Question

If $y={\tan }^{-1}\left[\frac{\sqrt{1+x^2}-1}{x}\right]$, then $\frac{dy}{dx}$?

Answer 1

We have,

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

Let $x=\tan \theta$

Therefore,

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

$y={\tan }^{-1}\left(\left(\frac{1-\cos \theta }{\cos \theta }\right)\frac{\cos \theta }{\sin \theta }\right)y={\tan }^{-1}\left(\frac{2{\sin }^2\left(\frac{\theta }{2}\right)}{2\sin \frac{\theta }{2}\cos \frac{\theta }{2}}\right)={\tan }^{-1}\left(\tan \frac{\theta }{2}\right)=\frac{\theta }{2}=\frac{1}{2}{\tan }^{-1}x\frac{dy}{dx}=\frac{1}{2\times \left(1+x^2\right)}$

Hence, this is the answer.