Class 12 RD Sharma Solutions – Chapter 11 Differentiation – Exercise 11.2 | Set 3

Question 50. Differentiate y = $3 e^{- 3x} \log \left( 1 + x \right)$ with respect to x.

Solution:

We have,
y = $3 e^{- 3x} \log \left( 1 + x \right)$
On differentiating y with respect to x we get,
$\frac{d y}{d x} = \frac{d}{dx}\left(3 e^{- 3x} \log \left( 1 + x \right) \right)$
$\frac{d y}{d x} = 3\frac{d}{dx}\left[ e^{- 3x} \log\left( 1 + x \right) \right]$
On using product rule and chain rule, we get
$\frac{d y}{d x} = 3\left\{ e^{- 3x} \frac{1}{1 + x} + \log\left( 1 + x \right)\left( - 3 e^{- 3x} \right) \right\}$
$\frac{d y}{d x} = 3\left\{ \frac{e^{- 3x}}{1 + x} - 3 e^{- 3x} \log\left( 1 + x \right) \right\}$
$\frac{d y}{d x} = 3 e^{- 3x} \left\{ \frac{1}{1 + x} - 3 \log\left( 1 + x \right) \right\}$

Question 51. Differentiate y = $\frac{x^2 + 2}{\sqrt{\cos x}}$ with respect to x.

Solution:

We have,
y = $\frac{x^2 + 2}{\sqrt{\cos x}}$
On differentiating y with respect to x we get,
$\frac{d y}{d x} = \frac{d}{dx}\left(\frac{x^2 + 2}{\sqrt{\cos x}}\right)$
On using quotient rule and chain rule, we get
$\frac{d y}{d x} = \frac{\sqrt{\cos x}\frac{d}{dx}\left( x^2 + 2 \right) - \left( x^2 + 2 \right)\frac{d}{dx}\left( \sqrt{\cos x} \right)}{\left( \sqrt{\cos x} \right)^2}$
$\frac{d y}{d x} = \frac{2x\sqrt{\cos x} - \left( x^2 + 2 \right)\left( - \frac{1}{2}\frac{\sin x}{\sqrt{\cos x}} \right)}{\cos x}$
$\frac{d y}{d x} = \frac{2x\sqrt{\cos x} + \frac{\left( x^2 + 2 \right)\sin x}{2\sqrt{\cos x}}}{\cos x}$
$\frac{d y}{d x} = \frac{4x \cos x + \left( x^2 + 2 \right)\sin x}{2 \left( \cos x \right)^\frac{3}{2}}$
$\frac{d y}{d x} = \frac{2x}{\sqrt{\cos x}} + \frac{1}{2}\frac{\left( x^2 + 2 \right)\sin x}{\left( \cos x \right)^\frac{3}{2}}$
$\frac{d y}{d x} = \frac{1}{\sqrt{\cos x}}\left\{ 2x + \frac{1}{2}\frac{\left( x^2 + 2 \right)\sin x}{\cos x} \right\}$
$\frac{d y}{d x} = \frac{1}{\sqrt{\cos x}}\left\{ 2x + \frac{\left( x^2 + 2 \right)\tan x}{2} \right\}$

Question 52. Differentiate y = $\frac{x^2 \left( 1 - x^2 \right)}{\cos 2x}$ with respect to x.

Solution:

We have,
y = $\frac{x^2 \left( 1 - x^2 \right)}{\cos 2x}$
On differentiating y with respect to x we get,
$\frac{d y}{d x} = \frac{d}{dx}\left(\frac{x^2 \left( 1 - x^2 \right)}{\cos 2x}\right)$
On using quotient rule and chain rule, we get
$\frac{d y}{d x} = \frac{\cos2x\frac{d}{dx}\left\{ x^2 \left( 1 - x^2 \right)^3 \right\} - x^2 \left( 1 - x^2 \right)^3 \frac{d}{dx}\cos2x}{\cos^2 2x}$
$\frac{d y}{d x} = \frac{\cos2x\left\{ x^2 \frac{d}{dx} \left( 1 - x^2 \right)^3 + \left( 1 - x^2 \right)^3 \frac{d}{dx}\left( x^2 \right) \right\} - x^2 \left( 1 - x^2 \right)^3 \left( - 2\sin2x \right)}{\cos^2 2x}$
$\frac{d y}{d x} = \frac{\cos2x\left\{ - 6 x^3 \left( 1 - x^2 \right)^2 + \left( 1 - x^2 \right)^3 2x \right\} + 2 x^2 \left( 1 - x^2 \right)^3 \sin2x}{\cos^2 2x}$
$\frac{d y}{d x} = \frac{2x \left( 1 - x^2 \right)^2}{\cos2x} - \frac{6 x^3 \left( 1 - x^2 \right)^2}{\cos2x} + \frac{2 x^2 \left( 1 - x^2 \right)^3 \sin2x}{\cos^2 2x}$
$\frac{d y}{d x} = 2x\left( 1 - x^2 \right)\sec2x\left\{ 1 - 4 x^2 + x\left( 1 - x^2 \right)\tan2x \right\}$

Question 53. Differentiate y = $\log\left\{ \cot\left( \frac{\pi}{4} + \frac{x}{2} \right) \right\}$ with respect to x.

Solution:

We have,
y = $\log\left\{ \cot\left( \frac{\pi}{4} + \frac{x}{2} \right) \right\}$
On differentiating y with respect to x we get,
$\frac{d y}{d x} = \frac{d}{dx}\left(\log\left\{ \cot\left( \frac{\pi}{4} + \frac{x}{2} \right) \right\}\right)$
On using chain rule, we get
$\frac{d y}{d x} = \frac{- {cosec}^2 \left( \frac{\pi}{4} + \frac{x}{2} \right)}{2\cot\left( \frac{\pi}{4} + \frac{x}{2} \right)}$
$\frac{d y}{d x} = \frac{- 1}{2\cos\left( \frac{\pi}{4} + \frac{x}{2} \right)\sin\left( \frac{\pi}{4} + \frac{x}{2} \right)}$
$\frac{d y}{d x} = \frac{- 1}{\sin\left( \frac{\pi}{2} + x \right)}$
$\frac{d y}{d x} = \frac{- 1}{\cos x}$
$\frac{d y}{d x} = - \sec x$

Question 54. Differentiate y = $e^{ax} \sec x \tan2x$ with respect to x.

