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

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

Let $y={\tan }^{-1}\left(\frac{\sqrt{1+x^2}-1}{x}\right)$ and $z={\tan }^{-1}\left(\frac{2x\sqrt{1-x^2}}{1-2x^2}\right)$
Substitutin $x=\tan \theta$ in $y$, we get
$y={\tan }^{-1}\left(\frac{\sec \theta -1}{\tan \theta }\right)={\tan }^{-1}\left(\tan \frac{\theta }{2}\right)=\frac{1}{2}{\tan }^{-1}x$
$\therefore \frac{dy}{dx}=\frac{1}{2(1+x^2)}$
Substituting $x=\sin \theta$ in $z$, we get
$z={\tan }^{-1}\left(\frac{2\sin \theta \cos \theta }{\cos 2\theta }\right)={\tan }^{-1}\left(\tan 2\theta \right)=2\theta =2si{n}^{-1}x$
$\therefore \frac{dz}{dx}=\frac{2}{\sqrt{1-x^2}}$
Thus, $\frac{dy}{dz}=\frac{\frac{dy}{dx}}{\frac{dz}{dx}}=\frac{1}{4(1+x^2)}\sqrt{1-x^2}$ or ${\left(\frac{dy}{dz}\right)}_{x=0}=\frac{1}{4}$