If $y=\sin ^{-1} x,$ show that $\left(1-x^{2}\right) \frac{d^{2} y}{d x^{2}}-x \frac{d y}{d x}=0$.

(N/A) Given $y=\sin ^{-1} x$.
Differentiating with respect to $x$,we get:
$\frac{d y}{d x}=\frac{1}{\sqrt{1-x^{2}}}$
This can be written as:
$\sqrt{1-x^{2}} \frac{d y}{d x}=1$
Differentiating both sides with respect to $x$ using the product rule:
$\frac{d}{d x}\left(\sqrt{1-x^{2}}\right) \cdot \frac{d y}{d x} + \sqrt{1-x^{2}} \cdot \frac{d}{d x}\left(\frac{d y}{d x}\right) = 0$
$\left(\frac{1}{2\sqrt{1-x^{2}}} \cdot (-2x)\right) \frac{d y}{d x} + \sqrt{1-x^{2}} \frac{d^{2} y}{d x^{2}} = 0$
Multiplying the entire equation by $\sqrt{1-x^{2}}$:
$-x \frac{d y}{d x} + (1-x^{2}) \frac{d^{2} y}{d x^{2}} = 0$
Rearranging the terms,we get:
$(1-x^{2}) \frac{d^{2} y}{d x^{2}} - x \frac{d y}{d x} = 0$.