Answer
$f^{\prime}(x)=18x-6$
Work Step by Step
Let $u(x)=x^{2},\qquad v(x)=3x-1$
Then,$\quad y=f(x)=u(v(x)$ and
$f^{\prime}(x)=\displaystyle \frac{dy}{dx}=\frac{du}{dv}\frac{dv}{dx}$
$\displaystyle \frac{du}{dx}=2x, \quad \frac{du}{dv}=2v=2(3x-1)$
$\displaystyle \frac{dv}{dx}=3$
$f^{\prime}(x)=\displaystyle \frac{du}{dv}\frac{dv}{dx}=2(3x-1)\cdot 3=6(3x-1)$
$f^{\prime}(x)=18x-6$
