Chapter 11 - Section 11.5 - Derivatives of Logarithmic and Exponential Functions - Exercises - Page 842: 2

$\displaystyle \frac{1}{x+3}$

(see p. 836, Generalized Rule)$\quad \displaystyle \frac{d}{dx}[\ln u]=\frac{1}{u}\cdot\frac{du}{dx}$ $u(x)=x+3$ $\displaystyle \frac{du}{dx}=1$ $ \displaystyle \frac{d}{dx}[\ln(x+3)]=\frac{\mathrm{l}}{(x+3)}\cdot 1 =\displaystyle \frac{1}{x+3}$