Let us assume that,

 $y={\tan }^{-1}\left[\frac{\sqrt{1+x^2}-1}{x}\right]$

Put :  $x=\tan \theta$

 $\begin{array}{l}\therefore \frac{\sqrt{1+x^2}-1}{x}=\frac{\sqrt{1+{\tan }^2\theta }-1}{\tan \theta }=\frac{\sec \theta -1}{\tan \theta }\\ =\frac{1-\cos \theta }{\sin \theta }=\frac{2{\sin }^2\frac{\theta }{2}}{2\sin \frac{\theta }{2}\cos \frac{\theta }{2}}\\ =\tan \frac{\theta }{2}\end{array}$

On differentiating with respect to x:

 $\begin{array}{l}\therefore y={\tan }^{-1}\left(\tan \frac{\theta }{2}\right)=\frac{1}{2}\theta =\frac{1}{2}{\tan }^{-1}x\\ \Rightarrow \frac{dy}{dx}=\frac{1}{2}.\frac{1}{1+x^2}.\end{array}$