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

$\frac{1}{4}$

Explanation for the correct answer:

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

Step 1: To find $\frac{du}{dx}$

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Step 2: To find $\frac{dv}{dx}$

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

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

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

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

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

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

$={\tan }^{-1}\tan 2\theta$

$=2\theta$

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

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

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

Step 3: To find $\frac{du}{dv}$

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

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

$=\frac{1}{4}\times \frac{{\displaystyle \sqrt{1-x^2}}}{{\displaystyle 1+x^2}}$

Therefore, $\frac{du}{dv}=\frac{1}{4}\times \frac{{\displaystyle \sqrt{1-x^2}}}{{\displaystyle 1+x^2}}$

Now at $x=0,\frac{du}{dv}=\frac{{\displaystyle 1}}{{\displaystyle 4}}$

Hence, option B is the correct answer.