Answer
$\displaystyle \frac{dy}{dx}=18x^{2}+6$
Work Step by Step
Set:
$f(x)=3x^{2}\ \ \quad g(x)=2x+1$
$f^{\prime}(x)=3(2x)=6x\qquad g^{\prime}(x)=2+0=2$
$\displaystyle \frac{dy}{dx}=\frac{d}{dx}[f(x)g(x)]$= ... product rule ...
$= f^{\prime}(x)g(x)+f(x)g^{\prime}(x)$
$=6x(2x+1)+3x^{2}(2)$
$=12x^{2}+6+6x^{2}$
$=18x^{2}+6$
