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(\sin2 \theta\right) = 2\theta$
$ z = \sin^{-1} \left(\cos2\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$