d d x sin 2 c o t - 1 1 + x 1 - x =

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

Put $x=\cos 2\theta$

$\frac{dx}{d\theta }=-2\sin 2\theta$

$\sqrt{\frac{1-\cos 2\theta }{1+\cos 2\theta }}=\tan \theta$

$y={\sin }^2{\cot }^{-1}\tan \theta$

$y={\sin }^2{\cot }^{-1}cot\left(\frac{\mathrm{\pi }}{2}-\theta \right)$

$y={\sin }^2\left(\frac{\mathrm{\pi }}{2}-\theta \right)$

$y={\cos }^2\theta$

$\frac{dy}{d\theta }=2\cos \theta .-\sin \theta$

$\frac{dy}{dx}=\frac{{\displaystyle \frac{dy}{d\theta }}}{{\displaystyle \frac{dx}{d\theta }}}=\frac{1}{2}$

$\frac{dy}{dx}=\frac{1}{2}$