Chapter 5 - Logarithmic, Exponential, and Other Transcendental Functions - 5.1 Exercises - Page 325: 41

$\displaystyle \frac{1}{x}$

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)$=3x$ $\displaystyle \frac{d}{dx}[\ln(3x)]=\frac{1}{3x}\cdot 3=\frac{1}{x}$