Answer
$\displaystyle \frac{1}{x}$
Work Step by Step
By theorem 5.3/2: $\ \ \ \displaystyle \frac{d}{dx}[\ln u]=\frac{1}{u}\cdot\frac{du}{dx}=\frac{u^{\prime}}{u},\ \ \ (u>0)$
Here, u(x)$=3x$
$\displaystyle \frac{d}{dx}[\ln(3x)]=\frac{1}{3x}\cdot 3=\frac{1}{x}$