F7 Parametric Form

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

💡 Tricks & Shortcuts

Speed techniques for board exams

Trick 1 — Continuity Flowchart

LHL = RHL = f(c)?

  • Compute all three values
  • If all equal → Continuous
  • If LHL = RHL but ≠ f(c) → Removable discontinuity
  • If LHL ≠ RHL → Jump discontinuity
Trick 2 — Differentiable implies Continuous

"D → C, but C ↛ D"

  • Differentiable → must be continuous
  • Continuous → may or may not be differentiable
  • To disprove differentiability, check for sharp corners, cusps, vertical tangents
Trick 3 — Chain Rule Memory

"Outside derivative × Inside derivative"

  • For $\sin(x^2)$: outside = $\cos$, inside = $x^2$
  • Result: $\cos(x^2) \cdot 2x$
  • Works for multiple nesting: keep multiplying inward
Trick 4 — When to use Logarithmic Differentiation

Use $\log y$ when you see:

  • Function of form $[f(x)]^{g(x)}$ (variable in base AND exponent)
  • Product of many factors (like 3+ terms multiplied)
  • Complicated fractions with powers
Trick 5 — Implicit Differentiation Step

Whenever you differentiate a $y$-term, multiply by $\frac{dy}{dx}$

  • $\frac{d}{dx}(y^2) = 2y \cdot \frac{dy}{dx}$
  • $\frac{d}{dx}(\sin y) = \cos y \cdot \frac{dy}{dx}$
  • Then group $\frac{dy}{dx}$ terms on one side
Trick 6 — Standard Substitutions for Inverse Trig

Common tricks to simplify expressions:

  • $x = \sin\theta$ for $\sqrt{1-x^2}$
  • $x = \tan\theta$ for $1+x^2$
  • $x = \sec\theta$ for $\sqrt{x^2-1}$
Trick 7 — Second Derivative Shortcut

For $y = f(x)$, $y_1 = \frac{dy}{dx}$, $y_2 = \frac{d^2y}{dx^2}$

  • If $y = \sin^{-1}x$, then $(1-x^2)y_2 - xy_1 = 0$
  • If $y = e^x$, then $y_2 = y_1 = y$
  • If $y = \log x$, then $y_2 = -\frac{1}{x^2}$
Trick 8 — Discontinuity Check

Common discontinuity points:

  • Where denominator = 0 (for rational functions)
  • At all integers (for $[x]$ greatest integer function)
  • Where $\cos x = 0$ (for $\tan x$, $\sec x$)
  • Where $\sin x = 0$ (for $\cot x$, $\csc x$)
Trick 9 — Parametric Quick

For $x = f(t)$, $y = g(t)$:

  • Compute $\frac{dx}{dt}$ and $\frac{dy}{dt}$ separately
  • Then $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$
  • Simplify — the parameter $t$ usually cancels nicely
Trick 10 — Sanity Check Answer

Always verify:

  • Is your answer in terms of $x$ only (unless parametric)?
  • Are all terms algebraically simplified?
  • For continuity problems — did you actually compute all three (LHL, RHL, f(c))?
Practice

📝 Solved Examples with Step Marking

Click "Show Answer" to reveal step-by-step solutions with marks

📋 Total: 2 Marks · Board Pattern

Q1. Check the continuity of $f(x) = 2x + 3$ at $x = 1$.

Step 1 — Find f(1)0.5 Mark

$f(1) = 2(1) + 3 = 5$

Step 2 — Find limit at x=11 Mark

$\lim_{x\to 1} f(x) = \lim_{x\to 1} (2x+3) = 2(1) + 3 = 5$

Step 3 — Verify condition0.5 Mark

Since $\lim_{x\to 1} f(x) = 5 = f(1)$, $f$ is continuous at $x = 1$.

$f$ is continuous at $x = 1$
📋 Total: 3 Marks

Q2. Show that $f(x) = |x|$ is continuous but not differentiable at $x = 0$.

