Answer
$\displaystyle \frac{dy}{dx}=x(8x+1)(1-x)+x(4x+1)(1-2x)$
Work Step by Step
$f(x)=4x^{2}+x, \ \ \quad g(x)=x-x^{2},\quad y=f(x)\cdot g(x)$
$f^{\prime}(x)=4(2x)+1=8x+1$
$g^{\prime}(x)=1-2x$
$\displaystyle \frac{dy}{dx}=\frac{d}{dx}[f(x)\cdot g(x)]$= ... product rule ...
=$f^{\prime}(x)g(x)+f(x)g^{\prime}(x)$
$=(8x+1)\cdot(x-x^{2})+(4x^{2}+x)\cdot(1-2x)$
... since we do not need to expand the answer, we will just factorize each term
$\displaystyle \frac{dy}{dx}=(8x+1)(x-x^{2})+(4x^{2}+x)(1-2x)$
$\displaystyle \frac{dy}{dx}=x(8x+1)(1-x)+x(4x+1)(1-2x)$
