# 022 Exam 2 Sample B, Problem 3

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Find the derivative of ${\displaystyle f(x)\,=\,2x^{3}e^{3x+5}}$.

Foundations:
This problem requires several advanced rules of differentiation. In particular, you need
The Chain Rule: If ${\displaystyle f}$ and ${\displaystyle g}$ are differentiable functions, then

${\displaystyle (f\circ g)'(x)=f'(g(x))\cdot g'(x).}$

The Product Rule: If ${\displaystyle f}$ and ${\displaystyle g}$ are differentiable functions, then

${\displaystyle (fg)'(x)=f'(x)\cdot g(x)+f(x)\cdot g'(x).}$
Additionally, we will need our power rule for differentiation:
${\displaystyle \left(x^{n}\right)'\,=\,nx^{n-1},}$ for ${\displaystyle n\neq 0}$,
as well as the derivative of the exponential function, ${\displaystyle e^{x}}$:
${\displaystyle (e^{x})'\,=\,e^{x}.}$

Solution:

Step 1:
We need to start by identifying the two functions that are being multiplied together so we can apply the product rule. Let's call ${\displaystyle g(x)\,=\,2x^{3},\,}$ and ${\displaystyle \,h(x)\,=\,e^{3x+5}}$, so ${\displaystyle f(x)=g(x)\cdot h(x)}$.
Step 2:
We can now apply the advanced techniques.This allows us to see that
${\displaystyle {\begin{array}{rcl}f'(x)&=&2(x^{3})'e^{3x+5}+2x^{3}(e^{3x+5})'\\&=&6x^{2}e^{3x+5}+2x^{3}(3e^{3x+5})\\&=&6x^{2}e^{3x+5}+6x^{3}e^{3x+5}.\end{array}}}$
${\displaystyle f'(x)\,=\,6x^{2}e^{3x+5}+6x^{3}e^{3x+5}.}$