Question

Differentiation of $\frac{d}{dx}\left({\sin }^{-1}\left(\frac{1-x^2}{1+x^2}\right)\right)$ equals, if $0<x<1$ :

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

Answer 1

The correct option is B $-\frac{2}{1+x^2}$

Let $y={\sin }^{-1}\left(\frac{1-x^2}{1+x^2}\right)$

Put $x=\tan t$ .....(i)

Therefore, $y={\sin }^{-1}\left(\frac{1-{\left(\tan t\right)}^2}{1+{\left(\tan t\right)}^2}\right)$

$\Rightarrow y={\sin }^{-1}\left(\frac{{\cos }^{2}t-{\sin }^{2}t}{{\cos }^{2}t+{\sin }^{2}t}\right)\Rightarrow y={\sin }^{-1}({\cos }^{2}t-{\sin }^{2}t)\Rightarrow y={\sin }^{-1}\left(\cos 2t\right)\Rightarrow y={\sin }^{-1}\sin (\frac{\pi }{2}-2t)\Rightarrow y=\frac{\pi }{2}-2t$

From (i), $t={\tan }^{-1}x$

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

$\Rightarrow \frac{dy}{dx}=\frac{d}{dx}\left(\frac{\pi }{2}\right)-2\frac{d}{dx}\left({\tan }^{-1}x\right)\Rightarrow \frac{dy}{dx}=0-2\frac{1}{1+x^2}\Rightarrow \frac{dy}{dx}=-\frac{2}{1+x^2}$

So, option B is correct.