Question

$\frac{d}{dx}\left({\tan }^{-1}\frac{\sqrt{1+x^2}-1}{x}\right)$

• A. $\frac{1}{2}.\frac{1}{1+x^2}$
• B. $\frac{1}{4}.\frac{1}{1+x^2}$
• C. $\frac{1}{2}.\frac{1}{1-x^2}$
• D. $-\frac{1}{2}.\frac{1}{1+x^2}$

Answer 1

The correct option is A $\frac{1}{2}.\frac{1}{1+x^2}$

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

$\therefore y={\tan }^{-1}\frac{\sec \theta -1}{\tan \theta }={\tan }^{-1}\frac{1-\cos \theta }{\sin \theta }$

or $y={\tan }^{-1}\tan \frac{\theta }{2}=\frac{\theta }{2}=\frac{1}{2}{\tan }^{-1}x$

$\therefore \frac{dy}{dx}=\frac{1}{2}.\frac{1}{1+x^2}$