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 $x=0$

• (A) $\frac{1}{8}$
• (B) $\frac{1}{4}$
• (C) $\frac{1}{2}$
• (D) 6

Answer 1

The Correct Option is B

Explanation:
$$\begin{array}{rl}\text{ Let }y& ={\tan }^{-1}\left(\frac{\sqrt{1+x^2}-1}{x}\right)\text{ and }\\ z& ={\tan }^{-1}\left(\frac{2x\sqrt{1-x^2}}{1-2x^2}\right)\end{array}$$Put $x=\tan \theta$ in $y,$ we get$$\begin{array}{rl}y& ={\tan }^{-1}\left(\frac{\sqrt{{\sec }^2\theta }-1}{\tan \theta }\right)\\ & ={\tan }^{-1}\left(\frac{\sec \theta -1}{\tan \theta }\right)={\tan }^{-1}(\tan \frac{\theta }{2})\\ & =\frac{1}{2}{\tan }^{-1}x\\ \Rightarrow \frac{dy}{dx}& =\frac{1}{2(1+x^2)}\end{array}$$Put $x=\sin \theta$ in $z,$ we get$$\begin{array}{l}\begin{array}{rl}z& ={\tan }^{-1}\left(\frac{2\sin \theta \sqrt{{\cos }^2\theta }}{1-2{\sin }^2\theta }\right)\\ & ={\tan }^{-1}\left(\frac{2\sin \theta \cos \theta }{\cos 2\theta }\right)\\ & ={\tan }^{-1}(\tan 2\theta )\\ & =2\theta \\ & z=2{\sin }^{-1}x\end{array}\\ \begin{array}{c}\Rightarrow \frac{dz}{dx}=\frac{2}{\sqrt{1-x^2}}\\ \text{ Thus, }\frac{dy}{dz}=\frac{dy/dx}{dz/dx}=\frac{1}{2(1+x^2)}\times \frac{\sqrt{1-x^2}}{2}\end{array}\end{array}$$At$$x=0,\frac{dy}{dz}=\frac{1}{4}$$