Step 1 — Continuity: LHL0.5 Mark

$f(x) = -x$ for $x < 0$

LHL = $\lim_{x\to 0^-} (-x) = 0$

Step 2 — Continuity: RHL and f(0)0.5 Mark

$f(x) = x$ for $x > 0$

RHL = $\lim_{x\to 0^+} x = 0$,   $f(0) = 0$

LHL = RHL = f(0) → Continuous ✓

Step 3 — Left Derivative at 01 Mark

$Lf'(0) = \lim_{h\to 0^-} \frac{f(0+h) - f(0)}{h} = \lim_{h\to 0^-} \frac{-h - 0}{h} = -1$

Step 4 — Right Derivative at 01 Mark

$Rf'(0) = \lim_{h\to 0^+} \frac{h - 0}{h} = 1$

Since $Lf'(0) \neq Rf'(0)$, $f$ is not differentiable at $x = 0$.

$f(x)=|x|$ is continuous but NOT differentiable at $x = 0$
📋 Total: 3 Marks

Q3. Find $\frac{dy}{dx}$ if $y = \sin(x^2)$.

Step 1 — Identify inner and outer0.5 Mark

Outer: $v(t) = \sin t$  |  Inner: $u(x) = x^2$

Step 2 — Apply Chain Rule1.5 Marks

$\frac{dy}{dx} = \cos(x^2) \cdot \frac{d}{dx}(x^2)$

$= \cos(x^2) \cdot 2x$

Step 3 — Simplify1 Mark

$\frac{dy}{dx} = 2x \cos(x^2)$

$\frac{dy}{dx} = 2x\cos(x^2)$
Test Yourself

✍️ MCQ Practice — 20 Questions

Select your answer and click "Check Answers"

1. A function $f$ is continuous at $x = c$ if:

Correct: B — By definition, continuity at $c$ means limit equals function value.

2. Every differentiable function is:

Correct: A — Differentiability → Continuity (but not vice versa).

3. $f(x) = |x|$ at $x = 0$ is:

Correct: C — Classic example: continuous but with a sharp corner at 0.

4. $\frac{d}{dx}(\sin^{-1}x)$ equals:

Correct: D — Standard derivative: $\frac{d}{dx}\sin^{-1}x = \frac{1}{\sqrt{1-x^2}}$.

5. $\frac{d}{dx}(e^x)$ equals:

Correct: A — $e^x$ is its own derivative.

6. If $f$ is continuous at $c$, then:

Correct: B — Continuity is necessary but not sufficient for differentiability.

7. The chain rule states $\frac{d}{dx}[f(g(x))]$ equals:

Correct: C — Chain rule: differentiate outer (keeping inner), multiply by derivative of inner.

8. $\frac{d}{dx}(\tan^{-1}x)$ equals:

Correct: D — Standard formula.

9. The function $f(x) = [x]$ is discontinuous at:

Correct: A — Greatest integer function has a jump at every integer.

10. $\frac{d}{dx}(\log x)$ equals:

Correct: B — Standard formula for natural log.

11. If $y = x^{\sin x}$, use what method?

Correct: C — Form $[f(x)]^{g(x)}$ requires log differentiation.

12. If $x^2 + y^2 = 25$, then $\frac{dy}{dx}$ =

Correct: A — Implicit: $2x + 2yy' = 0 \Rightarrow y' = -\frac{x}{y}$.

13. Second derivative of $y$ w.r.t. $x$ is denoted by:

Correct: D — Standard notation for second order derivative.

14. $\frac{d}{dx}(\cos^{-1}x)$ equals:

Correct: B — Derivative of $\cos^{-1}$ has a negative sign.

15. If $x = at^2, y = 2at$, then $\frac{dy}{dx}$ =

Correct: C — $\frac{dx}{dt} = 2at$, $\frac{dy}{dt} = 2a$, so $\frac{dy}{dx} = \frac{2a}{2at} = \frac{1}{t}$.

16. The function $f(x) = \sin x$ is:

Correct: A — $\sin x$ is smooth and continuous everywhere.

17. If $y = e^{2x}$, then $\frac{dy}{dx}$ =

Correct: D — Chain rule: $\frac{d}{dx}e^{2x} = e^{2x} \cdot 2 = 2e^{2x}$.

18. For $y = \sin^{-1}x$, which relation holds?

Correct: B — Standard result: $(1-x^2)y_2 - xy_1 = 0$.

19. If $f$ is differentiable at $c$, then LHL at $c$ equals:

Correct: C — Differentiability at $c$ implies continuity → LHL = RHL = f(c).

