RD Sharma Class 12 Ex 11.2 Solutions Chapter 11 Differentiation

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Textbook	NCERT
Class	Class 12th
Subject	Maths
Chapter	11
Exercise	11.2
Category	RD Sharma Solutions

Table of Contents

RD Sharma Class 12 Ex 11.2 Solutions Chapter 11 Differentiation

Question 1. Differentiate y = sin (3x + 5) with respect to x.

Solution:

We have,

y = sin (3x + 5)

On differentiating y with respect to x we get,

$\frac{d y}{d x} = \frac{d}{dx}\sin\left( 3x + 5 \right)$

On using chain rule, we have

$\frac{d y}{d x} = \cos\left( 3x + 5 \right)\frac{d}{dx}\left( 3x + 5 \right)$

$\frac{d y}{d x} = \cos\left( 3x + 5 \right) \times 3$

$\frac{d y}{d x} = 3\cos\left( 3x + 5 \right)$

Question 2. Differentiate y = tan² x with respect to x.

Solution:

We have,

y = tan² x

On differentiating y with respect to x we get,

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

On using chain rule, we have

$\frac{d y}{d x} = 2 \tan x\frac{d}{dx}\left( \tan x \right)$

$\frac{d y}{d x} = 2 \tan x \times \sec^2 x$

$\frac{d y}{d x} = 2 \tan x\sec^2 x$

Question 3. Differentiate y = tan (x + 45°) with respect to x.

Solution:

We have,

y = tan (x + 45°)

y = $\tan\left\{ \left( x + 45 \right)\frac{\pi}{180} \right\}$

On differentiating y with respect to x we get,

$\frac{d y}{d x} = \frac{d}{dx}\tan\left\{ \left( x + 45 \right)\frac{\pi}{180} \right\}$

On using chain rule, we have

$\frac{d y}{d x} = \sec^2 \left\{ \left( x + 45 \right)\frac{\pi}{180} \right\} \times \frac{d}{dx}\left( x + 45 \right)\frac{\pi}{180}$

$\frac{d y}{d x} = \frac{\pi}{180} \sec^2 \left( x^\circ + 45^\circ \right)$

Question 4. Differentiate y = sin (log x) with respect to x.

Solution:

We have,

y = sin (log x)

On differentiating y with respect to x we get,

$\frac{d y}{d x} = \frac{d}{dx}\sin\left( \log x \right)$

On using chain rule, we have

$\frac{d y}{d x} = \cos\left( \log x \right)\frac{d}{dx}\left( \log x \right)$

$\frac{d y}{d x} = \frac{1}{x}\cos\left( \log x \right)$

Question 5. Differentiate y = e^{sin √x} with respect to x.

Solution:

We have,

y = e^{sin √x}

On differentiating y with respect to x we get,

$\frac{d y}{d x} = \frac{d}{dx}\left( e^{\sin \sqrt{x}} \right)$

On using chain rule, we have

$\frac{d y}{d x} = e^{\sin \sqrt{x}} \frac{d}{dx}\left( \sin\sqrt{x} \right)$

On using chain rule again, we have

$\frac{d y}{d x} = e^{\sin \sqrt{x}} \times \cos\sqrt{x}\frac{d}{dx}\sqrt{x}$

$\frac{d y}{d x} = e^{\sin \sqrt{x}} \times \cos\sqrt{x} \times \frac{1}{2\sqrt{x}}$

$\frac{d y}{d x} = \frac{\cos\sqrt{x} e^{\sin}\sqrt{x}}{2\sqrt{x}}$

Question 6. Differentiate y = e^{tan x} with respect to x.

Solution:

We have,

y = e^{tan x}

On differentiating y with respect to x we get,

$\frac{d y}{d x} = \frac{d}{dx}\left( e^{\tan x} \right)$

On using chain rule, we have

$\frac{d y}{d x} = e^{\tan x} \frac{d}{dx}\left( \tan x \right)$

$\frac{d y}{d x} = e^{\tan x}\sec^2 x$

Question 7. Differentiate y = sin² (2x + 1) with respect to x.

Solution:

We have,

y = sin² (2x + 1)

On differentiating y with respect to x we get,

$\frac{d y}{d x} = \frac{d}{dx}\left[ \sin^2 \left( 2x + 1 \right) \right]$

On using chain rule, we have

$\frac{d y}{d x} = 2\sin\left( 2x + 1 \right)\frac{d}{dx}\sin\left( 2x + 1 \right)$

On using chain rule again, we have

$\frac{d y}{d x} = 2\sin\left( 2x + 1 \right) \cos\left( 2x + 1 \right) \frac{d}{dx}\left( 2x + 1 \right)$

$\frac{d y}{d x} = 4\sin\left( 2x + 1 \right) \cos\left( 2x + 1 \right)$

As sin 2A = 2 sin A cos A, we get

$\frac{d y}{d x} = 2\sin2\left( 2x + 1 \right)$

$\frac{d y}{d x} = 2 \sin\left( 4x + 2 \right)$

Question 8. Differentiate y = log₇ (2x − 3) with respect to x.

Solution:

We have,

y = log₇ (2x − 3)

As $\log_a b = \frac{\log b}{\log a}$ , we have

y = $\frac{\log\left( 2x - 3 \right)}{\log7}$

On differentiating y with respect to x we get,

$\frac{d y}{d x} = \frac{1}{\log7}\frac{d}{dx}\left\{ \log\left( 2x - 3 \right) \right\}$

On using chain rule, we have

$\frac{d y}{d x} = \frac{1}{\log7} \times \frac{1}{\left( 2x - 3 \right)}\frac{d}{dx}\left( 2x - 3 \right)$

$\frac{d y}{d x} = \frac{2}{\left( 2x - 3 \right)\log7}$

Question 9. Differentiate y = tan 5x° with respect to x.

Solution:

We have,

y = tan 5x°

y = $\tan\left( 5x \times \frac{\pi}{180} \right)$

On differentiating y with respect to x we get,

$\frac{d y}{d x} = \frac{d}{dx}\tan\left( 5x \times \frac{\pi}{180} \right)$

On using chain rule, we have

$\frac{d y}{d x} = \sec^2 \left( 5x \times \frac{\pi}{180} \right)\frac{d}{dx}\left( 5x \times \frac{\pi}{180} \right)$

$\frac{d y}{d x} = \left( \frac{5\pi}{180} \right) \sec^2 \left( 5x \times \frac{\pi}{180} \right)$

$\frac{d y}{d x} = \frac{5\pi}{180} \sec^2 \left( 5x^\circ\right)$

Question 10. Differentiate y = $2^{x^3}$ with respect to x.

Solution:

We have,

y = $2^{x^3}$

On differentiating y with respect to x we get,

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

On using chain rule, we have

$\frac{d y}{d x} = 2^{x^3} \times \log_e 2\frac{d}{dx}\left( x^3 \right)$

$\frac{d y}{d x} = 3 x^2 \times 2^{x^3} \times \log_e 2$

Question 11. Differentiate y = $3^{e^x}$ with respect to x.

Solution:

We have,

y = $3^{e^x}$

On differentiating y with respect to x we get,

$\frac{d y}{d x} = \frac{d}{dx}\left( 3^{e^x} \right)$

On using chain rule, we have

$\frac{d y}{d x} = 3^{e^x} \log3\frac{d}{dx}\left( e^x \right)$

$\frac{d y}{d x} = e^x \times 3^{e^x} \log3$

Question 12. Differentiate y = log_x 3 with respect to x.

Solution:

We have,

y = log_x 3

As $\log_a b = \frac{\log b}{\log a}$ , we get

y = $\frac{\log3}{\log x}$

On differentiating y with respect to x we get,

$\frac{d y}{d x} = \frac{d}{dx}\left( \frac{\log3}{\log x} \right)$

$\frac{d y}{d x} = \log3\frac{d}{dx} \left( \log x \right)^{- 1}$

On using chain rule, we have

$\frac{d y}{d x} = \log3 \times \left[ - 1 \left( \log x \right)^{- 2} \right]\frac{d}{dx}\left( \log x \right)$

$\frac{d y}{d x} = - \frac{\log3}{\left( \log x \right)^2} \times \frac{1}{x}$

$\frac{d y}{d x} = - \left( \frac{\log3}{\log x} \right)^2 \times \frac{1}{x} \times \frac{1}{\log3}$

As $\frac{\log b}{\log a} = \log_a b$ , we get

$\frac{d y}{d x} = - \frac{1}{x\log3 \left( \log_3 x \right)^2}$

Question 13. Differentiate y = $3^{x^2 + 2x}$ with respect to x.

