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

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

The derivative of inverse sine:

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

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To find the second derivative, rewrite the first derivative as:

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

Applying the chain rule:

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

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

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Computing $(1-x^2)\frac{d^2y}{dx^2}$:

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

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

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

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

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

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

$= x\frac{dy}{dx}$

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