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)
Since, $\sin^{- 1}{x = \frac{\pi}{2} - \sin^{- 1}y}$
$\Rightarrow \sin^{- 1}{x = \cos^{- 1}y}$
$\Rightarrow \ \ y = \ \sqrt{1 - x^{2}}$ $(\because\sin^{- 1}{x = \cos^{- 1}\sqrt{1 - x^{2}}})$
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.