Answer
18$x^{2}$-$\frac{3x}{2}$+12
Work Step by Step
According to the product rule, if f(x)=h(x)g(x), then f'(x)=h(x)g'(x)+g(x)h'(x). If we set h(x)=3$x^{2}$+6, and g(x)=2x-$\frac{1}{4}$, then using the product rule, f'(x)=(3$x^{2}$+6)$\frac{d}{dx}$[2x-$\frac{1}{4}$]+(2x-$\frac{1}{4}$)$\frac{d}{dx}$[3$x^{2}$+6]
$\frac{d}{dx}$[2x-$\frac{1}{4}$]=2, using the power rule
$\frac{d}{dx}$[3$x^{2}$+6]=6x, using the power rule
Therefore, f'(x)=(3$x^{2}$+6)(2)+(6x)(2x-$\frac{1}{4}$)
= 6$x^{2}$+12+12$x^{2}$-$\frac{3x}{2}$
=18$x^{2}$-$\frac{3x}{2}$+12
