Q. If $\sin^{- 1}{x + \sin^{- 1}{y = \frac{\pi}{2},}}$ then $\frac{dy}{dx}$ is equal to
A) $\frac{x}{y}$
B) $- \frac{x}{y}$
C) $\frac{y}{x}$
D) $- \frac{y}{x}$
Explanation
(b)
$\sin^{- 1}{x + \sin^{- 1}{y = \frac{\pi}{2}}}$
$\Rightarrow \sin^{- 1}{x = \frac{\pi}{2}} - \sin^{- 1}y$
${\Rightarrow \sin}^{- 1}{x = \cos^{- 1}y}$
$\Rightarrow y = \sqrt{1 - x^{2}}$
On differentiating w.r.t. $x$, we get
$\frac{dy}{dx} = \frac{1}{2\sqrt{1 - x^{2}}}( - 2x) = - \frac{x}{y}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.