Question

The derivative of ${\sin }^{-1}\left(\frac{2x}{1+x^2}\right)$ w.r.t ${\sin }^{-1}\left(\frac{1-x^2}{1+x^2}\right)$ is

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

Answer 1

The correct option is A $-1$

Let $y={\sin }^{-1}\frac{2x}{1+x^2},z={\sin }^{-1}\frac{1-x^2}{1+x^2}$

Put $x=\tan \theta$

$\therefore y={\sin }^{-1}\left(\sin 2\theta \right)=2\theta ,$

$z={\sin }^{-1}\left(\cos 2\theta \right)={\sin }^{-1}\left[\sin (\frac{\pi }{2}-2\theta )\right]=\frac{\pi }{2}-2\theta$

$\therefore \frac{dy}{d\theta }=2,\frac{dz}{d\theta }=-2$

$\therefore \frac{dy}{dz}=\frac{dy}{d\theta }\times \frac{d\theta }{dz}=\frac{2}{-2}=-1$