Question

If $y={\tan }^{-1}\frac{\sqrt{1+x^2}-1}{x}$, then $y^{\prime }\left(1\right)$ is equal to

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

Answer 1

Answer: (A)

$\frac{1}{4}$

Given $y={\tan }^{-1}\frac{\sqrt{1+x^2}-1}{x}$
Put $x=\tan \theta$
$\therefore y={\tan }^{-1}\frac{\sqrt{{\sec }^2\theta }-1}{\tan \theta }$

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

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

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

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

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

$\Rightarrow y^{\prime }=\frac{1}{2(1+x^2)}\Rightarrow y^{\prime }\left(1\right)=\frac{1}{2(1+1)}=\frac{1}{4}$