Chapter 11 - Section 11.5 - Derivatives of Logarithmic and Exponential Functions - Exercises - Page 842: 22

$ \displaystyle \frac{ 9}{ 9x-5 }$

(see p. 838, Generalized Rule)$\quad$ $\displaystyle \frac{d}{dx}[\ln|u|]=\frac{1}{u}\cdot\frac{du}{dx}$ $u(x)=5-9x$ $\displaystyle \frac{du}{dx}=-9$ $\displaystyle \frac{d}{dx}[\ln|5-9x|]=\frac{1}{5-9x}\cdot(-9)$ $=\displaystyle \frac{-9}{5-9x}=\frac{-9}{-(9x-5)}$ $=\displaystyle \frac{ 9}{ 9x-5 }$