Question

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

• A. $\sqrt{\frac{1-x^2}{1+x^2}}$
• B. $\frac{1}{\sqrt{1-x^2}}$
• C. $\frac{2}{1+x^2}$
• D. $\frac{2\sqrt{1-x^2}}{1+x^2}$$={\cos }^{-1}\sqrt{{\cos }^{2}x}$

Answer 1

The correct option is D

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

Explanation for the correct option:

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

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

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

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

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

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

$=2\theta$

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

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

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

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

Let $v={\cos }^{-1}\sqrt{1-x^2}$

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

Therefore, $v={\cos }^{-1}\sqrt{1-{\sin }^2\theta }$

$={\cos }^{-1}\sqrt{{\cos }^2\theta }$

$={\cos }^{-1}\cos \theta$

$=\theta$

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

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

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

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

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

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

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

Hence, Option D is the correct answer.