Difference between revisions of "022 Exam 2 Sample A, Problem 1"

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!Foundations:    
 
!Foundations:    
 
|-
 
|-
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|This problem requires several advanced rules of differentiation.  In particular, you need
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|This problem is best approached through properties of logarithms.  Remember that
 
|-
 
|-
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|'''The Chain Rule:''' If <math style="vertical-align: -25%;">f</math> and <math style="vertical-align: -15%;">g</math> are differentiable functions, then
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|<br>&nbsp;&nbsp;&nbsp;&nbsp; <math>\ln (xy) = \ln x + \ln y,</math>
 
|-
 
|-
−
 
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|and
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|<br>&nbsp;&nbsp;&nbsp;&nbsp; <math>(f\circ g)'(x) = f'(g(x))\cdot g'(x).</math>
 
 
|-
 
|-
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|<br>'''The Product Rule:'''  If <math style="vertical-align: -25%;">f</math> and <math style="vertical-align: -15%;">g</math> are differentiable functions, then
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|&nbsp;&nbsp;&nbsp;&nbsp; <math>\ln \left( \frac{x}{y}\right) = \ln x - \ln y,</math>
 
|-
 
|-
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|<br>&nbsp;&nbsp;&nbsp;&nbsp; <math>(fg)'(x) = f'(x)\cdot g(x)+f(x)\cdot g'(x).</math>
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|You will also need to apply
 
|-
 
|-
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|<br>'''The Quotient Rule:''' If <math style="vertical-align: -25%;">f</math> and <math style="vertical-align: -15%;">g</math> are differentiable functions and <math style="vertical-align: -21%;">g(x) \neq 0</math>&thinsp;, then
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|'''The Chain Rule:''' If <math style="vertical-align: -25%;">f</math> and <math style="vertical-align: -15%;">g</math> are differentiable functions, then
 
|-
 
|-
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|<br>&nbsp;&nbsp;&nbsp;&nbsp; <math>\left(\frac{f}{g}\right)'(x) = \frac{f'(x)\cdot g(x)-f(x)\cdot g'(x)}{\left(g(x)\right)^2}. </math>
+
 
 +
|&nbsp;&nbsp;&nbsp;&nbsp; <math>(f\circ g)'(x) = f'(g(x))\cdot g'(x).</math>
 
|-
 
|-
 +
 +
|Finally, recall that the derivative of natural log is
 +
|-
 +
|
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::<math>\left(\ln x\right)'\,=\,\frac{1}{x}.</math>
 
|<br>
 
|<br>
 
|}
 
|}
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!Step 1: &nbsp;
 
!Step 1: &nbsp;
 
|-
 
|-
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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: -20%">g(x)\,=\,\ln x</math>, and
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|We can use the log rules to rewrite our function as  
 
|-
 
|-
 
|
 
|
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::<math>f(x)\,=\,\frac{(x+5)(x-1)}{x},</math>
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::<math>y\,=\,\ln (x+5)+\ln(x-1)-\ln(x).</math>
 
|-
 
|-
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|we then have <math style="vertical-align: -22%">y\,=\,g\circ f(x)\,=\,g\left(f(x)\right).</math>
 
 
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|}
  
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!Step 2: &nbsp;
 
!Step 2: &nbsp;
 
|-
 
|-
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|We can now apply all three advanced techniques.  For example, to find the derivative <math>f'(x)</math>,
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|We can differentiate term-by-term, applying the chain rule to the first two terms to find
 +
|-
 +
|<br>
 +
::<math>\begin{array}{rcl}
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y' & = & \displaystyle{\frac{1}{x+5}\cdot(x+5)'+\frac{1}{x-1}\cdot(x-1)'+\frac{1}{x}}\\
 +
\\
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& = &  \displaystyle{\frac{1}{x+5}+\frac{1}{x-1}+\frac{1}{x}}.
 +
 
 +
\end{array}</math>
 
|
 
|
 
|}
 
|}
  
 
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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!Part (c): &nbsp;
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!Final Answer: &nbsp;
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|-
 
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|We can choose to expand the second term, finding
 
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|-
 
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|&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; <math>e(x^{2}+2)^{2}=ex^{4}+4ex^{2}+4e.</math>
 
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|-
 
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|We then only require the product rule on the first term, so
 
 
|-
 
|-
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|&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; <math>h'(x)\,=\,(4x)'\cdot\sin(x)+4x\cdot(\sin(x))'+\left(ex^{4}+4ex^{2}+4e\right)'\,=\,4\sin(x)+4x\cos(x)+4ex^{3}+8ex.</math>
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|<br><math>y'\,=\,\displaystyle{\frac{1}{x+5}+\frac{1}{x-1}+\frac{1}{x}}.</math>
 
|}
 
|}
  
 
[[022_Exam_2_Sample_A|'''<u>Return to Sample Exam</u>''']]
 
[[022_Exam_2_Sample_A|'''<u>Return to Sample Exam</u>''']]

Latest revision as of 21:42, 18 January 2017

Find the derivative of  

Foundations:  
This problem is best approached through properties of logarithms. Remember that

    
and
    
You will also need to apply
The Chain Rule: If and are differentiable functions, then
    
Finally, recall that the derivative of natural log is

 Solution:

Step 1:  
We can use the log rules to rewrite our function as
Step 2:  
We can differentiate term-by-term, applying the chain rule to the first two terms to find

Final Answer:  

Return to Sample Exam

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