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Question: The derivative of cos \(\left( 2\tan^{- 1}\left( \sqrt{\frac{1 - x}{1 + x}} \right) \right)\)– 2 cos...

The derivative of cos (2tan⁡−1(1−x1+x))\left( 2\tan^{- 1}\left( \sqrt{\frac{1 - x}{1 + x}} \right) \right)– 2 cos–1(1−x2)\left( \sqrt{\frac{1 - x}{2}} \right) w. r. to x is

A

1−11−x21 - \frac{1}{\sqrt{1 - x^{2}}}

B

1−11+x21 - \frac{1}{\sqrt{1 + x^{2}}}

C

2−11−x22 - \frac{1}{\sqrt{1 - x^{2}}}

D

2−11+x22 - \frac{1}{\sqrt{1 + x^{2}}}

Answer

1−11−x21 - \frac{1}{\sqrt{1 - x^{2}}}

Explanation

Solution

x = cosθ

⇒ θ = cos–1x

y = cos (2tan⁡−1(1−cos⁡θ1+cos⁡θ))\left( 2\tan^{- 1}\left( \sqrt{\frac{1 - \cos\theta}{1 + \cos\theta}} \right) \right)

– 2 cos–1 (1−cos⁡θ2)\left( \sqrt{\frac{1 - \cos\theta}{2}} \right)

= cos (2tan⁡−1(tan⁡θ2))\left( 2\tan^{- 1}\left( \tan\frac{\theta}{2} \right) \right)– 2 cos–1 (sin⁡θ2)\left( \sin\frac{\theta}{2} \right)

y = cosθ – 2 (π2−sin⁡−1(sin⁡θ2))\left( \frac{\pi}{2} - \sin^{- 1}\left( \sin\frac{\theta}{2} \right) \right)= cosθ – π + θ 

y = x – π + cos–1x

dydx=1−11−x2\frac{dy}{dx} = 1 - \frac{1}{{\sqrt{1 - x}}^{2}}