Answer
$ \displaystyle \frac{1}{x-1}+\frac{3}{3x-4}-\frac{1}{x-9}$
Work Step by Step
Apply the Product and Quotient Rules for logarithms,
$r(x)=\ln|-x+1|+\ln|3x-4\}-\ln|x-9|$
... $|-x+1|=|-(x-1)|=|x-1|$
$r(x)=\ln|x-1|+\ln|3x-4\}-\ln|x-9|$
Apply $\displaystyle \frac{d}{dx}[\ln|u|]=\frac{1}{u}\cdot\frac{du}{dx}$
$r^{\prime}(x)=\displaystyle \frac{1}{x-1}+\frac{3}{3x-4}-\frac{1}{x-9}$
