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If the function $y=\sin ^{-1} x$, then $\left(1-x^2\right) \frac{d^2 y}{d x^2}$ is equal to
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Verified Answer
The correct answer is:
$x \frac{d y}{d x}$
Again differentiating w.r.t. $x$, we get
$$
\begin{aligned}
\frac{d^2 y}{d x^2} & =\frac{0-\frac{1}{2} \cdot \frac{(-2 x)}{\sqrt{1-x^2}}}{\left(\sqrt{1-x^2}\right)^2} \\
\frac{d^2 y}{d x^2} & =\frac{1}{\left(1-x^2\right)} \cdot \frac{x}{\sqrt{1-x^2}}
\end{aligned}
$$
[From Eqs.(i)]
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