Answer
2
Work Step by Step
(the derivative evaluated at x=$1$)
Use the product rule,
$\displaystyle \frac{d}{dx}[f(x)g(x)]=f^{\prime}(x)g(x)+f(x)g^{\prime}(x)$.
$f(x)=x^{2}+x,\quad f^{\prime}(x)=2x+1$
$g(x)=x^{2}-x,\quad g^{\prime}(x)=2x-1$
$\displaystyle \frac{dy}{dx}=(2x+1)(x^{2}-x)+(x^{2}+x)(2x-1)$
$\left.\frac{dy}{dx}\right|_{1}=(2+1)(1-1)+(1+1)(2-1)$
$=0+2$
$=2$