20. If $y = \log(\sin x)$, then $\frac{dy}{dx}$ =

Correct: A — Chain rule: $\frac{1}{\sin x} \cdot \cos x = \cot x$.
Score: 0 / 20
New Pattern

🎯 Assertion & Reasoning — 10 Questions

Each question has an Assertion (A) and a Reason (R). Choose the correct option.

📖 Standard Options

(A) Both A and R are true, and R is the correct explanation of A.

(B) Both A and R are true, but R is NOT the correct explanation of A.

(C) A is true, but R is false.

(D) A is false, but R is true.

1
Assertion (A) $f(x) = |x|$ is continuous at $x = 0$.
Reason (R) LHL = RHL = $f(0) = 0$ for $f(x) = |x|$ at $x = 0$.
Correct: A — Both true. R is exactly the reason why A is true.
2
Assertion (A) Every differentiable function is continuous.
Reason (R) Differentiability at a point implies LHL = RHL = $f(c)$ at that point.
Correct: A — Both true. R is the underlying mathematical reason.
3
Assertion (A) Every continuous function is differentiable.
Reason (R) The function $f(x) = |x|$ is continuous but not differentiable at $x=0$.
Correct: D — A is false (converse not true). R is true — and R is actually the counter-example disproving A.
4
Assertion (A) $\frac{d}{dx}(e^x) = e^x$.
Reason (R) The natural exponential function $e^x$ has the unique property that its derivative equals itself.
Correct: A — Both true, R explains the fundamental property of $e^x$.
5
Assertion (A) $\frac{d}{dx}(\sin^{-1}x) = \frac{1}{\sqrt{1-x^2}}$ for $x \in (-1,1)$.
Reason (R) If $y = \sin^{-1}x$, then $x = \sin y$ and differentiating gives $1 = \cos y \cdot \frac{dy}{dx}$.
Correct: A — Both true, R derives the formula for A.
6
Assertion (A) $\frac{d}{dx}(x^x) = x \cdot x^{x-1}$.
Reason (R) Logarithmic differentiation is used for functions of the form $f(x)^{g(x)}$.
Correct: D — A is false (correct answer is $x^x(1+\log x)$). R is true.
7
Assertion (A) The function $f(x) = [x]$ is discontinuous at all integers.
Reason (R) The greatest integer function has a jump at each integer — LHL ≠ RHL.
Correct: A — Both true, R explains A well.
8
Assertion (A) Chain rule is used to differentiate composite functions.
Reason (R) If $f = v \circ u$, then $\frac{df}{dx} = \frac{dv}{dt} \cdot \frac{dt}{dx}$ where $t = u(x)$.
Correct: A — Both true, R is the formal statement of chain rule.
9
Assertion (A) $\frac{d}{dx}(a^x) = x \cdot a^{x-1}$.
Reason (R) $\frac{d}{dx}(a^x) = a^x \log a$ for $a > 0$.
Correct: D — A is false (correct derivative is $a^x \log a$, not power rule). R is true.
10
Assertion (A) Sum of two continuous functions is continuous.
Reason (R) If $f$ and $g$ are continuous at $c$, then $\lim_{x\to c}[f(x)+g(x)] = f(c)+g(c)$.
Correct: A — Both true, R is the algebraic proof of A.
Score: 0 / 10
Exam Focus

📖 Previous Year Questions

Frequently asked questions in board exams

2024

Q. Prove that $f(x) = 5x - 3$ is continuous at $x = 0$.

Step 1 — Find f(0)0.5 Mark

$f(0) = 5(0) - 3 = -3$

Step 2 — Find limit1 Mark

$\lim_{x\to 0}(5x-3) = -3$

Step 3 — Compare0.5 Mark

$\lim_{x\to 0} f(x) = -3 = f(0)$, so $f$ is continuous at $x = 0$.

