Question
If $y=\sin ^{-1} x$, then $\left(1-x^2\right) y_2$ is equal to

Answer

(b) $x y_1$
Explanation: $y=\sin ^{-1} x$
$\frac{d y}{d x}=\frac{1}{\sqrt{1-x^2}} \Rightarrow \sqrt{1-x^2} \cdot \frac{d y}{d x}=1$
Again, differentiating both sides w.r.to $x$, we get
$\sqrt{1-x^2} \cdot \frac{d^2 y}{d x^2}+\frac{d y}{d x} \cdot\left(\frac{-2 x}{2 \sqrt{1-x^2}}\right)=0$
Simplifying, we get $\left(1-x^2\right) y_2=x y_1$

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