Class 12 RD Sharma Solutions - Chapter 11 Differentiation - Exercise 11.2 | Set 1

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}}

Summary

Exercise 11.2 | Set 1 focuses on applications of derivatives. It covers topics such as finding the rate of change, velocity and acceleration problems, and optimization. Students are expected to apply differentiation techniques to solve real-world problems, interpret the meaning of derivatives in various contexts, and use derivatives to find maximum and minimum values of functions.