Answer
$ y'=y\left(\frac{1}{x+1}+\frac{2}{x+2}-\frac{1}{x+3}-\frac{1}{x+4}\right)$
Work Step by Step
Since $ y=\frac{(x+1)(x+2)^2}{(x+3)(x+4)}$, then by taking the $\ln $ on both sides we have
$$\ln y=\ln\left( \frac{(x+1)(x+2)^2}{(x+3)(x+4)}\right)=\ln((x+1)(x+2)^2)-\ln ((x+3)(x+4))\\
=\ln(x+1)+2\ln(x+2)-\ln (x+3)-\ln(x+4).$$
Now taking the derivative, we get
$$\frac{y'}{y}=\frac{1}{x+1}+\frac{2}{x+2}-\frac{1}{x+3}-\frac{1}{x+4}$$
hence $$ y'=y\left(\frac{1}{x+1}+\frac{2}{x+2}-\frac{1}{x+3}-\frac{1}{x+4}\right).$$
