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

$x\frac{dy}{dx}$

The correct answer is Option (2) → $x\frac{dy}{dx}$

Given: $y=\sin^{-1}x$

First derivative:

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

Second derivative:

$\frac{d^{2}y}{dx^{2}}=\frac{d}{dx}\left( (1-x^{2})^{-1/2} \right)$

$=\; -\frac{1}{2}(1-x^{2})^{-3/2}\cdot (-2x)$

$=\frac{x}{(1-x^{2})^{3/2}}$

Now compute:

$(1-x^{2})\frac{d^{2}y}{dx^{2}} = (1-x^{2})\cdot \frac{x}{(1-x^{2})^{3/2}}$

$=\frac{x}{\sqrt{1-x^{2}}}$

The expression equals $\frac{x}{\sqrt{1-x^{2}}} = x\frac{dy}{dx}$.