If $y=sin^{-1}x,$ then $(1-x^2)\frac{d^2y}{dx^2}-x\frac{dy}{dx}$ is :

0

The correct answer is Option (3) → 0

$y=sin^{-1}x⇒\frac{dy}{dx}=\frac{1}{\sqrt{1-x^2}}$

so $\frac{d^2y}{dx^2}=\frac{-1}{2\sqrt{1-x^2}}\frac{(-2x)}{(1-x^2)}=\frac{x}{\sqrt{1-x^2}(1-x^2)}$

so $(1-x^2)\frac{d^2y}{dx^2}=x\frac{dy}{dx}$

$⇒(1-x^2)\frac{d^2y}{dx^2}-x\frac{dy}{dx}=0$