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Question: If y = tan<sup>–1</sup>\(\sqrt{\frac{x + 1}{x - 1}}\) then \(\frac{dy}{dx}\) is equal to...

If y = tan–1x+1x−1\sqrt{\frac{x + 1}{x - 1}} then dydx\frac{dy}{dx} is equal to

A

−12∣x∣x2−1\frac{- 1}{2|x|\sqrt{x^{2} - 1}}

B

−12xx2−1\frac{- 1}{2x\sqrt{x^{2} - 1}}

C

12xx2−1\frac{1}{2x\sqrt{x^{2} - 1}}

D

None of these

Answer

−12∣x∣x2−1\frac{- 1}{2|x|\sqrt{x^{2} - 1}}

Explanation

Solution

Let x = sec θ

So y = tan–1sec⁡θ+1sec⁡θ−1\sqrt{\frac{\sec\theta + 1}{\sec\theta - 1}} = tan–11+cos⁡θ1−cos⁡θ\sqrt{\frac{1 + \cos\theta}{1 - \cos\theta}} = tan–1(cot⁡θ2)\left( \cot\frac{\theta}{2} \right)

= tan–1(tan⁡(π2−θ2))\left( \tan\left( \frac{\pi}{2} - \frac{\theta}{2} \right) \right) = π2\frac{\pi}{2} – 12\frac{1}{2} sec–1x

dydx\frac{dy}{dx} = –12\frac{1}{2}×1∣x∣x2−1\frac{1}{|x|\sqrt{x^{2} - 1}}