Solution:

We have,

y = $3^{x^2 + 2x}$

On differentiating y with respect to x we get,

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

On using chain rule, we have

$\frac{d y}{d x} = 3^{x^2 + 2x} \times \log_e 3\frac{d}{dx}\left( x^2 + 2x \right)$

$\frac{d y}{d x} = \left( 2x + 2 \right) 3^{x^2 + 2x} \log_e 3$

Question 14. Differentiate y = $\sqrt{\frac{a^2 - x^2}{a^2 + x^2}}$ with respect to x.

Solution:

We have,

y = $\sqrt{\frac{a^2 - x^2}{a^2 + x^2}}$

On differentiating y with respect to x we get,

$\frac{d y}{d x} = \frac{d}{dx}\left( \sqrt{\frac{a^2 - x^2}{a^2 + x^2}} \right)$

On using chain rule, we have

$\frac{d y}{d x} = \frac{1}{2} \left( \frac{a^2 - x^2}{a^2 + x^2} \right)^{\frac{1}{2} - 1} \times \frac{d}{dx}\left( \frac{a^2 - x^2}{a^2 + x^2} \right)$

$\frac{d y}{d x} = \frac{1}{2} \left( \frac{a^2 - x^2}{a^2 + x^2} \right)^\frac{- 1}{2} \times \left\{ \frac{\left( a^2 + x^2 \right)\frac{d}{dx}\left( a^2 - x^2 \right) - \left( a^2 - x^2 \right)\frac{d}{dx}\left( a^2 + x^2 \right)}{\left( a^2 + x^2 \right)^2} \right\}$

$\frac{d y}{d x} = \frac{1}{2} \left( \frac{a^2 + x^2}{a^2 - x^2} \right)^\frac{1}{2} \left\{ \frac{- 2x\left( a^2 + x^2 \right) - 2x\left( a^2 - x^2 \right)}{\left( a^2 + x^2 \right)^2} \right\}$

$\frac{d y}{d x} = \frac{1}{2} \left( \frac{a^2 + x^2}{a^2 - x^2} \right)^\frac{1}{2} \left\{ \frac{- 2x a^2 - 2 x^3 - 2x a^2 + 2 x^3}{\left( a^2 + x^2 \right)^2} \right\}$

$\frac{d y}{d x} = \frac{1}{2} \left( \frac{a^2 + x^2}{a^2 - x^2} \right)^\frac{1}{2} \left\{ \frac{- 4x a^2}{\left( a^2 + x^2 \right)^2} \right\}$

$\frac{d y}{d x} = \frac{- 2x a^2}{\sqrt{a^2 - x^2} \left( a^2 + x^2 \right)^\frac{3}{2}}$

Question 15. Differentiate y = $3^{x \log x}$ with respect to x.

Solution:

We have,

y = $3^{x \log x}$

On differentiating y with respect to x we get,

$\frac{d y}{d x} = \frac{d}{dx}\left( 3^{x \log x} \right)$

On using chain rule, we have

$\frac{d y}{d x} = 3^{x \log x} \times \log_e 3\frac{d}{dx}\left( x \log x \right)$

$\frac{d y}{d x} = 3^x \log x \times \log_e 3\left[ x\frac{d}{dx}\left( \log x \right) + \log x\frac{d}{dx}\left( x \right) \right]$

$\frac{d y}{d x} = 3^{x \log x} \times \log_e 3\left[ \frac{x}{x} + \log x \right]$

$\frac{d y}{d x} = 3^{x \log x} \left( 1 + \log x \right) \times \log_e 3$

$\frac{d y}{d x} = 3^{x \log x} \left( 1 + \log x \right)\log_e 3$

Question 16. Differentiate y = $\sqrt{\frac{1 + \sin x}{1 - \sin x}}$ with respect to x.

Solution:

We have,

y = $\sqrt{\frac{1 + \sin x}{1 - \sin x}}$

On differentiating y with respect to x we get,

$\frac{d y}{d x} = \frac{d}{dx} \left( \sqrt{\frac{1 + \sin x}{1 - \sin x}} \right)$

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

On using chain rule, we have

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

$\frac{d y}{d x} = \frac{1}{2} \left( \frac{1 - \sin x}{1 + \sin x} \right)^\frac{1}{2} \left[ \frac{\left( 1 - \sin x \right)\left( \cos x \right) - \left( 1 + \sin x \right)\left( - \cos x \right)}{\left( 1 - \sin x \right)^2} \right]$

$\frac{d y}{d x} = \frac{1}{2}\frac{\left( 1 - \sin x \right)^\frac{1}{2}}{\left( 1 + \sin x \right)^\frac{1}{2}}\left[ \frac{\cos x - \cos x \sin x + \cos x + \sin x \cos x}{\left( 1 - \sin x \right)^2} \right]$

$\frac{d y}{d x} = \frac{1}{2} \times \frac{2\cos x}{\sqrt{1 + \sin x}\left( 1 - \sin x \right)\frac{3}{2}}$

$\frac{d y}{d x} = \frac{\cos x}{\sqrt{1 + \sin x}\left( 1 - \sin x \right)\frac{3}{2}}$

$\frac{d y}{d x} = \frac{\cos x}{\sqrt{1 + \sin x}\sqrt{1 - \sin x}\left( 1 - \sin x \right)}$

$\frac{d y}{d x} = \frac{\cos x}{\sqrt{1 - \sin^2 x} \times \left( 1 - \sin x \right)}$

$\frac{d y}{d x} = \frac{\cos x}{\cos x\left( 1 - \sin x \right)}$

$\frac{d y}{d x} = \frac{1}{\left( 1 - \sin x \right)} \times \frac{\left( 1 + \sin x \right)}{\left( 1 + \sin x \right)}$

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

$\frac{d y}{d x} = \frac{1 + \sin x}{\cos^2 x}$

$\frac{d y}{d x} = \frac{1}{\cos x}\left( \frac{1}{\cos x} + \frac{\sin x}{\cos x} \right)$

$\frac{d y}{d x} = \sec x\left( \sec x + \tan x \right)$

Question 17. Differentiate y = $\sqrt{\frac{1 - x^2}{1 + x^2}}$ with respect to x.

Solution:

We have,

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

On differentiating y with respect to x we get,

$\frac{d y}{d x} = \frac{d}{dx} \left( \sqrt{\frac{1 - x^2}{1 + x^2}} \right)$

$\frac{d y}{d x} = \frac{d}{dx} \left( \frac{1 - x^2}{1 + x^2} \right)^\frac{1}{2}$

On using chain rule, we have

$\frac{d y}{d x} = \frac{1}{2} \left( \frac{1 - x^2}{1 + x^2} \right)^{\frac{1}{2} - 1} \times \frac{d}{dx}\left( \frac{1 - x^2}{1 + x^2} \right)$

$\frac{d y}{d x} = \frac{1}{2} \left( \frac{1 - x^2}{1 + x^2} \right)^\frac{- 1}{2} \times \left\{ \frac{\left( 1 + x^2 \right)\frac{d}{dx}\left( 1 - x^2 \right) - \left( 1 - x^2 \right)\frac{d}{dx}\left( 1 + x^2 \right)}{\left( 1 + x^2 \right)^2} \right\}$

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

$\frac{d y}{d x} = \frac{1}{2} \left( \frac{1 + x^2}{1 - x^2} \right)^\frac{1}{2} \left\{ \frac{- 2x - 2 x^3 - 2x + 2 x^3}{\left( 1 + x^2 \right)^2} \right\}$

$\frac{d y}{d x} = \frac{1}{2} \left( \frac{1 + x^2}{1 - x^2} \right)^\frac{1}{2} \left\{ \frac{- 4x}{\left( 1 + x^2 \right)^2} \right\}$

$\frac{d y}{d x} = \frac{- 2x}{\sqrt{1 - x^2} \left( 1 + x^2 \right)^\frac{3}{2}}$

Question 18. Differentiate y = (log sin x)² with respect to x.

