Differentiate the following functions with respect to x :

$y=\tan ^{-1}\left\{\frac{\sin x}{1+\cos x}\right\}$

Function $y$ is defined for all real numbers where $\cos x \neq-1$

Using $2 \cos ^{2} \theta=1+\cos 2 \theta$ and $2 \sin \theta \cos \theta=\sin 2 \theta$

$y=\tan ^{-1}\left\{\frac{2 \sin \frac{x}{2} \cos \frac{x}{2}}{2 \cos ^{2} \frac{x}{2}}\right\}$

$y=\tan ^{-1}\left\{\tan \frac{x}{2}\right\}$

$y=\frac{x}{2}$

Differentiating w.r.t $x$, we get

$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{\mathrm{x}}{2}\right)$

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