## Calculus (3rd Edition)

$$y' =\frac{4x+11}{(x+1)(2x+9)}$$
Recall that $(\ln x)'=\dfrac{1}{x}$ Recall the product rule: $(uv)'=u'v+uv'$ Since $y=\ln((x+1)(2x+9))$, by using the chain rule and the product rule, we have $$y'=\frac{(x+1)'(2x+9)+(x+1)(2x+9)'}{(x+1)(2x+9)}\\ =\frac{(1)(2x+9)+(x+1)(2)}{(x+1)(2x+9)}=\frac{4x+11}{(x+1)(2x+9)}$$