$f$ is continuous at $x = 0$
2024

Q. Find $\frac{dy}{dx}$ if $y = \sin(x^2 + 5)$.

Step 1 — Apply Chain Rule1 Mark

$\frac{dy}{dx} = \cos(x^2 + 5) \cdot \frac{d}{dx}(x^2+5)$

Step 2 — Simplify1 Mark

$= \cos(x^2+5) \cdot 2x = 2x\cos(x^2+5)$

$\frac{dy}{dx} = 2x\cos(x^2+5)$
2023

Q. Show that $f(x) = |x-1|$ is not differentiable at $x = 1$.

Step 1 — Left derivative1 Mark

For $x < 1$, $|x-1| = -(x-1) = 1-x$

$Lf'(1) = \lim_{h\to 0^-} \frac{f(1+h)-f(1)}{h} = \lim_{h\to 0^-} \frac{-h}{h} = -1$

Step 2 — Right derivative1 Mark

For $x > 1$, $|x-1| = x-1$

$Rf'(1) = \lim_{h\to 0^+} \frac{h}{h} = 1$

Step 3 — Compare1 Mark

$Lf'(1) \neq Rf'(1)$, so $f$ is not differentiable at $x = 1$.

$f$ is NOT differentiable at $x = 1$
2023

Q. Differentiate $\sqrt{\cot(x^2)}$ w.r.t. $x$.

Step 1 — Chain Rule on sqrt1 Mark

Let $y = \sqrt{\cot(x^2)} = [\cot(x^2)]^{1/2}$

$\frac{dy}{dx} = \frac{1}{2}[\cot(x^2)]^{-1/2} \cdot \frac{d}{dx}[\cot(x^2)]$

Step 2 — Derivative of cot1 Mark

$= \frac{1}{2\sqrt{\cot(x^2)}} \cdot [-\csc^2(x^2)] \cdot 2x$

Step 3 — Simplify1 Mark

$= \frac{-x \csc^2(x^2)}{\sqrt{\cot(x^2)}}$

$\frac{dy}{dx} = -\frac{x \csc^2(x^2)}{\sqrt{\cot(x^2)}}$
2022

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

Step 1 — Differentiate1 Mark

$2x + \left(y + x\frac{dy}{dx}\right) + 2y\frac{dy}{dx} = 0$

Step 2 — Collect dy/dx terms1 Mark

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

Step 3 — Solve1 Mark

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

$\frac{dy}{dx} = -\frac{2x+y}{x+2y}$
2022

Q. Differentiate $x^{\sin x}$ w.r.t. $x$ for $x > 0$.

Step 1 — Take log1 Mark

Let $y = x^{\sin x}$. Then $\log y = \sin x \cdot \log x$.

Step 2 — Differentiate1.5 Marks

$\frac{1}{y}\frac{dy}{dx} = \cos x \cdot \log x + \sin x \cdot \frac{1}{x}$

Step 3 — Solve for dy/dx0.5 Mark

$\frac{dy}{dx} = x^{\sin x}\left[\cos x \log x + \frac{\sin x}{x}\right]$

$\frac{dy}{dx} = x^{\sin x}\left(\cos x \log x + \frac{\sin x}{x}\right)$
2021

Q. If $y = 3\cos(\log x) + 4\sin(\log x)$, show that $x^2 y_2 + xy_1 + y = 0$.

Step 1 — Find y₁1 Mark

$y_1 = -3\sin(\log x) \cdot \frac{1}{x} + 4\cos(\log x) \cdot \frac{1}{x}$

$xy_1 = -3\sin(\log x) + 4\cos(\log x)$

Step 2 — Find y₂1 Mark

Differentiating $xy_1$: $y_1 + xy_2 = -3\cos(\log x)\cdot\frac{1}{x} - 4\sin(\log x)\cdot\frac{1}{x}$

Multiply by $x$: $xy_1 + x^2 y_2 = -3\cos(\log x) - 4\sin(\log x) = -y$

Step 3 — Rearrange1 Mark

$x^2 y_2 + xy_1 + y = 0$ ✓

Proved ✓
2020

Q. Find $\frac{dy}{dx}$ if $x = a(\theta + \sin\theta)$, $y = a(1 - \cos\theta)$.

Step 1 — Compute dx/dθ1 Mark

$\frac{dx}{d\theta} = a(1 + \cos\theta)$

Step 2 — Compute dy/dθ1 Mark

$\frac{dy}{d\theta} = a \sin\theta$

Step 3 — Divide1 Mark

$\frac{dy}{dx} = \frac{a\sin\theta}{a(1+\cos\theta)} = \frac{2\sin(\theta/2)\cos(\theta/2)}{2\cos^2(\theta/2)} = \tan\frac{\theta}{2}$

$\frac{dy}{dx} = \tan\frac{\theta}{2}$