Solution:

We have,

y = (log sin x)²

On differentiating y with respect to x we get,

$\frac{d y}{d x} = \frac{d}{dx} \left( \log \sin x \right)^2$

On using chain rule, we have

$\frac{d y}{d x} = 2\left( \log \sin x \right)\frac{d}{dx}\left( \log \sin x \right)$

$\frac{d y}{d x} = 2\left( \log \sin x \right) \times \frac{1}{\sin x}\frac{d}{dx}\left( \sin x \right)$

$\frac{d y}{d x} = 2\left( \log \sin x \right) \times \frac{1}{\sin x} \times \cos x$

$\frac{d y}{d x} = 2\left( \log \sin x \right)\cot x$

Question 19. Differentiate y = $\sqrt{\frac{1 + x}{1 - x}}$ with respect to x.

Solution:

We have,

y = $\sqrt{\frac{1 + x}{1 - x}}$

On differentiating y with respect to x we get,

$\frac{d y}{d x} = \frac{d}{dx} \left( \sqrt{\frac{1 + x}{1 - x}} \right)$

$\frac{d y}{d x} = \frac{d}{dx} \left( \frac{1 + x}{1 - x} \right)^\frac{1}{2}$

On using chain rule, we have

$\frac{d y}{d x} = \frac{1}{2} \left( \frac{1 + x}{1 - x} \right)^{\frac{1}{2} - 1} \times \frac{d}{dx}\left( \frac{1 + x}{1 - x} \right)$

On using quotient rule, we have

$\frac{d y}{d x} = \frac{1}{2} \left( \frac{1 + x}{1 - x} \right)^\frac{- 1}{2} \times \left\{ \frac{\left( 1 - x \right)\frac{d}{dx}\left( 1 + x \right) - \left( 1 + x \right)\frac{d}{dx}\left( 1 - x \right)}{\left( 1 - x \right)^2} \right\}$

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

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

$\frac{d y}{d x} = \frac{1}{2}\frac{\left( 1 - x \right)^\frac{1}{2}}{\left( 1 + x \right)^\frac{1}{2}} \times \frac{2}{\left( 1 - x \right)^2}$

$\frac{d y}{d x} = \frac{1}{\sqrt{1 + x} \left( 1 - x \right)^\frac{3}{2}}$

Question 20. Differentiate y = $\sin \left( \frac{1 + x^2}{1 - x^2} \right)$ with respect to x.

Solution:

We have,

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

On differentiating y with respect to x we get,

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

On using chain rule, we have

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

On using quotient rule, we have

$\frac{d y}{d x} = \cos x\left( \frac{1 + x^2}{1 - x^2} \right)\left[ \frac{\left( 1 - x^2 \right)\frac{d}{dx}\left( 1 + x^2 \right) - \left( 1 + x^2 \right)\frac{d}{dx}\left( 1 - x^2 \right)}{\left( 1 - x^2 \right)^2} \right]$

$\frac{d y}{d x} = \cos x\left( \frac{1 + x^2}{1 - x^2} \right)\left[ \frac{\left( 1 - x^2 \right)\left( 2x \right) - \left( 1 + x^2 \right)\left( - 2x \right)}{\left( 1 - x^2 \right)^2} \right]$

$\frac{d y}{d x} = \cos x\left( \frac{1 + x^2}{1 - x^2} \right)\left[ \frac{2x - 2 x^3 + 2x + 2 x^3}{\left( 1 - x^2 \right)^2} \right]$

$\frac{d y}{d x} = \frac{4x}{\left( 1 - x^2 \right)^2}\cos x\left( \frac{1 + x^2}{1 - x^2} \right)$

Question 21. Differentiate y = $e^{3x} \cos2x$ with respect to x.

Solution:

We have,

y = $e^{3x} \cos2x$

On differentiating y with respect to x we get,

$\frac{d y}{d x} = \frac{d}{dx} \left( e^{3x} \cos2x\right)$

On using product rule, we have

$\frac{d y}{d x} = e^{3x} \times \frac{d}{dx}\left( \cos2x \right) + \cos2x\frac{d}{dx}\left( e^{3x} \right)$

On using chain rule, we have

$\frac{d y}{d x} = e^{3x} \times \left( - \sin2x \right)\frac{d}{dx}\left( 2x \right) + \cos2x e^{3x} \frac{d}{dx}\left( 3x \right)$

$\frac{d y}{d x} = - 2 e^{3x} \sin2x + 3 e^{3x} \cos2x$

$\frac{d y}{d x} = e^{3x} \left( 3 \cos2x - 2 \sin2x \right)$

Question 22. Differentiate y = sin(log sin x) with respect to x.

Solution:

We have,

y = sin(log sin x)

On differentiating y with respect to x we get,

$\frac{d y}{d x} = \frac{d}{dx} \left(sin(log sin x)\right)$

On using chain rule, we have

$\frac{d y}{d x}=\cos(\log \sin x)\frac{d}{dx}(\log \sin x)$

On using chain rule again, we have

$\frac{d y}{d x}=\cos (\log \sin x)\frac{1}{sin x}\frac{d}{dx}(\sin x)$

$\frac{d y}{d x}=\cos (\log \sin x)\frac{cos x}{sin x}$

$\frac{d y}{d x}=\cos (\log \sin x) \cot x$

Question 23. Differentiate y = e^{tan 3x} with respect to x.

Solution:

We have,

y = e^{tan 3x}

On differentiating y with respect to x we get,

$\frac{d y}{d x} = \frac{d}{dx}\left(e^{\tan3 x} \right)$

On using chain rule, we have

$\frac{d y}{d x} = e^{\tan3x} \frac{d}{dx}\left( \tan3x \right)$

$\frac{d y}{d x} = e^{\tan3x} \sec^2 3x \times \frac{d}{dx}\left( 3x \right)$

$\frac{d y}{d x} = e^{\tan3x} \sec^2 3x \times 3$

$\frac{d y}{d x} = 3e^{\tan3x} \sec^2 3x$

Question 24. Differentiate y = $e^{\sqrt{\cot x}}$ with respect to x.

Solution:

We have,

y = $e^{\sqrt{\cot x}}$

On differentiating y with respect to x we get,

$\frac{d y}{d x} = \frac{d}{dx}\left(e^{\sqrt{\cot x}} \right)$

$\frac{d y}{d x} = \frac{d}{dx}\left( e^{\left( \cot x \right)^\frac{1}{2} }\right)$

On using chain rule, we have

$\frac{d y}{d x} = e^{\left( \cot x \right)^\frac{1}{2}} \times \frac{d}{dx} \left( \cot x \right)^\frac{1}{2}$

$\frac{d y}{d x} = e^{\sqrt{\cot x}} \times \frac{1}{2} \left( \cot x \right)^{\frac{1}{2} - 1} \frac{d}{dx}\left( \cot x \right)$

$\frac{d y}{d x} = - \frac{e^{\sqrt{\cot x}}{cosec}^2 x}{2\sqrt{\cot x}}$

Question 25. Differentiate y = $\log \left( \frac{\sin x}{1 + \cos x} \right)$ with respect to x.

Solution:

We have,

y = $\log \left( \frac{\sin x}{1 + \cos x} \right)$

On differentiating y with respect to x we get,

$\frac{d y}{d x} = \frac{d}{dx} \left( \log \left( \frac{\sin x}{1 + \cos x} \right) \right)$

On using chain rule, we have

$\frac{d y}{d x} = \frac{1}{\left( \frac{\sin x}{1 + \cos x} \right)} \times \frac{d}{dx}\left( \frac{\sin x}{1 + \cos x} \right)$

On using quotient rule, we have

$\frac{d y}{d x} = \left( \frac{1 + \cos x}{\sin x} \right)\left[ \frac{\left( 1 + \cos x \right)\frac{d}{dx}\left( \sin x \right) - \sin x\frac{d}{dx}\left( 1 + \cos x \right)}{\left( 1 + \cos x \right)^2} \right]$

$\frac{d y}{d x} = \left( \frac{1 + \cos x}{\sin x} \right)\left[ \frac{\left( 1 + \cos x \right)\left( \cos x \right) - \sin x\left( - \sin x \right)}{\left( 1 + \cos x \right)^2} \right]$

$\frac{d y}{d x} = \left( \frac{1 + \cos x}{\sin x} \right)\left[ \frac{\cos x + \cos^2 x + \sin^2 x}{\left( 1 + \cos x \right)^2} \right]$

$\frac{d y}{d x} = \left( \frac{1 + \cos x}{\sin x} \right)\left[ \frac{\left( 1 + \cos x \right)}{\left( 1 + \cos x \right)^2} \right]$

$\frac{d y}{d x} = \frac{1}{\sin x}$

$\frac{d y}{d x} = \cosec x$

Question 26. Differentiate y = $\log\sqrt{\frac{1 - \cos x}{1 + \cos x}}$ with respect to x.

