If $y=sin^{-1}x,$ then

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

The correct answer is Option (2) → $\frac{d^2y}{dx^2}=\frac{x}{1-x^2}\frac{dy}{dx}$

$y=sin^{-1}x$

$⇒\sin y=x⇒\cos y=\sqrt{1-x^2}$

so $\frac{dy}{dx}=\cos y⇒\frac{dy}{dx}=\sec y$

$\frac{d^2y}{dx^2}=\sec y\tan y\frac{dy}{dx}$

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