Question

If $y={({\sin }^{-1}x)}^2,$ then $(1-x^2)\frac{d^{2}y}{d{x}^2}-x\frac{dy}{dx}$ is equal to

• A. $0$
• B. $-1$
• C. $-2$
• D. $1$
• E. $2$

Answer 1

Answer: (E)

$y={({\sin }^{-1}x)}^2$
Differentiating on both sides, $\frac{dy}{dx}=2{\sin }^{-1}x.\frac{1}{\sqrt{1-x^2}}$
$\Rightarrow$ $\sqrt{1-x^2}=\frac{dy}{dx}=2{\sin }^{-1}x$
Differentiating on both sides, $\sqrt{(1-x^2)}\frac{d^{2}y}{d{x}^2}-\frac{2}{2\sqrt{1-x^2}}\frac{dy}{dx}=\frac{2}{\sqrt{1-x^2}}$ $2(1-x^2)\frac{d^{2}y}{d{x}^2}-2\frac{dy}{dx}.x=4$
$\Rightarrow$ $(1-x^2)\frac{d^{2}y}{d{x}^2}-x\frac{dy}{dx}=2$