Answer
$r^{\prime}(x)=\displaystyle \frac{1}{x+1}+3x^{2}e^{x}(3+x)$
Work Step by Step
First term: logarithm of a function
$\displaystyle \frac{d}{dx}[\ln(x+1)]=\frac{1}{x+1}\cdot(x+1)^{\prime} =\frac{1}{x+1}$
Second term is a product:
$u(x)=3x^{3}, \quad v(x)=e^{x},$
$\displaystyle \frac{d}{dx}[u(x)\cdot v(x)]=u(x)^{\prime}\cdot v(x)+u(x)\cdot v^{\prime}(x)$
$=(3\cdot 3x^{2})\cdot e^{x}+3x^{3}\cdot(e^{x})$
$=3x^{2}e^{x}(3+x)$
So,
$r^{\prime}(x)=\displaystyle \frac{1}{x+1}+3x^{2}e^{x}(3+x)$