Solution:

We have,
y = $e^{ax} \sec x \tan2x$
On differentiating y with respect to x we get,
$\frac{d y}{d x} = \frac{d}{dx}\left( e^{ax} sec x \tan2x \right)$
On using chain rule, we get
$\frac{d y}{d x} = e^{ax} \frac{d}{dx}\left\{ \sec x \tan2x \right\} + \sec x \tan2x\frac{d}{dx}\left\{ e^{ax} \right\}$
$\frac{d y}{d x} = e^{ax} \left[ \sec x \tan x \tan2x + 2 \sec^2 2x\sec x \right] + a e^{ax} \sec x \tan2x$
$\frac{d y}{d x} = e^{ax} \left[ \sec x \tan x \tan2x + 2 \sec^2 2x\sec x \right] + a e^{ax} sec x \tan2x$
$\frac{d y}{d x} = a e^{ax} \sec x \tan2x + e^{ax} \sec x \tan x \tan2x + 2 e^{ax} \sec x \sec^2 2x$
$\frac{d y}{d x} = e^{ax} \sec x\left\{ a \tan2x + \tan x \tan2x + 2 \sec^2 2x \right\}$

Question 55. Differentiate y = $\log \left( \cos x^2 \right)$ with respect to x.

Solution:

We have,
y = $\log \left( \cos x^2 \right)$
On differentiating y with respect to x we get,
$\frac{d y}{d x} = \frac{d}{dx}\left( \log \left( \cos x^2 \right) \right)$
On using chain rule, we get
$\frac{d y}{d x} = \frac{d}{dx}\left\{ \log\left( \cos x^2 \right) \right\}$
$\frac{d y}{d x} = \frac{- 2x \sin x^2}{\cos x^2}$
$\frac{d y}{d x} = - 2x \tan x^2$

Question 56. Differentiate y = $\cos \left( \log x \right)^2$ with respect to x.

Solution:

We have,
y = $\cos \left( \log x \right)^2$
On differentiating y with respect to x we get,
$\frac{d y}{d x} = \frac{d}{dx}\left( \log \left( \cos x^2 \right) \right)$
On using chain rule, we get
$\frac{d y}{d x} = - \sin \left( \log x \right)^2 \frac{d}{dx} \left( \log x \right)^2$
$\frac{d y}{d x} = - \sin \left( \log x \right)^2 \frac{2\log x}{x}$
$\frac{d y}{d x} = \frac{- 2\log x \sin \left( \log x \right)^2}{x}$

Question 57. Differentiate y = $\log \sqrt{\frac{x - 1}{x + 1}}$ with respect to x.

Solution:

We have,
y = $\log \sqrt{\frac{x - 1}{x + 1}}$
y = $\log \left( \frac{x - 1}{x + 1} \right)^\frac{1}{2}$
y = $\frac{1}{2}\log \left( \frac{x - 1}{x + 1} \right)$
y = $\frac{1}{2}\left[ \log\left( x - 1 \right) - \log\left( x + 1 \right) \right]$
On differentiating y with respect to x we get,
$\frac{d y}{d x} = \frac{d}{dx}\left( \frac{1}{2}\left[ \log\left( x - 1 \right) - \log\left( x + 1 \right) \right] \right)$
$\frac{d y}{d x} = \frac{1}{2}\left[ \frac{d}{dx}\left\{ \log\left( x - 1 \right) \right\} - \frac{d}{dx}\left\{ \log\left( x + 1 \right) \right\} \right]$
$\frac{d y}{d x} = \frac{1}{2}\left( \frac{1}{x - 1} - \frac{1}{x + 1} \right)$
$\frac{d y}{d x} = \frac{1}{2}\left( \frac{2}{x^2 - 1} \right)$
$\frac{d y}{d x} = \frac{1}{x^2 - 1}$

Question 58. If $y = \log \left\{ \sqrt{x - 1} - \sqrt{x + 1} \right\}$ , show that $\frac{dy}{dx} = \frac{- 1}{2\sqrt{x^2 - 1}}$ .

Solution:

We have,
y = $\log \left\{ \sqrt{x - 1} - \sqrt{x + 1} \right\}$
On differentiating y with respect to x we get,
$\frac{d y}{d x} = \frac{d}{dx}\log\left( \sqrt{x - 1} - \sqrt{x + 1} \right)$
On using chain rule, we get
$\frac{d y}{d x} = \frac{1}{\left( \sqrt{x - 1} - \sqrt{x + 1} \right)}\frac{d}{dx}\left( \sqrt{x - 1} - \sqrt{x + 1} \right)$
$\frac{d y}{d x} = \frac{1}{\left( \sqrt{x - 1} - \sqrt{x + 1} \right)}\left[ \frac{d}{dx}\sqrt{x - 1} - \frac{d}{dx}\sqrt{x + 1} \right]$
$\frac{d y}{d x} = \frac{1}{\left( \sqrt{x - 1} - \sqrt{x + 1} \right)}\left[ \frac{1}{2} \left( x - 1 \right)^\frac{- 1}{2} - \frac{1}{2} \left( x + 1 \right)^\frac{- 1}{2} \right]$
$\frac{d y}{d x} = \frac{1}{2}\frac{1}{\left( \sqrt{x - 1} - \sqrt{x + 1} \right)}\left( \frac{1}{\sqrt{x - 1}} - \frac{1}{\sqrt{x + 1}} \right)$
$\frac{d y}{d x} = \frac{1}{2}\frac{1}{\left( \sqrt{x - 1} - \sqrt{x + 1} \right)}\left\{ \frac{- \left( \sqrt{x - 1} - \sqrt{x + 1} \right)}{\left( \sqrt{x - 1} \right)\left( \sqrt{x + 1} \right)} \right\}$
$\frac{d y}{d x} = \frac{- 1}{2}\left( \frac{1}{\left( \sqrt{x - 1} \right)\left( \sqrt{x + 1} \right)} \right)$
$\frac{d y}{d x} = \frac{- 1}{2\sqrt{x^2 - 1}}$
Hence, proved.

