Answer
64
Work Step by Step
(the derivative evaluated at x=2)
Use the product rule,
$\displaystyle \frac{d}{dx}[f(x)g(x)]=f^{\prime}(x)g(x)+f(x)g^{\prime}(x)$.
$f(x)=x^{3}+2x,\quad f^{\prime}(x)=3x^{2}+2$
$g(x)=x^{2}-x,\quad g^{\prime}(x)=2x-1$
$\displaystyle \frac{dy}{dx}=(3x^{2}+2)(x^{2}-x)+(x^{3}+2x)(2x-1)$
$\left.\frac{dy}{dx}\right|_{2}=[3(4)+2](4-2)+(8+4)(4-1)$
$=28+36$
$=64$
