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. $0$
• D. $\frac{\sqrt{1-y^2}}{1-2y^2}$

Answer 1

The correct option is D

$\frac{\sqrt{1-y^2}}{1-2y^2}$

Explanation for the correct option.

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

Differentiate w.r.t $x$

$\begin{array}{rcl}1& =& y\left(\frac{-2y}{2\sqrt{1-y^2}}\right)\frac{dy}{dx}+\sqrt{1-y^2}\frac{dy}{dx}\\ \Rightarrow 1& =& \left[y\left(\frac{-2y}{2\sqrt{1-y^2}}\right)+\sqrt{1-y^2}\right]\frac{dy}{dx}\\ \Rightarrow 1& =& \left[\frac{-y^2+1-y^2}{\sqrt{1-y^2}}\right]\frac{dy}{dx}\\ \Rightarrow 1& =& \left(\frac{1-2y^2}{\sqrt{1-y^2}}\right)\frac{dy}{dx}\\ & \Rightarrow & \frac{dy}{dx}=\frac{\sqrt{1-y^2}}{1-2y^2}\end{array}$

Hence, option D is correct.