Question 59. If y = $\sqrt{x + 1} + \sqrt{x - 1}$ , prove that $\sqrt{x^2 - 1}\frac{dy}{dx} = \frac{1}{2}y$ .

Solution:

We have,
y = $\sqrt{x + 1} + \sqrt{x - 1}$
On differentiating y with respect to x we get,
$\frac{d y}{d x} = \frac{d}{dx}\log\left( \sqrt{x - 1} - \sqrt{x + 1} \right)$
$\frac{d y}{d x} = \frac{d}{dx}\left( \sqrt{x + 1} \right) + \frac{d}{dx}\left( \sqrt{x - 1} \right)$
$\frac{d y}{d x} = \frac{1}{2} \left( x + 1 \right)^\frac{- 1}{2} + \frac{1}{2} \left( x - 1 \right)^\frac{- 1}{2}$
$\frac{d y}{d x} = \frac{1}{2}\left( \frac{1}{\sqrt{x + 1}} + \frac{1}{\sqrt{x - 1}} \right)$
$\frac{d y}{d x} = \frac{1}{2}\left( \frac{\sqrt{x - 1} + \sqrt{x + 1}}{\left( \sqrt{x + 1} \right)\left( \sqrt{x - 1} \right)} \right)$
$\frac{d y}{d x} = \frac{1}{2}\left( \frac{y}{\left( \sqrt{x^2 - 1} \right)} \right)$
$\left( \sqrt{x^2 - 1} \right)\frac{d y}{d x} = \frac{1}{2}y$
Hence proved.

Question 60. If y = $\frac{x}{x + 2}$ , prove that $x\frac{dy}{dx} = \left( 1 - y \right) y$ .

Solution:

We have,
y = $\frac{x}{x + 2}$
On differentiating y with respect to x we get,
$\frac{d y}{d x} = \frac{d}{dx}\left( \frac{x}{x + 2} \right)$
$\frac{d y}{d x} = \frac{\left( x + 2 \right)\frac{d}{dx}\left( x \right) - x\frac{d}{dx}\left( x + 2 \right)}{\left( x + 2 \right)^2}$
$\frac{d y}{d x} = \frac{x + 2 - x}{\left( x + 2 \right)^2}$
$\frac{d y}{d x} = \frac{x + 2}{\left( x + 2 \right)^2} - \frac{x}{\left( x + 2 \right)^2}$
As $x + 2 = \frac{x}{y}$ , we have
$\frac{d y}{d x} = \frac{1}{x + 2} - \frac{x y^2}{x^2}$
$\frac{d y}{d x} = \frac{y}{x} - \frac{y^2}{x}$
$\frac{d y}{d x} = \frac{1}{x}y\left( 1 - y \right)$
$x\frac{d y}{d x} = \left( 1 - y \right)y$
Hence proved.

Question 61. If y = $\log \left( \sqrt{x} + \frac{1}{\sqrt{x}} \right)$ , prove that $\frac{dy}{dx} = \frac{x - 1}{2x \left( x + 1 \right)}$ .

Solution:

We have,
y = $\log \left( \sqrt{x} + \frac{1}{\sqrt{x}} \right)$
On differentiating y with respect to x we get,
$\frac{d y}{d x} = \frac{d}{dx}\log\left( \sqrt{x} + \frac{1}{\sqrt{x}} \right)$
$\frac{d y}{d x} = \frac{1}{\sqrt{x} + \frac{1}{\sqrt{x}}}\frac{d}{dx}\left( \sqrt{x} + \frac{1}{\sqrt{x}} \right)$
$\frac{d y}{d x} = \frac{\sqrt{x}}{x + 1}\left( \frac{1}{2\sqrt{x}} - \frac{1}{2x\sqrt{x}} \right)$
$\frac{d y}{d x} = \frac{1}{2}\frac{\sqrt{x}}{x + 1}\left( \frac{x - 1}{x\sqrt{x}} \right)$
$\frac{d y}{d x} = \frac{x - 1}{2x\left( x + 1 \right)}$
Hence proved.

Question 62. If y = $\log \sqrt{\frac{1 + \tan x}{1 - \tan x}}$ , prove that $\frac{dy}{dx} = \sec 2x$ .

Solution:

We have,
y = $\log \sqrt{\frac{1 + \tan x}{1 - \tan x}}$
y = $\log \left\{ \frac{1 + \tan x}{1 - \tan x} \right\}^\frac{1}{2}$
y = $\frac{1}{2}\log\left\{ \frac{1 + \tan x}{1 - \tan x} \right\}$
y = $\frac{1}{2}\left\{ \log\left( 1 + \tan x \right) - \log\left( 1 - \tan x \right) \right\}$
On differentiating y with respect to x we get,
$\frac{d y}{d x} = \frac{d}{dx}\log\left(\frac{1}{2}\left\{ \log\left( 1 + \tan x \right) - \log\left( 1 - \tan x \right) \right\} \right)$
$\frac{d y}{d x} = \frac{1}{2}\left\{ \frac{d}{dx}\left\{ \log\left( 1 + \tan x \right) \right\} - \frac{d}{dx}\left\{ \log\left( 1 - \tan x \right) \right\} \right\}$
$\frac{d y}{d x} = \frac{1}{2}\left\{ \frac{1}{1 + \tan x } \times \frac{d}{dx}\left( 1 + \tan x \right) - \frac{1}{1 - \tan x} \times \frac{d}{dx}\left( 1 - \tan x \right) \right\}$
$\frac{d y}{d x} = \frac{1}{2}\left\{ \frac{1}{1 + \tan x}\left( 0 + \sec^2 x \right) - \frac{1}{1 - \tan x}\left( 0 - \sec^2 x \right) \right\}$
$\frac{d y}{d x}= \frac{1}{2}\left\{ \frac{\sec^2 x}{1 + \tan x} + \frac{\sec^2 x}{1 - \tan x} \right\}$
$\frac{d y}{d x} = \frac{1}{2} \sec^2 x\left\{ \frac{1 - \tan x + 1 + \tan x}{1 - \tan^2 x} \right\}$
$\frac{d y}{d x} = \frac{1}{2} \sec^2 x\left( \frac{2}{1 - \tan^2 x} \right)$
$\frac{d y}{d x} = \frac{\sec^2 x}{1 - \tan^2 x}$
$\frac{d y}{d x} = \frac{1 + \tan^2 x}{1 - \tan^2 x}$
$\frac{d y}{d x} = \frac{1}{\frac{1 - \tan^2 x}{1 + \tan^2 x}}$
$\frac{d y}{d x} = \frac{1}{\cos2x}$
$\frac{d y}{d x} = \sec2x$
Hence proved.

