Question

Differentiate ${\sin }^{-1}\frac{1-x}{1+x}$ w.r.t. to $\sqrt{x}.$

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

Answer 1

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

Let $y={\sin }^{-1}\frac{1-x}{1+x}$ and $z=\sqrt{x}.$

Put $\sqrt{x}=\tan \theta \Rightarrow x={\tan }^2\theta$

$\therefore y={\sin }^{-1}\cos 2\theta =\frac{\pi }{2}-2\theta =\frac{\pi }{2}-2{\tan }^{-1}\sqrt{x}$

$\Rightarrow y=\frac{\pi }{2}-2{\tan }^{-1}z,$ where z = $\sqrt{x}$

$\therefore \frac{dy}{dz}=-\frac{2}{1+z^2}=-\frac{2}{1+x}$