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

Answer: (A)

Let $y={\sin }^{-1}\frac{2x}{1+x^2},z={\sin }^{-1}\frac{1-x^2}{1+x^2}$
Put $x=\tan \theta \therefore y={\sin }^{-1}\left(\sin 2\theta \right)=2\theta$
$z={\sin }^{-1}\left(\cos 2\theta \right)$
$={\sin }^{-1}\left(\sin \left(\frac{\pi }{2}-2\theta \right)\right)=\frac{\pi }{2}-2\theta$
$\therefore \frac{dy}{d\theta }=2,\frac{dz}{d\theta }=-2\therefore \frac{dy}{dz}=\frac{2}{-2}=-1$