Question 63. If y = $\sqrt{x} + \frac{1}{\sqrt{x}}$ , prove that $2 x\frac{dy}{dx} = \sqrt{x} - \frac{1}{\sqrt{x}}$ .

Solution:

We have,
y = $\sqrt{x} + \frac{1}{\sqrt{x}}$
On differentiating y with respect to x we get,
$\frac{d y}{d x} = \frac{d}{dx}\left( \sqrt{x} + \frac{1}{\sqrt{x}} \right)$
$\frac{d y}{d x} = \frac{d}{dx}\left( \sqrt{x} \right) + \frac{d}{dx}\left( \frac{1}{\sqrt{x}} \right)$
$\frac{d y}{d x} = \frac{1}{2\sqrt{x}} + \left( \frac{- \frac{1}{2\sqrt{x}}}{x} \right)$
$\frac{d y}{d x} = \frac{1}{2\sqrt{x}} - \frac{1}{2x\sqrt{x}}$
$\frac{d y}{d x} = \frac{x - 1}{2x\sqrt{x}}$
$2x\frac{d y}{d x} = \frac{x - 1}{\sqrt{x}}$
$2x\frac{d y}{d x} = \frac{x}{\sqrt{x}} - \frac{1}{\sqrt{x}}$
$2x\frac{d y}{d x} = \sqrt{x} - \frac{1}{\sqrt{x}}$
Hence proved.

Question 64. If y = $\frac{x \sin^{- 1} x}{\sqrt{1 - x^2}}$ , prove that $\left( 1 - x^2 \right) \frac{dy}{dx} = x + \frac{y}{x}$ .

Solution:

We have,
y = $\frac{x \sin^{- 1} x}{\sqrt{1 - x^2}}$
On differentiating y with respect to x we get,
$\frac{d y}{d x} = \frac{d}{dx}\left( \frac{x \sin^{- 1} x}{\sqrt{1 - x^2}} \right)$
$\frac{d y}{d x} = \left[ \frac{\sqrt{1 - x^2}\frac{d}{dx}\left( x \sin^{- 1} x \right) - \left( x \sin^{- 1} x \right)\frac{d}{dx}\left( \sqrt{1 - x^2} \right)}{\left( \sqrt{1 - x^2} \right)^2} \right]$
$\frac{d y}{d x} = \left[ \frac{\sqrt{1 - x^2}\left\{ x\frac{d}{dx}\left( \sin^{- 1} x \right) + \sin^{- 1} x\frac{d}{dx}\left( x \right) \right\} - \left( x \sin^{- 1} x \right)\frac{1}{2\sqrt{1 - x^2}}\frac{d}{dx}\left( 1 - x^2 \right)}{\left( 1 - x^2 \right)} \right]$
$\frac{d y}{d x} = \left[ \frac{\sqrt{1 - x^2}\left\{ \frac{x}{\sqrt{1 - x^2}} + \sin^{- 1} x \right\} - \frac{x \sin^{- 1} x\left( - 2x \right)}{2\sqrt{1 - x^2}}}{\left( 1 - x^2 \right)} \right]$
$\frac{d y}{d x} = \left[ \frac{x + \sqrt{1 - x^2} \sin^{- 1} x + \frac{x^2 \sin^{- 1} x}{\sqrt{1 - x^2}}}{\left( 1 - x^2 \right)} \right]$
$\left( 1 - x^2 \right)\frac{d y}{d x} = x + \frac{\sqrt{1 - x^2} \sin^{- 1} x}{1} + \frac{x^2 \sin^{- 1} x}{\sqrt{1 - x^2}}$
$\left( 1 - x^2 \right)\frac{d y}{d x} = x + \left( \frac{\left( 1 - x^2 \right) \sin^{- 1} x + x^2 \sin^{- 1} x}{\sqrt{1 - x^2}} \right)$
$\left( 1 - x^2 \right)\frac{d y}{d x} = x + \left( \frac{\sin^{- 1} x - x^2 \sin^{- 1} x + x^2 \sin^{- 1} x}{\sqrt{1 - x^2}} \right)$
$\left( 1 - x^2 \right)\frac{d y}{d x} = x + \left( \frac{\sin^{- 1} x}{\sqrt{1 - x^2}} \right)$
As we have $y = \frac{x \sin^{- 1} x}{\sqrt{1 - x^2}}$ , we get
$\left( 1 - x^2 \right)\frac{d y}{d x} = x + \frac{y}{x}$
Hence proved.

Question 65. If y = $\frac{e^x - e^{- x}}{e^x + e^{- x}}$ , prove that $\frac{dy}{dx} = 1 - y^2$ .

