Question

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

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

Answer 1

The correct option is E

$2$

Explanation for the correct option:

Finding the value of the expression:

$y={\left({\sin }^{-1}x\right)}^2$

Differentiating with respect to $x$ both sides:

$$\begin{gathered} \frac{dy}{dx}=2{\sin }^{-1}x\times \frac{1}{\sqrt{1-x^2}}\mathbf{[}\mathbf{\because }\mathbf{}\frac{\mathbf{d}}{\mathbf{dx}}\left({\sin }^{-1}x\right)\mathbf{=}\frac{\mathbf{1}}{\sqrt{\mathbf{1}\mathbf{-}{\mathbf{x}}^{\mathbf{2}}}}\mathbf{]} \\ =\frac{2{\sin }^{-1}x}{\sqrt{1-x^2}}...\left(1\right) \end{gathered}$$

Differentiating again with respect to $x$ both sides:

$$\begin{gathered} \frac{d^{2}y}{d{x}^2}=\frac{\sqrt{1-x^2}\times {\displaystyle \frac{2}{\sqrt{1-x^2}}}-2{\sin }^{-1}x\times {\displaystyle \frac{1}{2\sqrt{1-x^2}}}\times \left(-2x\right)}{1-x^2}\left[\because \frac{d}{\mathrm{dx}}\left(\frac{u}{v}\right)=\frac{v{\displaystyle \frac{\mathrm{du}}{\mathrm{dx}}}-u{\displaystyle \frac{\mathrm{du}}{\mathrm{dx}}}}{v^2}\right] \\ =\frac{2+{\displaystyle \frac{2x{\sin }^{-1}x}{\sqrt{1-x^2}}}}{1-x^2} \end{gathered}$$

So, $(1-x^2)\frac{d^{2}y}{d{x}^2}=2+\frac{2x{\sin }^{-1}x}{\sqrt{1-x^2}}...\left(2\right)$

Now , multiply by $x$ in the equation $\left(1\right)$ and subtract from $\left(2\right)$

$$\begin{gathered} (1-x^2)\frac{d^{2}y}{d{x}^2}-x\frac{dy}{dx}=2+\frac{2x{\sin }^{-1}x}{\sqrt{1-x^2}}-\frac{2x{\sin }^{-1}x}{\sqrt{1-x^2}} \\ \Rightarrow (1-x^2)\frac{d^{2}y}{d{x}^2}-x\frac{dy}{dx}=2 \end{gathered}$$

Hence, Option (E) is the correct answer.