Answer
$ 6|3x-1|$
Work Step by Step
$f(x)$ is a product of
$u(x)=3x-1$ and $v(x)=|3x-1|$.
$f^{\prime}(x)=(uv)^{\prime}$= ... product rule
$=u^{\prime}v+uv^{\prime}$
$\displaystyle \frac{du}{dx}=3$
For $\displaystyle \frac{dv}{dx}$, apply the Generalized Rule for absolute values,
$\displaystyle \frac{d}{dx}|w|=\frac{|w|}{w}\frac{dw}{dx}$
$\displaystyle \frac{dv}{dx}= \frac{d}{dx}|3x-1|=\frac{|3x-1|}{3x-1}\cdot 3=\frac{3|3x-1|}{3x-1}$
$f^{\prime}(x)=(uv)^{\prime}$
$=u^{\prime}v+uv^{\prime}$
$=3|3x-1|+(3x-1)\displaystyle \cdot\frac{3|3x-1|}{3x-1}$
$=3|3x-1|+3|3x-1|$
$=6|3x-1|$
