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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|-
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|as well as the derivative of the exponential function, <math>e^x</math>:
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|as well as the derivative of the exponential function, <math style="vertical-align: 5%">e^x</math>:
 
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::<math>\left(e^{f(x)}\right)'\,=\,\left(f(x)\right)'\cdot e^{f(x)}.</math>
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::<math>(e^x)'\,=\,e^x.</math>
 
|<br>
 
|<br>
 
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!Step 1: &nbsp;
 
!Step 1: &nbsp;
 
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|-
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|We need to identify the composed functions in order to apply the chain rule. Note that if we set <math style="vertical-align: -21%">g(x)\,=\,\ln x</math>, and
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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. 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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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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!Step 2: &nbsp;
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|-
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|We can now apply the advanced techniques.This allows us to see that
 
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::<math>f(x)\,=\,\frac{(x+5)(x-1)}{x},</math>
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::<math>\begin{array}{rcl}
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f'(x)&=&2(x^3)' e^{3x+5}+2x^3(e^{3x+5})' \\
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&=&6x^2e^{3x+5}+2x^3(3e^{3x+5})\\
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& = &6x^2e^{3x+5}+6x^3e^{3x+5}.
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\end{array}</math>
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|}
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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!Final Answer: &nbsp;
 
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|we then have&thinsp; <math style="vertical-align: -21%">y\,=\,g\circ f(x)\,=\,g\left(f(x)\right).</math>
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|
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::<math>f'(x)\,=\,6x^2e^{3x+5}+6x^3e^{3x+5}.
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</math>
 
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|}
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[[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:
for ,
as well as the derivative of the exponential function, :

 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:  
We can now apply the advanced techniques.This allows us to see that
Final Answer:  

Return to Sample Exam

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