Class 12 RD Sharma Solutions Chapter 11 Differentiation Exercise 11.2 | Set 2 (original) (raw)

Last Updated : 18 Sep, 2024

Chapter 11 of RD Sharma's Class 12 Mathematics textbook covers Differentiation, a fundamental concept in calculus. Exercise 11.2 Set 2 typically focuses on more advanced differentiation techniques, including the differentiation of composite functions, implicit differentiation, and applications of various differentiation rules.

**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}

Summary

Exercise 11.2 Set 2 in Chapter 11 of RD Sharma's Class 12 Mathematics provides a more challenging set of problems on differentiation. These questions typically involve composite functions, trigonometric functions, exponential and logarithmic functions, and combinations thereof. This set helps students apply multiple differentiation rules simultaneously and prepares them for more complex mathematical analysis.