Solution:

We have,
y = $\frac{e^x - e^{- x}}{e^x + e^{- x}}$
On differentiating y with respect to x we get,
$\frac{d y}{d x} = \frac{d}{dx}\left( \frac{e^x - e^{- x}}{e^x + e^{- x}} \right)$
$\frac{d y}{d x} = \left[ \frac{\left( e^x + e^{- x} \right)\frac{d}{dx}\left( e^x - e^{- x} \right) - \left( e^x - e^{- x} \right)\frac{d}{dx}\left( e^x + e^{- x} \right)}{\left( e^x + e^{- x} \right)^2} \right]$
$\frac{d y}{d x} = \left[ \frac{\left( e^x + e^{- x} \right)\left\{ e^x - e^{- x} \frac{d}{dx}\left( - x \right) \right\} - \left( e^x - e^{- x} \right)\left\{ e^x + e^{- x} \frac{d}{dx}\left( - x \right) \right\}}{\left( e^x + e^{- x} \right)^2} \right]$
$\frac{d y}{d x} = \left[ \frac{\left( e^x + e^{- x} \right)\left( e^x + e^{- x} \right) - \left( e^x - e^{- x} \right)\left( e^x - e^{- x} \right)}{\left( e^x + e^{- x} \right)^2} \right]$
$\frac{d y}{d x} = \frac{\left( e^x + e^{- x} \right)^2 - \left( e^x - e^{- x} \right)^2}{\left( e^x + e^{- x} \right)^2}$
$\frac{d y}{d x} = 1 - \frac{\left( e^x - e^{- x} \right)^2}{\left( e^x + e^{- x} \right)^2}$
$\frac{d y}{d x} = 1 - \left( \frac{e^x - e^{- x}}{e^x + e^{- x}} \right)^2$
As y = $\frac{e^x - e^{- x}}{e^x + e^{- x}}$ , we have
$\frac{d y}{d x} = 1 - y^2$
Hence proved.

Question 66. If y = $\left( x - 1 \right) \log \left( x - 1 \right) - \left( x + 1 \right) \log \left( x + 1 \right)$ , prove that $\frac{dy}{dc} = \log \left( \frac{x - 1}{1 + x} \right)$ .

Solution:

We have,
y = $\left( x - 1 \right) \log \left( x - 1 \right) - \left( x + 1 \right) \log \left( x + 1 \right)$
On differentiating y with respect to x we get,
$\frac{d y}{d x} = \frac{d}{dx}\left[ \left( x - 1 \right) \log\left( x - 1 \right) - \left( x + 1 \right) \log\left( x + 1 \right) \right]$
$\frac{d y}{d x} = \left[ \left( x - 1 \right)\frac{d}{dx}\log\left( x - 1 \right) + \log\left( x - 1 \right)\frac{d}{dx}\left( x - 1 \right) \right] - \left[ \left( x + 1 \right)\frac{d}{dx}\log\left( x + 1 \right) + \log\left( x + 1 \right)\frac{d}{dx}\left( x + 1 \right) \right]$
$\frac{d y}{d x} = \left[ \left( x - 1 \right) \times \frac{1}{\left( x - 1 \right)}\frac{d}{dx}\left( x - 1 \right) + \log\left( x - 1 \right) \times \left( 1 \right) \right] - \left[ \left( x + 1 \right) \times \frac{1}{\left( x + 1 \right)} \times \frac{d}{dx}\left( x + 1 \right) + \log\left( x + 1 \right)\left( 1 \right) \right]$
$\frac{d y}{d x} = \left[ 1 + \log\left( x - 1 \right) \right] - \left[ 1 + \log\left( x + 1 \right) \right]$
$\frac{d y}{d x} = \log\left( x - 1 \right) - \log\left( x + 1 \right)$
$\frac{d y}{d x} = \log\frac{\left( x - 1 \right)}{\left( x + 1 \right)}$
Hence proved.

Question 67. If y = $e^x \cos x$ , prove that $\frac{dy}{dx} = \sqrt{2} e^x \cdot \cos \left( x + \frac{\pi}{4} \right)$ .

Solution:

We have,
y = $e^x \cos x$
On differentiating y with respect to x we get,
$\frac{d y}{d x} = \frac{d}{dx}\left( e^x \cos x \right)$
$\frac{d y}{d x} = e^x \frac{d}{dx}\cos x + \cos x\frac{d}{dx} e^x$
$\frac{d y}{d x} = e^x \left( - \sin x \right) + e^x \cos x$
$\frac{d y}{d x} = e^x \left( \cos x - \sin x \right)$
$\frac{d y}{d x} = \sqrt{2} e^x \left( \frac{\cos x}{\sqrt{2}} - \frac{\sin x}{\sqrt{2}} \right)$
$\frac{d y}{d x} = \sqrt{2} e^x \left( \cos\frac{\pi}{4}\cos x - \sin\frac{\pi}{4}\sin x \right)$
$\frac{d y}{d x} = \sqrt{2} e^x \cos\left( x + \frac{\pi}{4} \right)$
Hence proved.

Question 68. If y = $\frac{1}{2} \log \left( \frac{1 - \cos 2x }{1 + \cos 2x} \right)$ , prove that $\frac{ dy }{ dx } = 2 \text{cosec }2x$ .

Solution:

We have,
y = $\frac{1}{2} \log \left( \frac{1 - \cos 2x }{1 + \cos 2x} \right)$
y = $\frac{1}{2}\log\left( \frac{2 \sin^2 x}{2 \cos^2 x} \right)$
y = $\frac{1}{2}\log\left( \tan^2 x \right)$
y = $\frac{2}{2}\log \tan x$
y = $\log \tan x$
On differentiating y with respect to x we get,
$\frac{d y}{d x} = \frac{d}{dx}\left(\log \tan x\right)$
$\frac{d y}{d x} = \frac{1}{\tan x} \times \frac{d}{dx}\left( \tan x \right)$
$\frac{d y}{d x} = \frac{\sec^2 x}{\tan x}$
$\frac{d y}{d x} = \frac{1}{\cos^2 x \times \frac{\sin x}{\cos x}}$
$\frac{d y}{d x} = \frac{1}{\sin x \cos x}$
$\frac{d y}{d x} = \frac{2}{2\sin x \cos x}$
$\frac{d y}{d x} = \frac{2}{\sin2x}$
$\frac{d y}{d x} = 2\cosec x$
Hence proved.

