Question

If $y={\tan }^{-1}\left[\frac{\sqrt{1+x^2}-1}{x}\right]$ prove that $\frac{dy}{dx}=\frac{1}{2(1+x^2)}$

Answer 1

We have,

$\begin{array}{c}y={\tan }^{-1}\left(\frac{\sqrt{1+x^2}-1}{x}\right)\\ Let,x=\tan \theta v={\tan }^{-1}x\\ y={\tan }^{-1}\left(\frac{\sec \theta -1}{\tan \theta }\right)\\ y={\tan }^{-1}\left(\frac{1-\cos \theta }{\sin \theta }\right)\\ y={\tan }^{-1}\left(\tan \frac{\theta }{2}\right)=\frac{\theta }{2}\\ y=\frac{{\tan }^{-1}x}{2}\\ \frac{dy}{dx}=\frac{1}{2\left(1+x^2\right)}\end{array}$.