If $y = cos^{-1} \left( \dfrac{1-x^2}{1+x^2} \right)$, 0 < $x$ < 1, then find $\dfrac{dy}{dx}$

The correct answer is $\dfrac{dy}{dx}$ = $\dfrac{2}{1+x^2}$

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

Let $x = tan \theta$


∴ $y = cos^{-1} \left( \dfrac{1-tan^2 \theta}{1+tan^2 \theta} \right)$


⇒ $y = cos^{-1} \left( cos2 \theta \right)$ = $2 \theta$


Replacing $\theta$ by $tan^{-1}x$

$y = 2tan^{-1}x$


Differentiating w.r.t. $x$

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