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

Question

Aicharya · Accepted Answer

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)]

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

Solution