Solution:

We have,

y = $\log\sqrt{\frac{1 - \cos x}{1 + \cos x}}$

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

As $\log a^b = b\log a$ , we get

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

On differentiating y with respect to x we get,

$\frac{d y}{d x} = \frac{d}{dx} \left( \log\sqrt{\frac{1 - \cos x}{1 + \cos x}} \right)$

$\frac{d y}{d x} = \frac{1}{\left( \frac{\sin x}{1 + \cos x} \right)} \times \frac{d}{dx}\left( \frac{\sin x}{1 + \cos x} \right)$

$\frac{d y}{d x} = \frac{d}{dx}\left\{ \frac{1}{2}\log\left( \frac{1 - \cos x}{1 + \cos x} \right) \right\}$

On using chain rule, we have

$\frac{d y}{d x} = \frac{1}{2} \times \frac{1}{\left( \frac{1 - \cos x}{1 + \cos x} \right)} \times \frac{d}{dx}\left( \frac{1 - \cos x}{1 + \cos x} \right)$

On using quotient rule, we have

$\frac{d y}{d x} = \frac{1}{2}\left( \frac{1 + \cos x}{1 - \cos x} \right)\left[ \frac{\left( 1 + \cos x \right)\frac{d}{dx}\left( 1 - \cos x \right) - \left( 1 - \cos x \right)\frac{d}{dx}\left( 1 + \cos x \right)}{\left( 1 + cos x \right)^2} \right]$

$\frac{d y}{d x} = \frac{1}{2}\left( \frac{1 + \cos x}{1 - \cos x} \right)\left[ \frac{\left( 1 + \cos x \right)\left( \sin x \right) - \left( 1 - \cos x \right)\left( - \sin x \right)}{\left( 1 + \cos x \right)^2} \right]$

$\frac{d y}{d x} = \frac{1}{2}\left( \frac{1 + \cos x}{1 - \cos x} \right)\left[ \frac{\sin x + \sin x \cos x + \sin x - \sin x \cos x}{\left( 1 + \cos x \right)^2} \right]$

$\frac{d y}{d x} = \frac{1}{2}\left( \frac{1 + \cos x}{1 - \cos x} \right)\left[ \frac{2\sin x}{\left( 1 + \cos x \right)^2} \right]$

$\frac{d y}{d x} = \frac{\sin x}{\left( 1 - \cos x \right)\left( 1 + \cos x \right)}$

$\frac{d y}{d x} = \frac{\sin x}{1 - \cos^2 x}$

$\frac{d y}{d x} = \frac{\sin x}{\sin^2 x}$

$\frac{d y}{d x} = \frac{1}{\sin x}$

$\frac{d y}{d x} = \cosec x$

Question 27. Differentiate y = $\tan \left( e^{\sin x }\right)$ with respect to x.

Solution:

We have,

y = $\tan \left( e^{\sin x }\right)$

On differentiating y with respect to x we get,

$\frac{d y}{d x} = \frac{d}{dx}\left[ \tan\left( e^{\sin x} \right) \right]$

On using chain rule, we have

$\frac{d y}{d x} = \sec^2 \left( e^{\sin x} \right)\frac{d}{dx}\left( e^{\sin x } \right)$

$\frac{d y}{d x} = \sec^2 \left( e^{\sin x} \right) \times e^{\sin x } \times \frac{d}{dx}\left( {\sin x} \right)$

$\frac{d y}{d x} = \cos x \sec^2 \left( e^{\sin x} \right) \times e^{\sin x}$

Question 28. Differentiate y = $\log\left( x + \sqrt{x^2 + 1} \right)$ with respect to x.

Solution:

We have,

I = $\log\left( x + \sqrt{x^2 + 1} \right)$

On differentiating y with respect to x we get,

$\frac{d y}{d x} = \frac{d}{dx}\log\left( x + \sqrt{x^2 + 1} \right)$

On using chain rule, we have

$\frac{d y}{d x} = \frac{1}{x + \sqrt{x^2 + 1}}\frac{d}{dx}\left( x + \left( x^2 + 1 \right)^\frac{1}{2} \right)$

$\frac{d y}{d x} = \frac{1}{x + \sqrt{x^2 + 1}}\left[ 1 + \frac{1}{2} \left( x^2 + 1 \right)^{\frac{1}{2} - 1} \frac{d}{dx}\left( x^2 + 1 \right) \right]$

$\frac{d y}{d x} = \frac{1}{x + \sqrt{x^2 + 1}}\left[ 1 + \frac{1}{2\sqrt{x^2 + 1}} \times 2x \right]$

$\frac{d y}{d x} = \frac{1}{x + \sqrt{x^2 + 1}}\left[ \frac{\sqrt{x^2 + 1} + x}{\sqrt{x^2 + 1}} \right]$

$\frac{d y}{d x} = \frac{1}{\sqrt{x^2 + 1}}$

Question 29. Differentiate y = $\frac{e^x \log x}{x^2}$ with respect to x.

Solution:

We have,

y = $\frac{e^x \log x}{x^2}$

On differentiating y with respect to x we get,

$\frac{d y}{d x} = \frac{x^2 \frac{d}{dx}\left( e^x \log x \right) - \left( e^x \log x \right)\frac{d}{dx} x^2}{\left( x^2 \right)^2}$

On using quotient rule, we have

$\frac{d y}{d x} = \frac{x^2 \left\{ e^x \frac{d}{dx}\left( \log x \right) + \log x\frac{d}{dx}\left( e^x \right) \right\} - e^x \log x \times 2x}{x^4}$

On using product rule, we have

$\frac{d y}{d x} = \frac{x^2 \left[ \frac{e^x}{x} + e^x \log x \right] - 2x e^x \log x}{x^4}$

$\frac{d y}{d x} = \frac{\frac{x^2 e^x \left( 1 + x\log x \right)}{x} - 2x e^x \log x}{x^4}$

$\frac{d y}{d x} = \frac{x e^x \left[ 1 + x\log x - 2\log x \right]}{x^4}$

$\frac{d y}{d x} = \frac{x e^x}{x^3}\left[ \frac{1}{x} + \frac{x \log x}{x} - \frac{2\log x}{x} \right]$

$\frac{d y}{d x} = e^x x^{- 2} \left[ \frac{1}{x} + \log x - \frac{2}{x}\log x \right]$

Question 30. Differentiate y = $\log \left( cosec x - \cot x \right)$ with respect to x.

Solution:

We have,

y = $\log \left( cosec x - \cot x \right)$

On differentiating y with respect to x we get,

$\frac{d y}{d x} = \frac{d}{dx}\log \left( cosec x - \cot x \right)$

On using chain rule, we have

$\frac{d y}{d x} = \frac{1}{\left( cosec x - \cot x \right)}\frac{d}{dx}\left( cosec x - \cot x \right)$

$\frac{d y}{d x} = \frac{1}{\left( cosec x - \cot x \right)} \times \left( - cosec x \cot x + {cosec}^2 x \right)$

$\frac{d y}{d x} = \frac{ cosec x\left( cosec x - \cot x \right) }{\left( cosec x - \cot x \right)}$

$\frac{d y}{d x} = \cosec x$

Question 31. Differentiate y = $\frac{e^{2x} + e^{- 2x}}{e^{2x} - e^{- 2x}}$ with respect to x.

