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

  • Last Updated : 14 Jul, 2021

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

Solution:

We have,

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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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