The derivative of ${\tan }^{-1}\frac{\sqrt{1+x^2}-1}{x}$ with respect to ${\tan }^{-1}x$ is

Let $y={\tan }^{-1}\frac{\sqrt{1+x^2}-1}{x}$

Put $x=\tan \theta$

$y={\tan }^{-1}\left(\frac{\sqrt{1+{\tan }^2\theta }-1}{\tan \theta }\right)$

$y={\tan }^{-1}\left(\frac{\sec \theta -1}{\tan \theta }\right)={\tan }^{-1}\left(\frac{1-\cos \theta }{\sin \theta }\right)$$={\tan }^{-1}\left(\frac{2{\sin }^2\frac{\theta }{2}}{2\sin \frac{\theta }{2}\cos \frac{\theta }{2}}\right)$

$y={\tan }^{-1}\left(\tan \frac{\theta }{2}\right)=\frac{\theta }{2}$

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

$u={\tan }^{-1}x\Rightarrow u={\tan }^{-1}\left(\tan \theta \right)$

$u=\theta \Rightarrow \frac{du}{d\theta }=1$

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