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

This problem requires several advanced rules of differentiation. In particular, you need

The Chain Rule: If $f$ and $g$ are differentiable functions, then

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

The Product Rule: If $f$ and $g$ are differentiable functions, then

$(fg)'(x)=f'(x)\cdot g(x)+f(x)\cdot g'(x).$

Additionally, we will need our power rule for differentiation:

 $\left(x^{n}\right)'\,=\,nx^{n1},$ for $n\neq 0$,

as well as the derivative of the exponential function, $e^{x}$:

 $(e^{x})'\,=\,e^{x}.$


Solution:
Step 1:

We need to start by identifying the two functions that are being multiplied together so we can apply the product rule.

 $g(x)\,=\,2x^{3},\,$ and $\,h(x)\,=\,e^{3x+5}$

Step 2:

We can now apply the three advanced techniques.This allows us to see that

${\begin{array}{rcl}f'(x)&=&2(x^{3})'e^{3x+5}+2x^{3}(e^{3x+5})'\\&=&6x^{2}e^{3x+5}+2x^{3}(3e^{3x+5})\\&=&6x^{2}e^{3x+5}+6x^{3}e^{3x+5}\end{array}}$

Final Answer:

$6x^{2}e^{3x+5}+6x^{3}e^{3x+5}$

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