Question

If $y={\tan }^{-1}\sqrt{\frac{1-x}{1+x}}$ for $0<x<1$ then $\frac{dy}{dx}=$

Answer 1

$y={\tan }^{-1}\sqrt{\frac{1-x}{1+x}}$

$\frac{dy}{dx}=\frac{1}{1+(\sqrt{\frac{1-x}{1+x}}{)}^2}.\frac{d}{dx}\sqrt{\frac{1-x}{1+x}}$

$\frac{dy}{dx}=\frac{1+x}{1+x+1-x}.\frac{1}{2}\sqrt{\frac{1+x}{1-x}}.\left[\frac{-(1+x)-(1-x)}{(1+x{)}^2}\right]$

$=\frac{1}{2}.\frac{1}{2}\sqrt{\frac{1+x}{1-x}}.\left[\frac{-2}{1+x}\right]$

$=\frac{-1}{2}\frac{1}{\sqrt{1-x}.\sqrt{1+x}}=\frac{-1}{2(\sqrt{1-x}{)}^2}-\frac{\sqrt{1-x}}{1+x}$

$\frac{dy}{dx}=\frac{-1}{2(1-x)}\tan y$

Therefore $\frac{dy}{dx}=\frac{-1}{2(1-x)}\tan y$