Q. The derivative of $\sin^{- 1}\left( \frac{2x}{1 + x^{2}} \right)$ with respect to $\cos^{- 1}\left( \frac{1 - x^{2}}{1 + x^{2}} \right)$ is
A) $- 1$
B) 1
C) 2
D) 4
Explanation
(b)
Let $p = \sin^{- 1}{\frac{2x}{1 + x^{2}} = 2\tan^{- 1}x}$
and $q = \cos^{- 1}{\frac{1 - x^{2}}{1 + x^{2}} = 2\tan^{- 1}x}$
$\Rightarrow \frac{dp}{dx} = \frac{2}{1 + x^{2}},\frac{dq}{dx} = \frac{2}{1 + x^{2}}$
$\therefore\frac{dp}{dq} = \frac{\frac{dp}{dx}}{\frac{dq}{dx}} = \frac{\frac{2}{(1 + x^{2})}}{\frac{2}{(1 + x^{2})}} = 1$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.