## Calculus (3rd Edition)

$$y'= 24x+47.$$
Taking the $\ln$ on both sides of the equation, we get $$\ln y= \ln (3x+5)(4x+9)$$ Then using the properties of $\ln$, we can write $$\ln y= \ln (3x+5)+\ln(4x+9).$$ Now taking the derivative for the above equation, we have $$\frac{y'}{y}= \frac{3}{3x+5}+ \frac{4}{4x+9},$$ Hence $y'$ is given by $$y'=y\left( \frac{3}{3x+5}+ \frac{4}{4x+9}\right)=(3x+5)(4x+9)\left( \frac{3}{3x+5}+ \frac{4}{4x+9}\right)\\ 12x+27+12x+20=24x+47.$$