Question

Differentiate ${\tan }^{-1}\frac{\sqrt{1+x^2}-1}{x}$ w.r.t. ${\tan }^{-1}x.$

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

Answer 1

The correct option is A $\frac{1}{2}$

Let $y={\tan }^{-1}\frac{\sqrt{1+x^2}-1}{x}$ and
$z={\tan }^{-1}x\Rightarrow x=\tan z$, Now we have to find $\frac{dy}{dz}.$
$\therefore y={\tan }^{-1}\frac{\sqrt{1+{\tan }^2}z-1}{\tan z}={\tan }^{-1}\frac{1-\cos z}{\sin z}={\tan }^{-1}\frac{2{\sin }^2\left(\frac{z}{2}\right)}{2\sin \left(\frac{z}{2}\right).\cos \left(\frac{z}{2}\right)}$
$={\tan }^{-1}\tan \left(\frac{z}{2}\right)\therefore y=\frac{1}{2}z.$
$\therefore \frac{dy}{dz}=\frac{1}{2}.$