Solution:

We have,

y = $\frac{e^{2x} + e^{- 2x}}{e^{2x} - e^{- 2x}}$

On differentiating y with respect to x we get,

$\frac{d y}{d x} = \frac{d}{dx}\left[ \frac{e^{2x} + e^{- 2x}}{e^{2x} - e^{- 2x}} \right]$

On using quotient rule and chain rule, we get

$\frac{d y}{d x} = \left[ \frac{\left( e^{2x} - e^{- 2x} \right)\frac{d}{dx}\left( e^{2x} + e^{- 2x} \right) - \left( e^{2x} + e^{- 2x} \right)\frac{d}{dx}\left( e^{2x} - e^{- 2x} \right)}{\left( e^{2x} - e^{- 2x} \right)^2} \right]$

$\frac{d y}{d x} = \frac{\left( e^{2x} - e^{- 2x} \right)\left[ e^{2x} \frac{d}{dx}\left( 2x \right) + e^{- 2x} \frac{d}{dx}\left( - 2x \right) \right] - \left( e^{2x} + e^{- 2x} \right)\left[ e^{2x} \frac{d}{dx}\left( 2x \right) - e^{- 2x} \frac{d}{dx}\left( - 2x \right) \right]}{\left( e^{2x} - e^{- 2x} \right)^2}$

$\frac{d y}{d x} = \frac{\left( e^{2x} - e^{- 2x} \right)\left( 2 e^{2x} - 2 e^{- 2x} \right) - \left( e^{2x} + e^{- 2x} \right)\left( 2 e^{2x} + 2 e^{- 2x} \right)}{\left( e^{2x} - e^{- 2x} \right)^2}$

$\frac{d y}{d x} = \frac{2 \left( e^{2x} - e^{- 2x} \right)^2 - 2 \left( e^{2x} + e^{- 2x} \right)^2}{\left( e^{2x} - e^{- 2x} \right)^2}$

$\frac{d y}{d x} = \frac{2\left[ e^{4x} + e^{- 4x} - 2 e^{2x} e^{- 2x} - e^{4x} - e^{- 4x} - 2 e^{2x} e^{- 2x} \right]}{\left( e^{2x} - e^{- 2x} \right)^2}$

$\frac{d y}{d x} = \frac{- 8}{\left( e^{2x} - e^{- 2x} \right)^2}$

Question 32. Differentiate y = $\log\left( \frac{x^2 + x + 1}{x^2 - x + 1} \right)$ with respect to x.

Solution:

We have,

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

On differentiating y with respect to x we get,

$\frac{d y}{d x} = \frac{d}{dx}\left[ \log\left( \frac{x^2 + x + 1}{x^2 - x + 1} \right) \right]$

$\frac{d y}{d x} = \frac{1}{\left( \frac{x^2 + x + 1}{x^2 - x + 1} \right)}\frac{d}{dx}\left( \frac{x^2 + x + 1}{x^2 - x + 1} \right)$

On using quotient rule and chain rule, we get

$\frac{d y}{d x} = \left( \frac{x^2 - x + 1}{x^2 + x + 1} \right)\left[ \frac{\left( x^2 - x + 1 \right)\frac{d}{dx}\left( x^2 + x + 1 \right) - \left( x^2 + x + 1 \right)\frac{d}{dx}\left( x^2 - x + 1 \right)}{\left( x^2 - x + 1 \right)^2} \right]$

$\frac{d y}{d x} = \left( \frac{x^2 - x + 1}{x^2 + x + 1} \right)\left[ \frac{\left( x^2 - x + 1 \right)\left( 2x + 1 \right) - \left( x^2 + x + 1 \right)\left( 2x - 1 \right)}{\left( x^2 - x + 1 \right)^2} \right]$

$\frac{d y}{d x} = \left( \frac{x^2 - x + 1}{x^2 + x + 1} \right)\left[ \frac{2 x^3 - 2 x^2 + 2x + x^2 - x + 1 - 2 x^3 - 2 x^2 - 2x + x^2 + x + 1}{\left( x^2 - x + 1 \right)^2} \right]$

$\frac{d y}{d x} = \frac{- 4 x^2 + 2 x^2 + 2}{\left( x^2 + x + 1 \right)\left( x^2 - x + 1 \right)}$

$\frac{d y}{d x} = \frac{- 4 x^2 + 2 x^2 + 2}{\left( x^2 + 1 \right)^2 - \left( x \right)^2}$

$\frac{d y}{d x} = \frac{- 2\left( x^2 - 1 \right)}{x^4 + 1 + 2 x^2 - x^2}$

$\frac{d y}{d x} = \frac{- 2\left( x^2 - 1 \right)}{x^4 + x^2 + 1}$

Question 33. Differentiate y = $\tan^{- 1} \left( e^x \right)$ with respect to x.

Solution:

We have,

y = $\tan^{- 1} \left( e^x \right)$

On differentiating y with respect to x we get,

$\frac{d y}{d x} = \frac{d}{dx}\left[ \tan^{- 1} \left( e^x \right) \right]$

On using chain rule, we have

$\frac{d y}{d x} = \frac{1}{1 + \left( e^x \right)^2}\frac{d}{dx}\left( e^x \right)$

$\frac{d y}{d x} = \frac{1}{1 + e^{2x}} \times e^x$

$\frac{d y}{d x} = \frac{e^x}{1 + e^{2x}}$

Question 34. Differentiate y = $e^{\sin^{- 1} 2x}$ with respect to x.

Solution:

We have,

y = $e^{\sin^{- 1} 2x}$

On differentiating y with respect to x we get,

$\frac{d y}{d x} = \frac{d}{dx}\left( e^{\sin^{- 1} 2x} \right)$

On using chain rule, we have

$\frac{d y}{d x} = e^{\sin^{- 1} 2x} \times \frac{d}{dx}\left( \sin^{- 1} 2x \right)$

$\frac{d y}{d x} = e^{\sin^{- 1} 2x} \times \frac{1}{\sqrt{1 - \left( 2x \right)^2}}\frac{d}{dx}\left( 2x \right)$

$\frac{d y}{d x} = \frac{2 e^{\sin^{- 1} 2x}}{\sqrt{1 - 4 x^2}}$

Question 35. Differentiate y = $\sin \left( 2 \sin^{- 1} x \right)$ with respect to x.

Solution:

We have,

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

On differentiating y with respect to x we get,

$\frac{d y}{d x} = \frac{d}{dx}\left[ \sin\left( 2 \sin^{- 1} x \right) \right]$

On using chain rule, we have

$\frac{d y}{d x} = \cos\left( 2 \sin^{- 1} x \right)\frac{d}{dx}\left( 2 \sin^{- 1} x \right)$

$\frac{d y}{d x} = \cos\left( 2 \sin^{- 1} x \right) \times 2\frac{1}{\sqrt{1 - x^2}}$

$\frac{d y}{d x} = \frac{2\cos\left( 2 \sin^{- 1} x \right)}{\sqrt{1 - x^2}}$

Question 36. Differentiate y = $e^{\tan^{- 1} \sqrt{x}}$ with respect to x.

Solution:

We have,

y = $e^{\tan^{- 1} \sqrt{x}}$

On differentiating y with respect to x we get,

$\frac{d y}{d x} = \frac{d}{dx}\left( e^{\tan^{- 1}} \sqrt{x} \right)$

On using chain rule, we have

$\frac{d y}{d x} = e^{{\tan^{- 1}} \sqrt{x}} \frac{d}{dx}\left( \tan^{- 1} \sqrt{x} \right)$

$\frac{d y}{d x} = e^{{\tan^{- 1}} \sqrt{x}} \times \frac{1}{1 + \left( \sqrt{x} \right)^2}\frac{d}{dx}\left( \sqrt{x} \right)$

$\frac{d y}{d x} = \frac{e^{{\tan^{- 1}} \sqrt{x}}}{1 + x} \times \frac{1}{2\sqrt{x}}$

$\frac{d y}{d x} = \frac{e^{{\tan^{- 1}} \sqrt{x}}}{2\sqrt{x}\left( 1 + x \right)}$

Question 37. Differentiate y = $\sqrt{\tan^{- 1} \left( \frac{x}{2} \right)}$ with respect to x.

Solution:

We have,

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

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

On differentiating y with respect to x we get,

$\frac{d y}{d x} = \frac{d}{dx} \left\{ \tan^{- 1} \left( \frac{x}{2} \right) \right\}^\frac{1}{2}$

On using chain rule, we have

$\frac{d y}{d x} = \frac{1}{2} \left\{ \tan^{- 1} \left( \frac{x}{2} \right) \right\}^{\frac{1}{2} - 1} \frac{d}{dx}\left( \tan^{- 1} \frac{x}{2} \right)$

$\frac{d y}{d x} = \frac{1}{2} \left\{ \tan^{- 1} \left( \frac{x}{2} \right) \right\}^\frac{- 1}{2} \times \frac{1}{1 + \left( \frac{x}{2} \right)^2} \times \frac{d}{dx}\left( \frac{x}{2} \right)$

$\frac{d y}{d x} = \frac{4}{4\left( 4 + x^2 \right)\sqrt{\tan^{- 1} \left( \frac{x}{2} \right)}}$

$\frac{d y}{d x} = \frac{1}{\left( 4 + x^2 \right)\sqrt{\tan^{- 1} \left( \frac{x}{2} \right)}}$

Question 38. Differentiate y = $\log\left( \tan^{- 1} x \right)$ with respect to x.

