Answer
$ 8(x-1)\cdot(2x^{2}-4x+1)$
Work Step by Step
$f(x)=2x^{2}-4x+1, \ \ \quad g(x)=2x^{2}-4x+1,\quad y=f(x)\cdot g(x)$
$f^{\prime}(x)=2(2x)-4=4x-4$
$g^{\prime}(x)=4x-4$
$\displaystyle \frac{dy}{dx}=\frac{d}{dx}[f(x)\cdot g(x)]$= ... product rule ...
=$f^{\prime}(x)g(x)+f(x)g^{\prime}(x)$
$=(4x-4)\cdot(2x^{2}-4x+1)+(2x^{2}-4x+1)\cdot(4x-4)$
$=2(4x-4)\cdot(2x^{2}-4x+1)$
$=2\cdot 4(x-1)\cdot(2x^{2}-4x+1)$
$=8(x-1)\cdot(2x^{2}-4x+1)$
