Answer
$ \displaystyle \frac{2}{x+1}-\frac{9}{3x-4}-\frac{1}{x-9}$
Work Step by Step
Apply the product and quatient rules for logarithms
$s(x)=\ln|(x+1)^{2}|-\ln|(3x-4)^{3}|-\ln|(x-9)|$
Apply the power rule for logarithms
$s(x)=2\ln|(x+1)|-3\ln|(3x-4)|-\ln|(x-9)|$
Apply $\displaystyle \frac{d}{dx}[\ln|u|]=\frac{1}{u}\cdot\frac{du}{dx}$,
with constant multiple,
$s^{\prime}(x)=2\displaystyle \cdot\frac{1}{x+1}-3\cdot\frac{3}{3x-4}-\frac{1}{x-9}$
$s^{\prime}(x)=\displaystyle \frac{2}{x+1}-\frac{9}{3x-4}-\frac{1}{x-9}$
