## Calculus (3rd Edition)

$y'=y\left(\frac{1}{x+1}+\frac{2}{x+2}-\frac{1}{x+3}-\frac{1}{x+4}\right)$
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 $$y'=y\left(\frac{1}{x+1}+\frac{2}{x+2}-\frac{1}{x+3}-\frac{1}{x+4}\right).$$