Question

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

Answer 1

$x=y\sqrt{1-y^2}$

$\Rightarrow \frac{dx}{dx}=y\frac{d}{dx}\left(\sqrt{1-y^2}\right)+\sqrt{1-y^2}\frac{d}{dx}\left(y\right)$

$\Rightarrow 1=y\left(\frac{-2y}{\sqrt{1-y^2}}\frac{dy}{dx}\right)+\sqrt{1-y^2}\frac{dy}{dx}$

$\Rightarrow 1=(\frac{-2y^2}{\sqrt{1-y^2}}+\sqrt{1-y^2})\frac{dy}{dx}$

$\Rightarrow \frac{dy}{dx}=\frac{1}{\frac{-2y^2}{\sqrt{1-y^2}}+\sqrt{1-y^2}}$

$\Rightarrow \frac{dy}{dx}=\frac{\sqrt{1-y^2}}{-2y^2+1-y^2}$

$\therefore \frac{dy}{dx}=\frac{\sqrt{1-y^2}}{1-3y^2}$