If $y = \cos^{-1}(\frac{1-x^2}{1+x^2}),0<x<1$, then $\frac{dy}{dx}$ is equal to

$\frac{2}{1+x^2}$

The correct answer is Option (1) → $\frac{2}{1+x^2}$

$y=\cos^{-1}\left(\frac{1-x^2}{1+x^2}\right)$

Let $\cos y=\frac{1-x^2}{1+x^2}$

Differentiate implicitly with respect to $x$

$-\sin y\frac{dy}{dx} =\frac{(1+x^2)(-2x)-(1-x^2)(2x)}{(1+x^2)^2}$

$=-\frac{4x}{(1+x^2)^2}$

Now

$\sin y=\sqrt{1-\cos^2 y}$

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

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

$=\frac{\sqrt{4x^2}}{1+x^2}$

$=\frac{2x}{1+x^2}$ since $0<\text{ x }<1$

Hence

$-\frac{2x}{1+x^2}\frac{dy}{dx} =-\frac{4x}{(1+x^2)^2}$

$\frac{dy}{dx}=\frac{2}{1+x^2}$

The value of $\frac{dy}{dx}$ is $\frac{2}{1+x^2}$.