Solution:

We have,

y = $\log\left( \tan^{- 1} x \right)$

On differentiating y with respect to x we get,

$\frac{d y}{d x} = \frac{d}{dx}\log\left( \tan^{- 1} x \right)$

On using chain rule, we have

$\frac{d y}{d x} = \frac{1}{\tan^{- 1} x} \times \frac{d}{dx}\left( \tan^{- 1} x \right)$

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

Question 39. Differentiate y = $\frac{2^x \cos x}{\left( x^2 + 3 \right)^2}$ with respect to x.

Solution:

We have,

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

On differentiating y with respect to x we get,

$\frac{d y}{d x} = \frac{d}{dx}\left[ \frac{2^x \cos x}{\left( x^2 + 3 \right)^2} \right]$

On using quotient rule, we have

$\frac{d y}{d x} = \left[ \frac{\left( x^2 + 3 \right)^2 \frac{d}{dx}\left( 2^x \cos x \right) - \left( 2^x \cos x \right)\frac{d}{dx} \left( x^2 + 3 \right)^2}{\left[ \left( x^2 + 3 \right)^2 \right]^2} \right]$

On using product rule and chain rule, we have

$\frac{d y}{d x} = \left[ \frac{\left( x^2 + 3 \right)^2 \left\{ 2^x \frac{d}{dx}\cos x + \cos x\frac{d}{dx} 2^x \right\} - \left( 2^x \cos x \right)2\left( x^2 + 3 \right)\frac{d}{dx}\left( x^2 + 3 \right)}{\left( x^2 + 3 \right)^4} \right]$

$\frac{d y}{d x} = \left[ \frac{\left( x^2 + 3 \right)^2 \left\{ - 2^x \sin x + \cos x 2^x \log_e 2 \right\} - 2\left( 2^x \cos x \right)\left( x^2 + 3 \right)\left( 2x \right)}{\left( x^2 + 3 \right)^4} \right]$

$\frac{d y}{d x} = \left[ \frac{2^x \left( x^2 + 3 \right)\left\{ \left( x^2 + 3 \right)\left( \cos x \log_e 2 - \sin x \right) - 4x \cos x \right\}}{\left( x^2 + 3 \right)^4} \right]$

$\frac{d y}{d x} = \frac{2^x}{\left( x^2 + 3 \right)^2}\left[ \cos x \log_e 2 - \sin x - \frac{4x \cos x}{\left( x^2 + 3 \right)} \right]$

Question 40. Differentiate y = $x \sin 2x + 5^x + k^k + \left( \tan^2 x \right)^3$ with respect to x.

Solution:

We have,

y = $x \sin 2x + 5^x + k^k + \left( \tan^2 x \right)^3$

On differentiating y with respect to x we get,

$\frac{d y}{d x} = \frac{d}{dx}\left[ x \sin2x + 5^x + k^k + \left( \tan^6 x \right) \right]$

$\frac{d y}{d x} = \frac{d}{dx}\left( x \sin2x \right) + \frac{d}{dx}\left( 5^x \right) + \frac{d}{dx}\left( k^k \right) + \frac{d}{dx}\left( \tan^6 x \right)$

On using product rule and chain rule, we have

$\frac{d y}{d x} = \left[ x\frac{d}{dx}\left( \sin2x \right) + \sin2x\frac{d}{dx}\left( x \right) \right] + 5^x \log_e 5 + 0 + 6 \tan^5 x \times \frac{d}{dx}\left( \tan x \right)$

$\frac{d y}{d x} = \left[ x \cos2x\frac{d}{dx}\left( 2x \right) + \sin2x \right] + 5^x \log_e 5 + 6 \tan^5 x \sec^2 x$

$\frac{d y}{d x} = 2x \cos2x + \sin2x + 5^x \log_e 5 + 6 \tan^5 x \sec^2 x$

Question 41. Differentiate y = $\log \left( 3x + 2 \right) - x^2 \log \left( 2x - 1 \right)$ with respect to x.

Solution:

We have,

y = $\log \left( 3x + 2 \right) - x^2 \log \left( 2x - 1 \right)$

On differentiating y with respect to x we get,

$\frac{d y}{d x} = \frac{d}{dx}\left[ \log\left( 3x + 2 \right) - x^2 \log\left( 2x - 1 \right) \right]$

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

On using product rule and chain rule, we have

$\frac{d y}{d x} = \frac{1}{\left( 3x + 2 \right)}\frac{d}{dx}\left( 3x + 2 \right) - \left[ x^2 \frac{d}{dx}\log\left( 2x - 1 \right) + \log\left( 2x - 1 \right)\frac{d}{dx}\left( x^2 \right) \right]$

$\frac{d y}{d x} = \frac{3}{3x + 2} - \frac{2 x^2}{\left( 2x - 1 \right)} - 2x \log\left( 2x - 1 \right)$

Question 42. Differentiate y = $\frac{3 x^2 \sin x}{\sqrt{7 - x^2}}$ with respect to x.

Solution:

We have,

y = $\frac{3 x^2 \sin x}{\sqrt{7 - x^2}}$

On differentiating y with respect to x we get,

$\frac{d y}{d x} = \frac{d}{dx}\left\{ \frac{3 x^2 sinx}{\left( 7 - x^2 \right)^\frac{1}{2}} \right\}$

On using quotient rule, chain rule and product rule we get,

$\frac{d y}{d x} = \frac{\left( 7 - x^2 \right)^\frac{1}{2} \times \frac{d}{dx}\left( 3 x^2 \sin x \right) - \left( 3 x^2 \sin x \right)\frac{d}{dx} \left( 7 - x^2 \right)^\frac{1}{2}}{\left[ \left( 7 - x^2 \right)^\frac{1}{2} \right]^2} \left[ \text{} \right]$

$\frac{d y}{d x} = \left[ \frac{\left( 7 - x^2 \right)^\frac{1}{2} \times 3\left[ x^2 \frac{d}{dx}\left( \sin x \right) + \sin x\frac{d}{dx}\left( x^2 \right) \right] - 3 x^2 \sin x \times \frac{1}{2}\left( 7 - x^2 \right) \times \frac{d}{dx}\left( 7 - x^2 \right)}{\left( 7 - x^2 \right)} \right]$

$\frac{d y}{d x} = \left[ \frac{\left( 7 - x^2 \right)^\frac{1}{2} 3\left( x^2 \cos x + 2x \sin x \right) - 3 x^2 \sin x \times \frac{1}{2} \left( 7 - x^2 \right)^\frac{- 1}{2} \left( - 2x \right)}{\left( 7 - x^2 \right)} \right]$

$\frac{d y}{d x} = \left[ \frac{\left( 7 - x^2 \right)^\frac{1}{2} \times 3\left( x^2 \cos x + 2x \sin x \right)}{\left( 7 - x^2 \right)} + \frac{3 x^3 \sin x \left( 7 - x^2 \right)^\frac{- 1}{2}}{\left( 7 - x^2 \right)} \right]$

$\frac{d y}{d x} = \left[ \frac{6x \sin x + 3 x^2 \cos x}{\sqrt{\left( 7 - x^2 \right)}} + \frac{3 x^3 \sin x}{\left( 7 - x^2 \right)^\frac{3}{2}} \right]$

Question 43. Differentiate y = $\sin^2 \left\{ \log \left( 2x + 3 \right) \right\}$ with respect to x.

