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 $x=\frac{1}{2}$ is :

• A. $\frac{2\sqrt{3}}{3}$
• B. $\frac{2\sqrt{3}}{5}$
• C. $\frac{\sqrt{3}}{12}$
• D. $\frac{\sqrt{3}}{10}$

Answer 1

The correct option is D $\frac{\sqrt{3}}{10}$

Let $x=\tan \theta$
$u={\tan }^{-1}\left(\frac{\sec \theta -1}{\tan \theta }\right)={\tan }^{-1}\left(\tan \frac{\theta }{2}\right)=\frac{\theta }{2}$
$\Rightarrow u=\frac{{\tan }^{-1}x}{2}$
$\Rightarrow \frac{du}{dx}=\frac{1}{2(1+x^2)}\cdots \left(i\right)$

Let $x=\sin \theta$
$v={\tan }^{-1}\left(\frac{2\sin \theta \cos \theta }{\cos 2\theta }\right)=2\theta$
$\Rightarrow v=2{\sin }^{-1}x$
$\Rightarrow \frac{dv}{dx}=\frac{2}{\sqrt{1-x^2}}\cdots \left(ii\right)$

Equation $\left(i\right)/\left(ii\right)$
$\frac{du}{dv}=\frac{1}{2(1+x^2)}\times \frac{\sqrt{1-x^2}}{2}$
${\left[\frac{du}{dv}\right]}_{x=1/2}=\frac{\sqrt{3}}{2\times 2}\times \frac{4}{5\times 2}$
$\therefore {\left[\frac{du}{dv}\right]}_{x=1/2}=\frac{\sqrt{3}}{10}$