Answer
$2(x^{-0.5}-2x)\cdot(2x^{0.5}-x^{2})$
Work Step by Step
$f(x)=2x^{0.5}-x^{2}, \ \ \quad g(x)=2x^{0.5}-x^{2},\quad y=f(x)\cdot g(x)$
$f^{\prime}(x)=2(0.5x^{-0.5})-2x=x^{-0.5}-2x$
$g^{\prime}(x)=x^{-0.5}-2x$
$\displaystyle \frac{dy}{dx}=\frac{d}{dx}[f(x)\cdot g(x)]$= ... product rule ...
=$f^{\prime}(x)g(x)+f(x)g^{\prime}(x)$
$=(x^{-0.5}-2x)\cdot(2x^{0.5}-x^{2})+(2x^{0.5}-x^{2})\cdot(x^{-0.5}-2x)$
$=2(x^{-0.5}-2x)\cdot(2x^{0.5}-x^{2})$
... we do not need to expand the answer...