Solution:

We have,

y = $\sin^2 \left\{ \log \left( 2x + 3 \right) \right\}$

On differentiating y with respect to x we get,

$\frac{d y}{d x} = \frac{d}{dx}\left[ \sin^2 \left\{ \log\left( 2x + 3 \right) \right\} \right]$

On using chain rule, we get

$\frac{d y}{d x} = 2 \sin\left\{ \log\left( 2x + 3 \right) \right\}\frac{d}{dx}\sin\left\{ \log\left( 2x + 3 \right) \right\}$

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

As 2 sin A cos A = sin 2A, we get

$\frac{d y}{d x} = \sin\left\{ 2\log\left( 2x + 3 \right) \right\} \times \frac{1}{\left( 2x + 3 \right)}\frac{d}{dx}\left( 2x + 3 \right)$

$\frac{d y}{d x} = \sin\left\{ 2\log\left( 2x + 3 \right) \right\}\left( \frac{2}{\left( 2x + 3 \right)} \right)$

Question 44. Differentiate y = $e^x \log \sin 2x$ with respect to x.

Solution:

We have,

y = $e^x \log \sin 2x$

On differentiating y with respect to x we get,

$\frac{d y}{d x} = \frac{d}{dx}\left[ e^x \log \sin2x \right]$

On using product rule and chain rule, we have

$\frac{d y}{d x} = e^x \frac{d}{dx}\left( \log \sin2x \right) + \left( \log \sin2x \right)\frac{d}{dx}\left( e^x \right)$

$\frac{d y}{d x} = e^x \frac{1}{\sin2x}\frac{d}{dx}\left( \sin2x \right) + \log \sin2x\left( e^x \right)$

$\frac{d y}{d x} = \frac{e^x}{\sin2x}\cos2x \frac{d}{dx}\left( 2x \right) + e^x \log \sin2x$

$\frac{d y}{d x} = \frac{2\cos2x e^x}{\sin2x} + e^x \log \sin2x$

$\frac{d y}{d x} = 2 e^x \cot2x + e^x \log \sin2x$

Question 45. Differentiate y = $\frac{\sqrt{x^2 + 1} + \sqrt{x^2 - 1}}{\sqrt{x^2 + 1} - \sqrt{x^2 - 1}}$ with respect to x.

Solution:

We have,

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

On rationalizing we get,

y = $\frac{\sqrt{x^2 + 1} + \sqrt{x^2 - 1}}{\sqrt{x^2 + 1} - \sqrt{x^2 - 1}} \times \frac{\sqrt{x^2 + 1} + \sqrt{x^2 - 1}}{\sqrt{x^2 + 1} + \sqrt{x^2 - 1}}$

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

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

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

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

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

On differentiating y with respect to x we get,

$\frac{dy}{dx} = \frac{d}{dx}\left( x^2 + \sqrt{x^4 - 1} \right)$

$\frac{d y}{d x} = 2x + \frac{1}{2\sqrt{x^4 - 1}} \times \frac{d}{dx}\left( x^4 - 1 \right)$

$\frac{d y}{d x} = 2x + \frac{1}{2\sqrt{x^4 - 1}} \times \left( 4 x^3 \right)$

$\frac{d y}{d x} = 2x + \frac{2 x^3}{\sqrt{x^4 - 1}}$

Question 46. Differentiate y = $\log [x+2+\sqrt{x^2+4x+1}]$ with respect to x.

Solution:

We have,

y = $\log [x+2+\sqrt{x^2+4x+1}]$

On differentiating y with respect to x we get,

$\frac{d y}{d x}=\frac{d}{dx}\log[x+2+\sqrt{x^2+4x+1}]$

On using chain rule, we have

$\frac{d y}{d x}=\frac{1}{([x+2+sqrt(x^4+4x+1)])}\frac{d}{dx}[x+2+(x^2+4x+1)^{\frac{1}{2}}]$

$\frac{d y}{d x}=\frac{1}{x+2+sqrt(x^4+4x+1)}[1+0+\frac{1}{2}(x^2+4x+1)^{-1/2}\frac{d}{dx}(x^2+4x+1)]$

$\frac{d y}{d x}=\frac{1+\frac{2x+4}{2\sqrt{x^2+4x+1}}}{[x+2+\sqrt{x^4+4x+1}]}$

$\frac{d y}{d x}=\frac{\sqrt{x^4+4x+1}+x+2}{[x+2+\sqrt{x^4+4x+1}]\sqrt{x^4+4x+1}}$

$\frac{d y}{d x}=\frac{1}{\sqrt{x^2+4x+1}}$

Question 47. Differentiate y = $\left( \sin^{- 1} x^4 \right)^4$ with respect to x.

Solution:

We have,

y = $\left( \sin^{- 1} x^4 \right)^4$

On differentiating y with respect to x we get,

$\frac{d y}{d x} = \frac{d}{dx} \left( \sin^{- 1} x^4 \right)^4$

On using chain rule, we have

$\frac{d y}{d x} = 4 \left( \sin^{- 1} x^4 \right)^3 \frac{d}{dx}\left( \sin^{- 1} x^4 \right)$

On using chain rule again, we have

$\frac{d y}{d x} = 4 \left( \sin^{- 1} x^4 \right)^3 \frac{1}{\sqrt{1 - \left( x^4 \right)^2}}\frac{d}{dx}\left( x^4 \right)$

$\frac{d y}{d x} = 4 \left( \sin^{- 1} x^4 \right)^3 \frac{4 x^3}{\sqrt{1 - x^8}}$

$\frac{d y}{d x} = \frac{16 x^3 \left( \sin^{- 1} x^4 \right)^3}{\sqrt{1 - x^8}}$

Question 48. Differentiate y = $\sin^{- 1} \left( \frac{x}{\sqrt{x^2 + a^2}} \right)$ with respect to x.

Solution:

We have,

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

On differentiating y with respect to x we get,

$\frac{d y}{d x} = \frac{d}{dx}\left\{ \sin^{- 1} \left( \frac{x}{\sqrt{x^2 + a^2}} \right) \right\}$

On using chain rule and quotient rule, we get

$\frac{d y}{d x} = \frac{1}{\sqrt{1 - \left( \frac{x}{\sqrt{x^2 + a^2}} \right)^2}} \times \frac{d}{dx}\left( \frac{x}{\sqrt{x^2 + a^2}} \right)$

$\frac{d y}{d x} = \frac{1}{\sqrt{1 - \left( \frac{x}{\sqrt{x^2 + a^2}} \right)^2}} \times \left[ \frac{\left( x^2 + a^2 \right)^\frac{1}{2} \frac{d}{dx}\left( x \right) - x\frac{d}{dx} \left( x^2 + a^2 \right)^\frac{1}{2}}{\left[ \left( x^2 + a^2 \right)^\frac{1}{2} \right]^2} \right]$

$\frac{d y}{d x} = \frac{\sqrt{x^2 + a^2}}{\sqrt{x^2 + a^2 - x^2}}\left[ \frac{\sqrt{x^2 + a^2} - \frac{x}{2\sqrt{x^2 + a^2}}\frac{d}{dx}\left( x^2 + a^2 \right)}{\left( x^2 + a^2 \right)} \right]$

$\frac{d y}{d x} = \frac{\sqrt{x^2 + a^2}}{a\left( x^2 + a^2 \right)}\left[ \sqrt{x^2 + a^2} - \frac{x}{2\sqrt{x^2 + a^2}} \times 2x \right]$

$\frac{d y}{d x} = \frac{\sqrt{x^2 + a^2}}{a\left( x^2 + a^2 \right)}\left[ \frac{x^2 + a^2 - x^2}{\sqrt{x^2 + a^2}} \right]$

$\frac{d y}{d x} = \frac{a^2}{a\left( x^2 + a^2 \right)}$

$\frac{d y}{d x} = \frac{a}{\left( x^2 + a^2 \right)}$

Question 49. Differentiate y = $\frac{e^x \sin x}{\left( x^2 + 2 \right)^3}$ with respect to x.

