Question

The derivative of ${\tan }^{-1}\left(\frac{\sqrt{1+x^2}-1}{x}\right)$ w.r.t. ${\tan }^{-1}\left(\frac{2x\sqrt{1-x^2}}{1-2x^2}\right)$ at x = 0 is

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

Answer 1

Answer: (B)

Let $y={\tan }^{-1}\left(\frac{\sqrt{1+x^2}-1}{x}\right)$ Put $x=\tan \theta$
$\therefore y={\tan }^{-1}\left(\frac{\sqrt{1+{\tan }^2\theta }-1}{\tan \theta }\right)$
$={\tan }^{-1}\left(\frac{\sec \theta -1}{\frac{\sin \theta }{\cos \theta }}\right)$
$={\tan }^{-1}\left(\frac{1-\cos \theta }{\sin \theta }\right)$
$={\tan }^{-1}\left(\frac{2{\sin }^2\frac{\theta }{2}}{2\sin \frac{\theta }{2}\cos \frac{\theta }{2}}\right)$
$={\tan }^{-1}\left(\tan \frac{\theta }{2}\right)=\frac{\theta }{2}=\frac{1}{2}{\tan }^{-1}x$
$\therefore \frac{dy}{dx}=\frac{1}{2{\left(1+x\right)}^2}$
Let $z={\tan }^{-1}\left(\frac{2x\sqrt{1-x^2}}{1-2x^2}\right)$ Put $x=\sin \varphi$
$\therefore z={\tan }^{-1}\left(\frac{2\sin \varphi \cos \varphi }{1-2{\sin }^2\varphi }\right)$
$={\tan }^{-1}\left(\frac{\sin 2\varphi }{\cos 2\varphi }\right)=2\varphi$
$\therefore z=2{\sin }^{-1}x\therefore \frac{dz}{dx}=\frac{2}{\sqrt{1-x^2}}$
$\therefore \frac{dy}{dz}=\frac{\frac{1}{2\left(1+x^2\right)}}{\frac{2}{\sqrt{1-x^2}}}=\frac{\sqrt{1-x^2}}{4\left(1+x^2\right)}$
At $x=0,\frac{dy}{dz}=\frac{\sqrt{1-0}}{4\left(1+0\right)}=\frac{1}{4}$