Question 69. If y = $x \sin^{- 1} x + \sqrt{1 - x^2}$ , prove that $\frac{dy}{dx} = \sin^{- 1} x$ .

Solution:

We have,
y = $x \sin^{- 1} x + \sqrt{1 - x^2}$
On differentiating y with respect to x we get,
$\frac{d y}{d x} = \frac{d}{dx}\left[ x \sin^{- 1} x + \sqrt{1 - x^2} \right]$
$\frac{d y}{d x} = \frac{d}{dx}\left( x \sin^{- 1} x \right) + \frac{d}{dx}\left( \sqrt{1 - x^2} \right)$
$\frac{d y}{d x} = \left[ x \frac{d}{dx} \sin^{- 1} x + \sin^{- 1} x\frac{d}{dx}\left( x \right) \right] + \frac{1}{2\sqrt{1 - x^2}}\frac{d}{dx}\left( 1 - x^2 \right)$
$\frac{d y}{d x} = \left[ \frac{x}{\sqrt{1 - x^2}} + \sin^{- 1} x \right] - \frac{2x}{2\sqrt{1 - x^2}}$
$\frac{d y}{d x} = \frac{x}{\sqrt{1 - x^2}} + \sin^{- 1} x - \frac{x}{\sqrt{1 - x^2}}$
$\frac{d y}{d x} = \sin^{- 1} x$
Hence proved.

Question 70. If y = $\sqrt{x^2 + a^2}$ , prove that $y\frac{dy}{dx} - x = 0$ .

Solution:

We have,
y = $\sqrt{x^2 + a^2}$
On squaring both sides we get,
$y^2 = x^2 + a^2$
On differentiating both sides with respect to x we get,
$2y\frac{d y}{d x} = \frac{d}{dx}\left( x^2 + a^2 \right)$
$2y\frac{d y}{d x} = \left( 2x \right)$
$2y\frac{d y}{d x} = 2x$
$y\frac{d y}{d x} = x$
$y\frac{d y}{d x} - x = 0$
Hence proved.

Question 71. If y = $e^x + e^{- x}$ , prove that $\frac{dy}{dx} = \sqrt{y^2 - 4}$ .

Solution:

We have,
y = $e^x + e^{- x}$
On differentiating both sides with respect to x we get,
$\frac{d y}{d x} = \frac{d}{dx}\left( e^x + e^{- x} \right)$
$\frac{d y}{d x} = \frac{d}{dx} e^x + \frac{d}{dx} e^{- x}$
On using chain rule, we get
$\frac{d y}{d x} = e^x + e^{- x} \frac{d}{dx}\left( - x \right)$
$\frac{d y}{d x} = e^x + e^{- x} \left( - 1 \right)$
$\frac{d y}{d x} = \left( e^x - e^{- x} \right)$
$\frac{d y}{d x} = \sqrt{\left( e^x + e^{- x} \right)^2 - 4 e^x \times e^{- x}}$
As $e^x + e^{- x} = y$ , we get
$\frac{d y}{d x} = \sqrt{y^2 - 4}$
Hence proved.

Question 72. If y = $\sqrt{a^2 - x^2}$ , prove that $y\frac{dy}{dx} + x = 0$ .

Solution:

We have,
y = $\sqrt{a^2 - x^2}$
On squaring both sides we get,
$y^2 = a^2 - x^2$
On differentiating both sides with respect to x we get,
$2y\frac{d y}{d x} = \frac{d}{dx}\left( a^2 - x^2 \right)$
$2y\frac{d y}{d x} = 0 - 2x$
$y\frac{d y}{d x} = - x$
$y\frac{d y}{d x} + x = 0$
Hence proved.

Question 73. If xy = 4, prove that $x\left( \frac{dy}{dx} + y^2 \right) = 3 y$ .

Solution:

We have,
xy = 4
y = $\frac{4}{x}$
On differentiating both sides with respect to x we get,
$\frac{d y}{d x} = \frac{d}{dx}\left( \frac{4}{x} \right)$
$\frac{d y}{d x} = 4\frac{d}{dx}\left( x^{- 1} \right)$
$\frac{d y}{d x} = 4\left( - 1 \times x^{- 1 - 1} \right)$
$\frac{d y}{d x} = 4\left( - \frac{1}{x^2} \right)$
$\frac{d y}{d x} = \frac{- 4}{x^2}$
As x = $\frac{4}{y}$ , we get
$\frac{d y}{d x} = - \frac{4}{\left( \frac{4}{y} \right)^2}$
$\frac{d y}{d x} = - \frac{4 y^2}{16}$
$\frac{d y}{d x} = - \frac{y^2}{4}$
$4\frac{d y}{d x} = - y^2$
$4\frac{d y}{d x} = 3 y^2 - 4 y^2$
$4\frac{d y}{d x} + 4 y^2 = 3 y^2$
$4\left( \frac{d y}{d x} + y^2 \right) = 3 y^2$
On dividing both sides by x, we get
$\frac{4}{x}\left( \frac{d y}{d x} + y^2 \right) = \frac{3 y^2}{x}$
$y\left( \frac{d y}{d x} + y^2 \right) = \frac{3 y^2}{x}$
$x\left( \frac{d y}{d x} + y^2 \right) = \frac{3 y^2}{y}$
$x\left( \frac{d y}{d x} + y^2 \right) = 3y$
Hence proved.

Question 74. Prove that $\frac{d}{dx} \left\{ \frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2} \sin^{- 1} \frac{x}{a} \right\} = \sqrt{a^2 - x^2}$ .

