Question

The derivative of ${\tan }^{-1}\left(\frac{\sqrt{1+x^2}-1}{x}\right)$ with respect to ${\tan }^{-1}\left(\frac{2x\sqrt{\left(1-x^2\right)}}{\left(1-2x^2\right)}\right)$ 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

$\frac{\sqrt{3}}{10}$

Explanation for the correct option:

Step 1: Differentiate ${\tan }^{-1}\left(\frac{\sqrt{1+x^2}-1}{x}\right)$with respect to $x$

Let $u={\tan }^{-1}\left(\frac{\sqrt{1+x^2}-1}{x}\right)$

Put $x=\tan \theta$. Then $\theta ={\tan }^{-1}x$

Therefore, $u={\tan }^{-1}\left(\frac{\sqrt{1+{\tan }^2\theta }-1}{\tan \theta }\right)$

$={\tan }^{-1}\left(\frac{\sqrt{se{c}^2\theta }-1}{\tan \theta }\right)$

$={\tan }^{-1}\left(\frac{sec\theta -1}{\tan \theta }\right)$

$={\tan }^{-1}\left(\frac{{\displaystyle \frac{1}{\cos \theta }}-1}{{\displaystyle \frac{\sin \theta }{\cos \theta }}}\right)$

$={\tan }^{-1}\left(\frac{{\displaystyle \frac{1-\cos \theta }{\cos \theta }}}{{\displaystyle \frac{\sin \theta }{\cos \theta }}}\right)$

$={\tan }^{-1}\left(\frac{{\displaystyle 1-\cos \theta }}{{\displaystyle \sin \theta }}\right)$

$={\tan }^{-1}\left(\tan \frac{\theta }{2}\right)$

$=\frac{\theta }{2}$

$=\frac{{\tan }^{-1}x}{2}$

Now,$\frac{du}{dx}=\frac{d}{dx}\left(\frac{{\tan }^{-1}x}{2}\right)$

$=\frac{1}{2}\frac{1}{1+x^2}$

$=\frac{1}{2\left(1+x^2\right)}$

Step 2: Differentiate ${\tan }^{-1}\left(\frac{2x\sqrt{\left(1-x^2\right)}}{\left(1-2x^2\right)}\right)$with respect to $x$

Let $v={\tan }^{-1}\left(\frac{2x\sqrt{\left(1-x^2\right)}}{\left(1-2x^2\right)}\right)$

Put $x=\sin \theta$ Then $\theta ={\sin }^{-1}x$

Therefore, $v={\tan }^{-1}\left(\frac{2\sin \theta \sqrt{\left(1-{\sin }^2\theta \right)}}{\left(1-2{\sin }^2\theta \right)}\right)$

$={\tan }^{-1}\left(\frac{2\sin \theta \sqrt{{\cos }^2\theta }}{\cos 2\theta }\right)$

$={\tan }^{-1}\left(\frac{2\sin \theta \cos \theta }{\cos 2\theta }\right)$

$={\tan }^{-1}\left(\frac{\sin 2\theta }{\cos 2\theta }\right)$

$={\tan }^{-1}\left(\tan 2\theta \right)$

$=2\theta$

$=2{\sin }^{-1}x$

Now, $\frac{dv}{dx}=\frac{d}{dx}\left(2{\sin }^{-1}x\right)$

$=2\times \frac{1}{\sqrt{1-x^2}}$

$=\frac{2}{\sqrt{1-x^2}}$

Step 3: Differentiate ${\tan }^{-1}\left(\frac{\sqrt{1+x^2}-1}{x}\right)$ with respect to ${\tan }^{-1}\left(\frac{2x\sqrt{\left(1-x^2\right)}}{\left(1-2x^2\right)}\right)$

$\frac{du}{dv}=\raisebox{1ex}{${\displaystyle \frac{du}{dx}}$}\!\left/ \!\raisebox{-1ex}{$\frac{dv}{dx}$}\right.$

$=\raisebox{1ex}{${\displaystyle \frac{1}{2\left(1+x^2\right)}}$}\!\left/ \!\raisebox{-1ex}{$\frac{2}{\sqrt{1-x^2}}$}\right.$

$=\frac{\sqrt{1-x^2}}{4\left(1+x^2\right)}$

Therefore, $\frac{du}{dv}=\frac{\sqrt{1-x^2}}{4\left(1+x^2\right)}$

At $x=\frac{1}{2},\frac{du}{dv}=\frac{\sqrt{1-{\displaystyle \frac{1}{4}}}}{4\left(1+{\displaystyle \frac{1}{4}}\right)}$

$=\frac{\sqrt{{\displaystyle \frac{3}{4}}}}{4\left({\displaystyle \frac{5}{4}}\right)}$

$=\frac{\sqrt{{\displaystyle 3}}}{2}\times \frac{1}{5}$

$=\frac{\sqrt{{\displaystyle 3}}}{10}$

Hence option D is the correct option.