Difference between revisions of "022 Exam 2 Sample B, Problem 3"
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::<math style="vertical-align: -21%;">\left(x^n\right)'\,=\,nx^{n-1},</math> for <math style="vertical-align: -25%;">n\neq 0</math>, | ::<math style="vertical-align: -21%;">\left(x^n\right)'\,=\,nx^{n-1},</math> for <math style="vertical-align: -25%;">n\neq 0</math>, | ||
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| − | |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>: |
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| − | ::<math> | + | ::<math>(e^x)'\,=\,e^x.</math> |
|<br> | |<br> | ||
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!Step 1: | !Step 1: | ||
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| − | |We need to start by identifying the two functions that are being multiplied together so we can apply the product rule. | + | |We need to start by identifying the two functions that are being multiplied together so we can apply the product rule. Let's call <math style="vertical-align: -20%">g(x)\,=\,2x^3,\,</math> and <math style="vertical-align: -20%">\,h(x) \, = \, e^{3x + 5}</math>, so <math style="vertical-align: -20%">f(x)=g(x)\cdot h(x)</math>. |
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| − | : | ||
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!Step 2: | !Step 2: | ||
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| − | |We can now apply the | + | |We can now apply the advanced techniques.This allows us to see that |
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| − | <math>\begin{array}{rcl} | + | ::<math>\begin{array}{rcl} |
f'(x)&=&2(x^3)' e^{3x+5}+2x^3(e^{3x+5})' \\ | f'(x)&=&2(x^3)' e^{3x+5}+2x^3(e^{3x+5})' \\ | ||
&=&6x^2e^{3x+5}+2x^3(3e^{3x+5})\\ | &=&6x^2e^{3x+5}+2x^3(3e^{3x+5})\\ | ||
| − | & = &6x^2e^{3x+5}+6x^3e^{3x+5} | + | & = &6x^2e^{3x+5}+6x^3e^{3x+5}. |
\end{array}</math> | \end{array}</math> | ||
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| − | <math>6x^2e^{3x+5}+6x^3e^{3x+5} | + | ::<math>f'(x)\,=\,6x^2e^{3x+5}+6x^3e^{3x+5}. |
</math> | </math> | ||
|} | |} | ||
[[022_Exam_2_Sample_B|'''<u>Return to Sample Exam</u>''']] | [[022_Exam_2_Sample_B|'''<u>Return to Sample Exam</u>''']] | ||
Latest revision as of 06:50, 17 May 2015
Find the derivative of .
| Foundations: | |
|---|---|
| This problem requires several advanced rules of differentiation. In particular, you need | |
| The Chain Rule: If and are differentiable functions, then | |
The Product Rule: If and are differentiable functions, then | |
| Additionally, we will need our power rule for differentiation: | |
| |
| as well as the derivative of the exponential function, : | |
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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 and , so . |
| Step 2: |
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| We can now apply the advanced techniques.This allows us to see that |
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| Final Answer: |
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