Exercise 5.3 Class 12 Maths Continuity and Differentiability

NCERT Solutions Class 12 Maths Chapter 5 Ex 5.3 Detailed | LearnCBSEHub

💡 Key Concept: Implicit Differentiation

When $y$ cannot be explicitly expressed in terms of $x$ (e.g., $x^2 + y^2 = r^2$), we differentiate both sides of the equation with respect to $x$.

Remember: $\frac{d}{dx}(y^2) = 2y \cdot \frac{dy}{dx}$ (Chain Rule).

Question 01

Find $\frac{dy}{dx}$ for $2x + 3y = \sin x$

Step 1: Differentiate Both Sides

$$\frac{d}{dx}(2x) + \frac{d}{dx}(3y) = \frac{d}{dx}(\sin x)$$ $$2 + 3\frac{dy}{dx} = \cos x$$

Step 2: Isolate dy/dx

$$3\frac{dy}{dx} = \cos x – 2$$ $$\frac{dy}{dx} = \frac{\cos x – 2}{3}$$

Answer: $\frac{\cos x – 2}{3}$

Question 02

Find $\frac{dy}{dx}$ for $2x + 3y = \sin y$

Step 1: Differentiate w.r.t x

$$\frac{d}{dx}(2x) + \frac{d}{dx}(3y) = \frac{d}{dx}(\sin y)$$ $$2 + 3\frac{dy}{dx} = \cos y \cdot \frac{dy}{dx}$$

Step 2: Group Terms

$$3\frac{dy}{dx} – \cos y \frac{dy}{dx} = -2$$ $$\frac{dy}{dx}(3 – \cos y) = -2$$

Answer: $\frac{-2}{3 – \cos y}$

Question 03

Find $\frac{dy}{dx}$ for $ax + by^2 = \cos y$

Step 1: Differentiate

$$\frac{d}{dx}(ax) + \frac{d}{dx}(by^2) = \frac{d}{dx}(\cos y)$$ $$a + b(2y)\frac{dy}{dx} = -\sin y \cdot \frac{dy}{dx}$$

Step 2: Solve for dy/dx

$$2by\frac{dy}{dx} + \sin y \frac{dy}{dx} = -a$$ $$\frac{dy}{dx}(2by + \sin y) = -a$$

Answer: $\frac{-a}{2by + \sin y}$

Question 04

Find $\frac{dy}{dx}$ for $xy + y^2 = \tan x + y$

Step 1: Product Rule & Chain Rule

$$\frac{d}{dx}(xy) + \frac{d}{dx}(y^2) = \frac{d}{dx}(\tan x) + \frac{d}{dx}(y)$$ $$\left(x\frac{dy}{dx} + y\cdot 1\right) + 2y\frac{dy}{dx} = \sec^2 x + \frac{dy}{dx}$$

Step 2: Collect dy/dx Terms

$$x\frac{dy}{dx} + 2y\frac{dy}{dx} – \frac{dy}{dx} = \sec^2 x – y$$ $$\frac{dy}{dx}(x + 2y – 1) = \sec^2 x – y$$

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

Question 05

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

Step 1: Differentiate

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

Step 2: Rearrange

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

Answer: $-\frac{2x + y}{x + 2y}$

Question 06

Find $\frac{dy}{dx}$ for $x^3 + x^2y + xy^2 + y^3 = 81$

Step 1: Apply Product Rule

$$3x^2 + \left(x^2\frac{dy}{dx} + y(2x)\right) + \left(x(2y)\frac{dy}{dx} + y^2\right) + 3y^2\frac{dy}{dx} = 0$$

Step 2: Group Terms

$$\frac{dy}{dx}(x^2 + 2xy + 3y^2) = -(3x^2 + 2xy + y^2)$$

Answer: $-\frac{3x^2 + 2xy + y^2}{x^2 + 2xy + 3y^2}$

Question 07

Find $\frac{dy}{dx}$ for $\sin^2 y + \cos(xy) = k$

Step 1: Chain Rule

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

Step 2: Solve

$$\frac{dy}{dx}(\sin 2y – x\sin(xy)) = y\sin(xy)$$

Answer: $\frac{y\sin(xy)}{\sin 2y – x\sin(xy)}$

Question 08

Find $\frac{dy}{dx}$ for $\sin^2 x + \cos^2 y = 1$

Step 1: Differentiate

$$2\sin x \cos x + 2\cos y (-\sin y)\frac{dy}{dx} = 0$$ $$\sin 2x – \sin 2y \frac{dy}{dx} = 0$$

