Question

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

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

Answer 1

The correct option is B $\frac{\sqrt{1-y^2}}{1-2y^2}$

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

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

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

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

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

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

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