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