Question

If $y={\tan }^{-1}\frac{\sqrt{x+1}}{\sqrt{x-1}}$ then $\frac{dy}{dx}$ is equal to

• A. $\frac{-1}{2\left|x\right|\sqrt{x^2-1}}$
• B. $\frac{-1}{2x\sqrt{x^2-1}}$
• C. $\frac{1}{2x\sqrt{x^2-1}}$
• D. none of these

Answer 1

Answer: (B)

$\frac{-1}{2x\sqrt{x^2-1}}$

$y={\tan }^{-1}\frac{\sqrt{x+1}}{\sqrt{x-1}}$
$\frac{dy}{dx}=\frac{1}{(1+\frac{x+1}{x-1})}\frac{d}{dx}\left(\frac{\sqrt{x+1}}{\sqrt{x-1}}\right)$
$=\frac{x-1}{2x}\left(\frac{-2}{2(x-1)\sqrt{x^2-1}}\right)$
$=\frac{-1}{2x\sqrt{x^2-1}}$