Difference between revisions of "022 Exam 2 Sample B, Problem 3"

Revision as of 06:46, 17 May 2015

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:
$\left(x^{n}\right)'\,=\,nx^{n-1},$ 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:
$6x^{2}e^{3x+5}+6x^{3}e^{3x+5}$

@@ Line 21: / Line 21: @@
 ::<math style="vertical-align: -21%;">\left(x^n\right)'\,=\,nx^{n-1},</math> for <math style="vertical-align: -25%;">n\neq 0</math>,
 |-
-|as well as the derivative of the exponential function, <math>e^x</math>:
+|as well as the derivative of the exponential function, <math style="vertical-align: 5%">e^x</math>:
 |-
 |
-::<math>\left(e^{f(x)}\right)'\,=\,f'(x)\cdot e^{f(x)}.</math>
+::<math>(e^x)'\,=\,e^x.</math>
 |<br>
 |}