Solution:

We have,

y = $\frac{e^x \sin x}{\left( x^2 + 2 \right)^3}$

On differentiating y with respect to x we get,

$\frac{d y}{d x} = \frac{d}{dx}\left\{ \frac{e^x \sin x}{\left( x^2 + 2 \right)^3} \right\}$

On using quotient rule, we get

$\frac{d y}{d x} = \frac{\left( x^2 + 2 \right)^3 \frac{d}{dx}\left( e^x \sin x \right) - e^x \sin x\frac{d}{dx} \left( x^2 + 2 \right)^3}{\left[ \left( x^2 + 2 \right)^3 \right]^2}$

On using product rule, we get

$\frac{d y}{d x} = \frac{\left( x^2 + 2 \right)^3 \left[ e^x \cos x + \sin x e^x \right] - e^x \sin x 3 \left( x^2 + 2 \right)^2 \left( 2x \right)}{\left( x^2 + 2 \right)^6}$

$\frac{d y}{d x} = \frac{\left( x^2 + 2 \right)^3 \left[ e^x \cos x + e^x \sin x \right] - 6x e^x \sin x \left( x^2 + 2 \right)^2}{\left( x^2 + 2 \right)^6}$

$\frac{d y}{d x} = \frac{\left( x^2 + 2 \right)^2 \left[ \left( x^2 + 2 \right)\left( e^x \cos x + e^x \sin x \right) - 6x e^x \sin x \right]}{\left( x^2 + 2 \right)^6}$

$\frac{d y}{d x} = \frac{\left( x^2 + 2 \right)\left( e^x \cos x + e^x \sin x \right) - 6x e^x \sin x}{\left( x^2 + 2 \right)^4}$

$\frac{d y}{d x} = \frac{e^x \sin x + e^x \cos x}{\left( x^2 + 2 \right)^3} - \frac{6x e^x \sin x}{\left( x^2 + 2 \right)^4}$

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RD Sharma Class 12 Ex 11.2 Solutions Chapter 11 Differentiation

Question 1. Differentiate y = sin (3x + 5) with respect to x.

Question 2. Differentiate y = tan2 x with respect to x.

Question 3. Differentiate y = tan (x + 45°) with respect to x.

Question 4. Differentiate y = sin (log x) with respect to x.

Question 5. Differentiate y = esin √x with respect to x.

Question 6. Differentiate y = etan x with respect to x.

Question 7. Differentiate y = sin2 (2x + 1) with respect to x.

Question 8. Differentiate y = log7 (2x − 3) with respect to x.

Question 9. Differentiate y = tan 5x° with respect to x.

Question 10. Differentiate y = with respect to x.

Question 11. Differentiate y = with respect to x.

Question 12. Differentiate y = logx 3 with respect to x.

Question 13. Differentiate y = with respect to x.

Question 14. Differentiate y = with respect to x.

Question 15. Differentiate y = with respect to x.

Question 16. Differentiate y = with respect to x.

Question 17. Differentiate y = with respect to x.

Question 18. Differentiate y = (log sin x)2 with respect to x.

Question 19. Differentiate y = with respect to x.

Question 20. Differentiate y = with respect to x.

Question 21. Differentiate y = with respect to x.

Question 22. Differentiate y = sin(log sin x) with respect to x.

Question 23. Differentiate y = etan 3x with respect to x.

Question 24. Differentiate y = with respect to x.

Question 25. Differentiate y = with respect to x.

Question 26. Differentiate y = with respect to x.

Question 27. Differentiate y = with respect to x.

Question 28. Differentiate y = with respect to x.

Question 29. Differentiate y = with respect to x.

Question 30. Differentiate y = with respect to x.

Question 31. Differentiate y = with respect to x.

Question 32. Differentiate y = with respect to x.

Question 33. Differentiate y = with respect to x.

Question 34. Differentiate y = with respect to x.

Question 35. Differentiate y = with respect to x.

Question 36. Differentiate y = with respect to x.

Question 37. Differentiate y = with respect to x.

Question 38. Differentiate y = with respect to x.

Question 39. Differentiate y = with respect to x.

Question 40. Differentiate y = with respect to x.

Question 41. Differentiate y = with respect to x.

Question 42. Differentiate y = with respect to x.

Question 43. Differentiate y = with respect to x.

Question 44. Differentiate y = with respect to x.

Question 45. Differentiate y = with respect to x.

Question 46. Differentiate y = with respect to x.

Question 47. Differentiate y = with respect to x.

Question 48. Differentiate y = with respect to x.

Question 49. Differentiate y = with respect to x.

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Question 2. Differentiate y = tan² x with respect to x.

Question 5. Differentiate y = e^{sin √x} with respect to x.

Question 6. Differentiate y = e^{tan x} with respect to x.

Question 7. Differentiate y = sin² (2x + 1) with respect to x.

Question 8. Differentiate y = log₇ (2x − 3) with respect to x.

Question 10. Differentiate y = $2^{x^3}$ with respect to x.

Question 11. Differentiate y = $3^{e^x}$ with respect to x.

Question 12. Differentiate y = log_x 3 with respect to x.

Question 13. Differentiate y = $3^{x^2 + 2x}$ with respect to x.

Question 14. Differentiate y = $\sqrt{\frac{a^2 - x^2}{a^2 + x^2}}$ with respect to x.

Question 15. Differentiate y = $3^{x \log x}$ with respect to x.

Question 16. Differentiate y = $\sqrt{\frac{1 + \sin x}{1 - \sin x}}$ with respect to x.

Question 17. Differentiate y = $\sqrt{\frac{1 - x^2}{1 + x^2}}$ with respect to x.

Question 18. Differentiate y = (log sin x)² with respect to x.

Question 19. Differentiate y = $\sqrt{\frac{1 + x}{1 - x}}$ with respect to x.

Question 20. Differentiate y = $\sin \left( \frac{1 + x^2}{1 - x^2} \right)$ with respect to x.

Question 21. Differentiate y = $e^{3x} \cos2x$ with respect to x.

Question 23. Differentiate y = e^{tan 3x} with respect to x.

Question 24. Differentiate y = $e^{\sqrt{\cot x}}$ with respect to x.

Question 25. Differentiate y = $\log \left( \frac{\sin x}{1 + \cos x} \right)$ with respect to x.

Question 26. Differentiate y = $\log\sqrt{\frac{1 - \cos x}{1 + \cos x}}$ with respect to x.

Question 27. Differentiate y = $\tan \left( e^{\sin x }\right)$ with respect to x.

Question 28. Differentiate y = $\log\left( x + \sqrt{x^2 + 1} \right)$ with respect to x.

Question 29. Differentiate y = $\frac{e^x \log x}{x^2}$ with respect to x.

Question 30. Differentiate y = $\log \left( cosec x - \cot x \right)$ with respect to x.

Question 31. Differentiate y = $\frac{e^{2x} + e^{- 2x}}{e^{2x} - e^{- 2x}}$ with respect to x.

Question 32. Differentiate y = $\log\left( \frac{x^2 + x + 1}{x^2 - x + 1} \right)$ with respect to x.

Question 33. Differentiate y = $\tan^{- 1} \left( e^x \right)$ with respect to x.

Question 34. Differentiate y = $e^{\sin^{- 1} 2x}$ with respect to x.

Question 35. Differentiate y = $\sin \left( 2 \sin^{- 1} x \right)$ with respect to x.

Question 36. Differentiate y = $e^{\tan^{- 1} \sqrt{x}}$ with respect to x.

Question 37. Differentiate y = $\sqrt{\tan^{- 1} \left( \frac{x}{2} \right)}$ with respect to x.

Question 38. Differentiate y = $\log\left( \tan^{- 1} x \right)$ with respect to x.

Question 39. Differentiate y = $\frac{2^x \cos x}{\left( x^2 + 3 \right)^2}$ with respect to x.

Question 40. Differentiate y = $x \sin 2x + 5^x + k^k + \left( \tan^2 x \right)^3$ with respect to x.

Question 41. Differentiate y = $\log \left( 3x + 2 \right) - x^2 \log \left( 2x - 1 \right)$ with respect to x.

Question 42. Differentiate y = $\frac{3 x^2 \sin x}{\sqrt{7 - x^2}}$ with respect to x.

Question 43. Differentiate y = $\sin^2 \left\{ \log \left( 2x + 3 \right) \right\}$ with respect to x.

Question 44. Differentiate y = $e^x \log \sin 2x$ with respect to x.

Question 45. Differentiate y = $\frac{\sqrt{x^2 + 1} + \sqrt{x^2 - 1}}{\sqrt{x^2 + 1} - \sqrt{x^2 - 1}}$ with respect to x.

Question 46. Differentiate y = $\log [x+2+\sqrt{x^2+4x+1}]$ with respect to x.

Question 47. Differentiate y = $\left( \sin^{- 1} x^4 \right)^4$ with respect to x.

Question 48. Differentiate y = $\sin^{- 1} \left( \frac{x}{\sqrt{x^2 + a^2}} \right)$ with respect to x.

Question 49. Differentiate y = $\frac{e^x \sin x}{\left( x^2 + 2 \right)^3}$ with respect to x.