Answer
$\displaystyle \frac{dy}{dx}=3x^{2}-5x^{4}$
Work Step by Step
Set:
$f(x)=x^{3}\ \ \quad g(x)=1-x^{2}$
$f^{\prime}(x)=3x^{2}\qquad g^{\prime}(x)=0-2x=-2x$
$\displaystyle \frac{dy}{dx}=\frac{d}{dx}[f(x)g(x)]$= ... product rule ...
$= f^{\prime}(x)g(x)+f(x)g^{\prime}(x)$
$=3x^{2}(1-x^{2})+x^{3}(-2x)$
$=3x^{2}-3x^{4}-2x^{4}$
$=3x^{2}-5x^{4}$
