Q. If the function $y = \sin^{- 1}x$, then $\left( 1 - x^{2} \right)\frac{d^{2}y}{dx^{2}}$ is equal to
A) $- x\ \frac{dy}{dx}$
B) $0$
C) $x\ \frac{dy}{dx}$
D) $x\left( \frac{dy}{dx} \right)^{2}$
Explanation
(c)
Given, $y = \sin^{- 1}x$
$\Rightarrow$ $\frac{dy}{dx} = \frac{1}{\sqrt{1 - x^{2}}}$ …(i)
$\Rightarrow$ $\frac{d^{2}y}{dx^{2}} = \frac{0 - \frac{1}{2}.\ \frac{( - 2x)}{\sqrt{1 - x^{2}}}}{\left( \sqrt{1 - x^{2}} \right)^{2}}$
$\Rightarrow$ $\left( 1 - x^{2} \right)\ \ \frac{d^{2}y}{dx^{2}} = x\ \frac{dy}{dx}\$….[from Eq.(i)]
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.