Answer
$$ y'= 24x+47.$$
Work Step by Step
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
$$ y'=y\left( \frac{3}{3x+5}+ \frac{4}{4x+9}\right)=(3x+5)(4x+9)\left( \frac{3}{3x+5}+ \frac{4}{4x+9}\right)\\
12x+27+12x+20=24x+47.$$
