Differentiate the following with respect to x :

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

Put $x=\tan \theta$

$y=\cot ^{-1}\left(\frac{1-\tan \theta}{1+\tan \theta}\right)$

$y=\cot ^{-1}\left(\frac{\tan \left(\frac{\pi}{4}\right)-\tan \theta}{1+\tan \left(\frac{\pi}{4}\right) \tan \theta}\right)$

Using, $\tan (x-y)=\left(\frac{\tan x-\tan y}{1+\tan x \tan y}\right)$

$y=\cot ^{-1}\left(\tan \left(\frac{\pi}{4}-\theta\right)\right)$

$y=\cot ^{-1}\left(\cot \left(\frac{\pi}{2}-\frac{\pi}{4}+\theta\right)\right)$

$y=\frac{\pi}{4}+\theta$

$y=\frac{\pi}{4}+\tan ^{-1} x$

Differentiating w.r.t $x$ we get

$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{\pi}{4}+\tan ^{-1} \mathrm{x}\right)$

$\frac{\mathrm{dy}}{\mathrm{dx}}=0+\frac{1}{1+\mathrm{x}^{2}}$

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