Step 2: Isolate

$$\sin 2y \frac{dy}{dx} = \sin 2x$$

Answer: $\frac{\sin 2x}{\sin 2y}$

Question 09

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

Step 1: Substitution

Put $x = \tan \theta \implies \theta = \tan^{-1} x$

$$y = \sin^{-1}\left(\frac{2\tan\theta}{1+\tan^2\theta}\right)$$ $$y = \sin^{-1}(\sin 2\theta) = 2\theta$$ $$y = 2\tan^{-1} x$$

Step 2: Differentiate

$$\frac{dy}{dx} = 2 \cdot \frac{1}{1+x^2}$$

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

Question 10

Find $\frac{dy}{dx}$ for $y = \tan^{-1}\left(\frac{3x – x^3}{1 – 3x^2}\right)$

Step 1: Substitution

Put $x = \tan \theta$

$$y = \tan^{-1}\left(\frac{3\tan\theta – \tan^3\theta}{1 – 3\tan^2\theta}\right)$$ $$y = \tan^{-1}(\tan 3\theta) = 3\theta$$ $$y = 3\tan^{-1} x$$

Step 2: Differentiate

Answer: $\frac{3}{1+x^2}$

Question 11

Find $\frac{dy}{dx}$ for $y = \cos^{-1}\left(\frac{1-x^2}{1+x^2}\right)$

Step 1: Substitution

Put $x = \tan \theta$

$$y = \cos^{-1}\left(\frac{1-\tan^2\theta}{1+\tan^2\theta}\right)$$ $$y = \cos^{-1}(\cos 2\theta) = 2\theta$$ $$y = 2\tan^{-1} x$$

Step 2: Differentiate

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

Question 12

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

Step 1: Substitution

Put $x = \tan \theta$. We know $\frac{1-\tan^2\theta}{1+\tan^2\theta} = \cos 2\theta$.

$$y = \sin^{-1}(\cos 2\theta) = \sin^{-1}\left(\sin\left(\frac{\pi}{2} – 2\theta\right)\right)$$ $$y = \frac{\pi}{2} – 2\theta = \frac{\pi}{2} – 2\tan^{-1} x$$

Step 2: Differentiate

$$\frac{dy}{dx} = 0 – 2 \cdot \frac{1}{1+x^2}$$

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

Question 13

Find $\frac{dy}{dx}$ for $y = \cos^{-1}\left(\frac{2x}{1+x^2}\right)$

Step 1: Substitution

Put $x = \tan \theta$. We know $\frac{2\tan\theta}{1+\tan^2\theta} = \sin 2\theta$.

$$y = \cos^{-1}(\sin 2\theta) = \cos^{-1}\left(\cos\left(\frac{\pi}{2} – 2\theta\right)\right)$$ $$y = \frac{\pi}{2} – 2\theta = \frac{\pi}{2} – 2\tan^{-1} x$$

Step 2: Differentiate

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

Question 14

Find $\frac{dy}{dx}$ for $y = \sin^{-1}(2x\sqrt{1-x^2})$

Step 1: Substitution

Put $x = \sin \theta$. Then $\sqrt{1-x^2} = \cos \theta$.

$$y = \sin^{-1}(2\sin\theta\cos\theta) = \sin^{-1}(\sin 2\theta)$$ $$y = 2\theta = 2\sin^{-1} x$$

Step 2: Differentiate

$$\frac{dy}{dx} = 2 \cdot \frac{1}{\sqrt{1-x^2}}$$

Answer: $\frac{2}{\sqrt{1-x^2}}$

Question 15

Find $\frac{dy}{dx}$ for $y = \sec^{-1}\left(\frac{1}{2x^2 – 1}\right)$

Step 1: Simplify Inverse Function

Use identity $\sec^{-1} x = \cos^{-1}(1/x)$.

$$y = \cos^{-1}(2x^2 – 1)$$

Step 2: Substitution

Put $x = \cos \theta$. We know $2\cos^2\theta – 1 = \cos 2\theta$.

$$y = \cos^{-1}(\cos 2\theta) = 2\theta = 2\cos^{-1} x$$

Step 3: Differentiate

$$\frac{dy}{dx} = 2 \cdot \left(-\frac{1}{\sqrt{1-x^2}}\right)$$

Answer: $-\frac{2}{\sqrt{1-x^2}}$