Solution:

We have,
L.H.S. = $\frac{d}{dx} \left\{ \frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2} \sin^{- 1} \frac{x}{a} \right\}$
= $\frac{d}{dx}\left( \frac{x}{2}\sqrt{a^2 - x^2} \right) + \frac{d}{dx}\left( \frac{a^2}{2} \sin^{- 1} \frac{x}{a} \right)$
= $\frac{1}{2}\left[ x\frac{d}{dx}\sqrt{a^2 - x^2} + \sqrt{a^2 - x^2}\frac{d}{dx}\left( x \right) \right] + \frac{a^2}{2} \times \frac{1}{\sqrt{1 - \left( \frac{x}{a} \right)^2}} \times \frac{d}{dx}\left( \frac{x}{a} \right)$
= $\frac{1}{2}\left[ x \times \frac{1}{2\sqrt{a^2 - x^2}}\frac{d}{dx}\left( a^2 - x^2 \right) + \sqrt{a^2 - x^2} \right] + \left[ \frac{a^2}{2} \right] \times \frac{1}{\sqrt{\frac{a^2 - x^2}{a^2}}} \times \left( \frac{1}{a} \right)$
= $\frac{1}{2}\left[ \frac{x\left( - 2x \right)}{2\sqrt{a^2 - x^2}} + \sqrt{a^2 - x^2} \right] + \left( \frac{a^2}{2} \right)\frac{a}{\sqrt{a^2 - x^2}} \times \left( \frac{1}{a} \right)$
= $\frac{1}{2}\left[ \frac{- 2 x^2 + 2\left( a^2 - x^2 \right)}{2\sqrt{a^2 - x^2}} \right] + \frac{a^2}{2\sqrt{a^2 - x^2}}$
= $\frac{1}{2}\left[ \frac{2\left( a^2 - 2 x^2 \right)}{2\sqrt{a^2 - x^2}} \right] + \frac{a^2}{2\sqrt{a^2 - x^2}}$
= $\frac{a^2 - 2 x^2}{2\sqrt{a^2 - x^2}} + \frac{a^2}{2\sqrt{a^2 - x^2}}$
= $\frac{a^2 - 2 x^2 + a^2}{2\sqrt{a^2 - x^2}}$
= $\frac{2 a^2 - 2 x^2}{2\sqrt{a^2 - x^2}}$
= $\frac{2\left( a^2 - x^2 \right)}{2\sqrt{a^2 - x^2}}$
= $\frac{\left( a^2 - x^2 \right)}{\sqrt{a^2 - x^2}}$
= $\sqrt{a^2 - x^2}$
= R.H.S.
Hence proved.

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Last Updated : 14 Jul, 2021

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Class 12 RD Sharma Solutions – Chapter 11 Differentiation – Exercise 11.2 | Set 3

Question 50. Differentiate y = $3 e^{- 3x} \log \left( 1 + x \right)$ with respect to x.

Question 51. Differentiate y = $\frac{x^2 + 2}{\sqrt{\cos x}}$ with respect to x.

Question 52. Differentiate y = $\frac{x^2 \left( 1 - x^2 \right)}{\cos 2x}$ with respect to x.

Question 53. Differentiate y = $\log\left\{ \cot\left( \frac{\pi}{4} + \frac{x}{2} \right) \right\}$ with respect to x.

Question 54. Differentiate y = $e^{ax} \sec x \tan2x$ with respect to x.

Question 55. Differentiate y = $\log \left( \cos x^2 \right)$ with respect to x.

Question 56. Differentiate y = $\cos \left( \log x \right)^2$ with respect to x.

Question 57. Differentiate y = $\log \sqrt{\frac{x - 1}{x + 1}}$ with respect to x.

Question 58. If $y = \log \left\{ \sqrt{x - 1} - \sqrt{x + 1} \right\}$ , show that $\frac{dy}{dx} = \frac{- 1}{2\sqrt{x^2 - 1}}$ .

Question 59. If y = $\sqrt{x + 1} + \sqrt{x - 1}$ , prove that $\sqrt{x^2 - 1}\frac{dy}{dx} = \frac{1}{2}y$ .

Question 60. If y = $\frac{x}{x + 2}$ , prove that $x\frac{dy}{dx} = \left( 1 - y \right) y$ .

Question 61. If y = $\log \left( \sqrt{x} + \frac{1}{\sqrt{x}} \right)$ , prove that $\frac{dy}{dx} = \frac{x - 1}{2x \left( x + 1 \right)}$ .

Question 62. If y = $\log \sqrt{\frac{1 + \tan x}{1 - \tan x}}$ , prove that $\frac{dy}{dx} = \sec 2x$ .

Question 63. If y = $\sqrt{x} + \frac{1}{\sqrt{x}}$ , prove that $2 x\frac{dy}{dx} = \sqrt{x} - \frac{1}{\sqrt{x}}$ .

Question 64. If y = $\frac{x \sin^{- 1} x}{\sqrt{1 - x^2}}$ , prove that $\left( 1 - x^2 \right) \frac{dy}{dx} = x + \frac{y}{x}$ .

Question 65. If y = $\frac{e^x - e^{- x}}{e^x + e^{- x}}$ , prove that $\frac{dy}{dx} = 1 - y^2$ .

Question 66. If y = $\left( x - 1 \right) \log \left( x - 1 \right) - \left( x + 1 \right) \log \left( x + 1 \right)$ , prove that $\frac{dy}{dc} = \log \left( \frac{x - 1}{1 + x} \right)$ .

Question 67. If y = $e^x \cos x$ , prove that $\frac{dy}{dx} = \sqrt{2} e^x \cdot \cos \left( x + \frac{\pi}{4} \right)$ .

Question 68. If y = $\frac{1}{2} \log \left( \frac{1 - \cos 2x }{1 + \cos 2x} \right)$ , prove that $\frac{ dy }{ dx } = 2 \text{cosec }2x$ .

Question 69. If y = $x \sin^{- 1} x + \sqrt{1 - x^2}$ , prove that $\frac{dy}{dx} = \sin^{- 1} x$ .

