Question

The derivative of ${\tan }^{-1}\left(\frac{\sqrt{1+x^2}-1}{x}\right)$ with respect to ${\tan }^{-1}\left(\frac{2x\sqrt{1-x^2}}{1-2x^2}\right)$ at $x=0$, is

• A. $\frac{1}{8}$
• B. $\frac{1}{4}$
• C. $\frac{1}{2}$
• D. 1

Answer 1

Answer: (B)

Let $y={\tan }^{-1}\left(\frac{\sqrt{1+x^2}-1}{x}\right)$
Putting $x=\tan \theta$, we get
$y={\tan }^{-1}\left(\frac{\sec \theta -1}{\tan \theta }\right)={\tan }^{-1}\left(\tan \frac{\theta }{2}\right)$
$=\frac{1}{2}{\tan }^{-1}x$
On differentiating both sides w.r.t. $x$, we get
$\frac{dy}{dx}=\frac{1}{2\left(1+x^2\right)}$
and $z={\tan }^{-1}\left(\frac{2x\sqrt{1-x^2}}{1-2x^2}\right)$
Putting $x=\sin \theta$, we get
$z={\tan }^{-1}\left(\frac{2\sin \theta \cos \theta }{\cos 2\theta }\right)={\tan }^{-1}(\tan 2\theta )$
$\Rightarrow z=2\theta =2{\sin }^{-1}x$
On differentiating both sides w.r.t. $x$, we get
$\frac{dz}{dx}=\frac{2}{\sqrt{1-x^2}}$
Thus, $\frac{dy}{dz}=\frac{dy/dx}{dz/dx}=\frac{1}{4\left(1+x^2\right)}\sqrt{1-x^2}$
$\therefore {\left[\frac{dy}{dz}\right]}_{x=0}=\frac{1}{4}$