Question

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

• A. $x+y$
• B. $1+xy$
• C. $1-xy$
• D. $xy-2$

Answer 1

The correct option is B

$1+xy$

Explanation for the correct option:

$$\begin{gathered} y=\frac{{\sin }^{-1}x}{\sqrt{1-x^2}} \\ \Rightarrow y\sqrt{1-x^2}={\sin }^{-1}x \end{gathered}$$

Differentiating with respect to $x$ both sides:

$\begin{array}{rcl}\left(\sqrt{1-x^2}\right)\frac{dy}{dx}+y\times \frac{1}{2\sqrt{1-x^2}}\times (-2x)& =& \frac{1}{\sqrt{1-x^2}}\left[\because \frac{d\left(u·v\right)}{dx}=u\frac{dv}{dx}+v\frac{du}{dx}\right]\\ & \Rightarrow & \left(\sqrt{1-x^2}\right)\frac{dy}{dx}-\frac{xy}{\sqrt{1-x^2}}=\frac{1}{\sqrt{1-x^2}}\\ & \Rightarrow & \left(\sqrt{1-x^2}\right)\frac{dy}{dx}=\frac{1+xy}{\sqrt{1-x^2}}\\ & \Rightarrow & \sqrt{1-x^2}\times \left(\sqrt{1-x^2}\right)\frac{dy}{dx}=1+xy\\ & \Rightarrow & \left(1-x^2\right)\frac{dy}{dx}=1+xy\end{array}$

Hence, Option (B) is the correct answer.