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Question: If \(y = \sec^{- 1}\left( \frac{\sqrt{x} + 1}{\sqrt{x} - 1} \right) + \sin^{- 1}\left( \frac{\sqrt{x...

If y=sec⁡−1(x+1x−1)+sin⁡−1(x−1x+1)y = \sec^{- 1}\left( \frac{\sqrt{x} + 1}{\sqrt{x} - 1} \right) + \sin^{- 1}\left( \frac{\sqrt{x} - 1}{\sqrt{x} + 1} \right), then dydx=\frac{dy}{dx} =

A

0

B

1x+1\frac{1}{\sqrt{x} + 1}

C

1

D

None of these

Answer

0

Explanation

Solution

y=sec⁡−1(x+1x−1)+sin⁡−1(x−1x+1)y = \sec^{- 1}\left( \frac{\sqrt{x} + 1}{\sqrt{x} - 1} \right) + \sin^{- 1}\left( \frac{\sqrt{x} - 1}{\sqrt{x} + 1} \right)

= cos⁡−1(x−1x+1)+sin⁡−1(x−1x+1)=π2\cos^{- 1}\left( \frac{\sqrt{x} - 1}{\sqrt{x} + 1} \right) + \sin^{- 1}\left( \frac{\sqrt{x} - 1}{\sqrt{x} + 1} \right) = \frac{\pi}{2}

⇒ dydx=0\frac { d y } { d x } = 0 {∵sin⁡−1x+cos⁡−1x=π2}\left\{ \because\sin^{- 1}x + \cos^{- 1}x = \frac{\pi}{2} \right\}