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

Answer 1

The correct option is C $\frac{1}{4}$

Let $y={\tan }^{-1}\left(\frac{\sqrt{1+x^2}-1}{x}\right)$ and $z={\tan }^{-1}\left(\frac{2x\sqrt{1-x^2}}{1-2x^2}\right)$
Putting $x=\tan \theta$ in $y$, we get
$y={\tan }^{-1}\left(\frac{\sec \theta -1}{\tan \theta }\right)$
$\Rightarrow y={\tan }^{-1}\left(\frac{1-\cos \theta }{\sin \theta }\right)$
$\Rightarrow y={\tan }^{-1}\left(\tan \frac{\theta }{2}\right)=\frac{\theta }{2}$
$\to y=\frac{1}{2}{\tan }^{-1}x$
$\Rightarrow \frac{dy}{dx}=\frac{1}{2(1+x^2)}$

Putting $x=\sin \theta$ in $z$, we get
$z={\tan }^{-1}\left(\frac{2\sin \theta \cos \theta }{1-2{\sin }^2\theta }\right)$
$\Rightarrow z={\tan }^{-1}\frac{\sin 2\theta }{\cos 2\theta }\Rightarrow z=2\theta =2{\sin }^{-1}x$
$\Rightarrow \frac{dz}{dx}=\frac{2}{\sqrt{1-x^2}}$

$\frac{dy}{dz}=\frac{\sqrt{1-x^2}}{4(1+x^2)}$
at $x=0$, $\Rightarrow \frac{dy}{dz}=\frac{1}{4}$