Question

Derivative of ${\tan }^{-1}\left(\frac{\sqrt{1+x^2}-1}{x}\right)$ w.r.t. ${\tan }^{-1}x$is, (where $x\ne 0$)

• A. $\frac{1}{3}$
• B. $\frac{1}{2}$
• C. $\frac{1}{4}$
• D. $-\frac{1}{2}$

Answer 1

Answer: (B)

$\frac{1}{2}$

Let $u={\tan }^{-1}\left(\frac{\sqrt{1+x^2}-1}{x}\right)$ and $v={\tan }^{-1}x$.

Putting $x=\tan \theta$, we get

$u={\tan }^{-1}\left(\frac{\sqrt{1+{\tan }^2\theta }-1}{\tan \theta }\right)$
$u={\tan }^{-1}\left(\frac{\sec \theta -1}{\tan \theta }\right)$
$u={\tan }^{-1}\left(\frac{1-\cos \theta }{\sin \theta }\right)$
$u={\tan }^{-1}\left(\frac{2{\sin }^2\theta /2}{2\sin \frac{\theta }{2}\cos \frac{\theta }{2}}\right)$
$\therefore u={\tan }^{-1}\left(\tan \frac{\theta }{2}\right)=\frac{1}{2}\theta =\frac{1}{2}{\tan }^{-1}x$

Differentiate w.r. to $x$

$\frac{du}{dx}=\frac{1}{2(1+x^2)}$ and $\frac{dv}{dx}=\frac{1}{1+x^2}$
$\therefore \frac{du}{dv}=\frac{\frac{du}{dx}}{\frac{dv}{dx}}=\frac{1}{2(1+x^2)}.(1+x^2)=\frac{1}{2}$