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

• A

  $-x \frac{d y}{d x}$
• B

  $0$
• Correct answer
  C

  $x \frac{d y}{d x}$
• D

  $x\left(\frac{d y}{d x}\right)^2$

Solution

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)]

Hence, the correct answer is option: 'C'.