Question

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

• A. $x$
• B. $\frac{\sqrt{1-y^2}}{1+2y^2}$
• C. $\frac{\sqrt{1-y^2}}{1-2y^2}$
• D. $0$

Answer 1

Answer: (C)

We have $x=y\sqrt{1-y^2}$
On differentiating both sides w.r.t. $x$, we get
$1=\frac{dy}{dx}\sqrt{1-y^2}+\frac{y(-y)}{\sqrt{1-y^2}}\frac{dy}{dx}$
$\Rightarrow \frac{dy}{dx}\left[\sqrt{1-y^2}-\frac{y^2}{\sqrt{1-y^2}}\right]=1$
$\Rightarrow \frac{dy}{dx}\left[\frac{1-y^2-y^2}{\sqrt{1-y^2}}\right]=1$
$\Rightarrow \frac{dy}{dx}=\frac{\sqrt{1-y^2}}{1-2y^2}$