Question 70. If y = $\sqrt{x^2 + a^2}$ , prove that $y\frac{dy}{dx} - x = 0$ .

Question 71. If y = $e^x + e^{- x}$ , prove that $\frac{dy}{dx} = \sqrt{y^2 - 4}$ .

Question 72. If y = $\sqrt{a^2 - x^2}$ , prove that $y\frac{dy}{dx} + x = 0$ .

Question 73. If xy = 4, prove that $x\left( \frac{dy}{dx} + y^2 \right) = 3 y$ .

Question 74. Prove that $\frac{d}{dx} \left\{ \frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2} \sin^{- 1} \frac{x}{a} \right\} = \sqrt{a^2 - x^2}$ .

Data Structures and Algorithms - Self Paced

CBSE Class 12 Computer Science

GATE CS & IT 2024

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Class 12 RD Sharma Solutions – Chapter 11 Differentiation – Exercise 11.2 | Set 3

Question 50. Differentiate y = with respect to x.

Question 51. Differentiate y = with respect to x.

Question 52. Differentiate y = with respect to x.

Question 53. Differentiate y = with respect to x.

Question 54. Differentiate y = with respect to x.

Question 55. Differentiate y = with respect to x.

Question 56. Differentiate y = with respect to x.

Question 57. Differentiate y = with respect to x.

Question 58. If , show that .

Question 59. If y = , prove that .

Question 60. If y = , prove that .

Question 61. If y = , prove that .

Question 62. If y = , prove that .

Question 63. If y = , prove that .

Question 64. If y = , prove that .

Question 65. If y = , prove that .

Question 66. If y = , prove that .

Question 67. If y = , prove that .

Question 68. If y = , prove that .

Question 69. If y = , prove that .

Question 70. If y = , prove that .

Question 71. If y = , prove that .

Question 72. If y = , prove that .

Question 73. If xy = 4, prove that .

Question 74. Prove that .

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Data Structures and Algorithms - Self Paced

CBSE Class 12 Computer Science

GATE CS & IT 2024

Question 50. Differentiate y = $3 e^{- 3x} \log \left( 1 + x \right)$ with respect to x.

Question 51. Differentiate y = $\frac{x^2 + 2}{\sqrt{\cos x}}$ with respect to x.

Question 52. Differentiate y = $\frac{x^2 \left( 1 - x^2 \right)}{\cos 2x}$ with respect to x.

Question 53. Differentiate y = $\log\left\{ \cot\left( \frac{\pi}{4} + \frac{x}{2} \right) \right\}$ with respect to x.

Question 54. Differentiate y = $e^{ax} \sec x \tan2x$ with respect to x.

Question 55. Differentiate y = $\log \left( \cos x^2 \right)$ with respect to x.

Question 56. Differentiate y = $\cos \left( \log x \right)^2$ with respect to x.

Question 57. Differentiate y = $\log \sqrt{\frac{x - 1}{x + 1}}$ with respect to x.

Question 58. If $y = \log \left\{ \sqrt{x - 1} - \sqrt{x + 1} \right\}$ , show that $\frac{dy}{dx} = \frac{- 1}{2\sqrt{x^2 - 1}}$ .

Question 59. If y = $\sqrt{x + 1} + \sqrt{x - 1}$ , prove that $\sqrt{x^2 - 1}\frac{dy}{dx} = \frac{1}{2}y$ .

Question 60. If y = $\frac{x}{x + 2}$ , prove that $x\frac{dy}{dx} = \left( 1 - y \right) y$ .

Question 61. If y = $\log \left( \sqrt{x} + \frac{1}{\sqrt{x}} \right)$ , prove that $\frac{dy}{dx} = \frac{x - 1}{2x \left( x + 1 \right)}$ .

Question 62. If y = $\log \sqrt{\frac{1 + \tan x}{1 - \tan x}}$ , prove that $\frac{dy}{dx} = \sec 2x$ .

Question 63. If y = $\sqrt{x} + \frac{1}{\sqrt{x}}$ , prove that $2 x\frac{dy}{dx} = \sqrt{x} - \frac{1}{\sqrt{x}}$ .

Question 64. If y = $\frac{x \sin^{- 1} x}{\sqrt{1 - x^2}}$ , prove that $\left( 1 - x^2 \right) \frac{dy}{dx} = x + \frac{y}{x}$ .

Question 65. If y = $\frac{e^x - e^{- x}}{e^x + e^{- x}}$ , prove that $\frac{dy}{dx} = 1 - y^2$ .

Question 66. If y = $\left( x - 1 \right) \log \left( x - 1 \right) - \left( x + 1 \right) \log \left( x + 1 \right)$ , prove that $\frac{dy}{dc} = \log \left( \frac{x - 1}{1 + x} \right)$ .

Question 67. If y = $e^x \cos x$ , prove that $\frac{dy}{dx} = \sqrt{2} e^x \cdot \cos \left( x + \frac{\pi}{4} \right)$ .

Question 68. If y = $\frac{1}{2} \log \left( \frac{1 - \cos 2x }{1 + \cos 2x} \right)$ , prove that $\frac{ dy }{ dx } = 2 \text{cosec }2x$ .

Question 69. If y = $x \sin^{- 1} x + \sqrt{1 - x^2}$ , prove that $\frac{dy}{dx} = \sin^{- 1} x$ .

Question 70. If y = $\sqrt{x^2 + a^2}$ , prove that $y\frac{dy}{dx} - x = 0$ .

Question 71. If y = $e^x + e^{- x}$ , prove that $\frac{dy}{dx} = \sqrt{y^2 - 4}$ .

Question 72. If y = $\sqrt{a^2 - x^2}$ , prove that $y\frac{dy}{dx} + x = 0$ .

Question 73. If xy = 4, prove that $x\left( \frac{dy}{dx} + y^2 \right) = 3 y$ .

Question 74. Prove that $\frac{d}{dx} \left\{ \frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2} \sin^{- 1} \frac{x}{a} \right\} = \sqrt{a^2 - x^2}$ .