Question

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

• A. $\frac{\sqrt{1+x^2}-1}{x^2}$
• B. $1$
• C. $\frac{1}{1+x^2}$
• D. none of these

Answer 1

The correct option is D none of these

Let $u={\tan }^{-1}\frac{\sqrt{1+x^2}-1}{x}$
Substitute $x=\tan \theta$
$u={\tan }^{-1}\left(\frac{\sec \theta -1}{\tan \theta }\right)$
$={\tan }^{-1}\left(\frac{1-\cos \theta }{\sin \theta }\right)$
$={\tan }^{-1}\left(\tan \frac{\theta }{2}\right)$
$\Rightarrow u=\frac{\theta }{2}$
$\Rightarrow u=\frac{1}{2}{\tan }^{-1}x$ ....(1)
Let $v={\tan }^{-1}x$ .....(2)
From (1) and (2),it follows
$\Rightarrow u=\frac{v}{2}$
$\Rightarrow \frac{du}{dv}=\frac{1}{2}$