Difference between revisions of "022 Exam 2 Sample B, 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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|while
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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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|and
 
|-
 
|-
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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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|&nbsp;&nbsp;&nbsp;&nbsp; <math>\ln \left( x^n \right) = n\ln x,</math>
 
|-
 
|-
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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>
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|You will also need to apply
 
|-
 
|-
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|Additionally, we will need our power rule for differentiation:
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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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::<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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|&nbsp;&nbsp;&nbsp;&nbsp; <math>(f\circ g)'(x) = f'(g(x))\cdot g'(x).</math>
 
|-
 
|-
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|as well as the derivative of natural log:
+
 
 +
|Finally, recall that the derivative of natural log is
 
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|-
 
|
 
|
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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: -21%">g(x)\,=\,\ln x</math>, and
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|We can use the log rules to rewrite our function as  
 
|-
 
|-
 +
|<br>
 +
::<math>\begin{array}{rcl}
 +
y & = & \displaystyle{\ln \frac{(x+1)^4}{(2x - 5)(x + 4)}}\\
 +
\\
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& = &  4\ln (x+1)-\ln(2x-5)-\ln (x+4).
 +
 +
\end{array}</math>
 
|
 
|
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::<math>f(x)\,=\,\frac{(x+1)^4}{(2x - 5)(x + 4)},</math>
 
 
|-
 
|-
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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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|<br>
 +
|-
 
|}
 
|}
  
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!Step 2: &nbsp;
 
!Step 2: &nbsp;
 
|-
 
|-
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|We can now apply all three advanced techniques.  For <math style="vertical-align: -20%">f'(x)</math>, we can use both the quotient and product rule to find
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|We can differentiate term-by-term, applying the chain rule to each term to find
 
|-
 
|-
 
|<br>
 
|<br>
 
::<math>\begin{array}{rcl}
 
::<math>\begin{array}{rcl}
−
f'(x)&=&\displaystyle{\frac{((x+1)^4)'(2x-5)(x+4)-((2x-5)(x+4))'(x+1)^4}{(2x-5)^2(x+4)^2}} \\
+
y' & = & \displaystyle{4 \cdot \frac{1}{x+1}\cdot(x+1)' - \frac{1}{2x-5}\cdot(2x-5)' - \frac{1}{x+4} \cdot (x+4)'}\\
 
\\
 
\\
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&=&\displaystyle{\frac{(4(x+1)^3)(2x-5)(x+4)-(2(x+4)+(2x-5))(x+1)^4}{(2x-5)^2(x+4)^2}.}
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& = & \displaystyle{\frac{4}{x+1}-\frac{2}{2x-5}-\frac{1}{x+4}}.
 +
 
 
\end{array}</math>
 
\end{array}</math>
 
|
 
|
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|}
 
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
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!Step 3: &nbsp;
 
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|-
 
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|We can now use the chain rule to find<br>
 
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|-
 
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|
 
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::<math>\begin{array}{rcl}
 
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y' & = & \left(g\circ f\right)'(x)\\
 
−
\\
 
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& = & g'\left(f(x)\right)\cdot f'(x)\\
 
−
\\& = &\displaystyle{\left[\frac{(2x-5)(x+4)}{(x+1)^4} \right]\frac{((x+1)^4)'(2x-5)(x+4)-((2x-5)(x+4))'(x+1)^4}{(2x-5)^2(x+4)^2}} \\
 
−
\\
 
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&=&\displaystyle{\left[\frac{(2x-5)(x+4)}{(x+1)^4} \right]\frac{(4(x+1)^3)(2x-5)(x+4)-(2(x+4)+(2x-5))(x+1)^4}{(2x-5)^2(x+4)^2}. }
 
−
\end{array}</math>
 
−
Note that many teachers <u>'''do not'''</u> prefer a cleaned up answer, and may request that you <u>'''do not simplify'''</u>.  This problem seems like it would be of that type, as it doesn't simplify too well.  Nevertheless, it's always a good idea to ask the teacher if you aren't sure of his or her intent.
 
 
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!Final Answer: &nbsp;
 
!Final Answer: &nbsp;
 
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|-
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|<math>y'=\left[\frac{(2x-5)(x+4)}{(x+1)^4} \right]\frac{(4(x+1)^3)(2x-5)(x+4)-(2(x+4)+(2x-5))(x+1)^4}{(2x-5)^2(x+4)^2}. </math>
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|<br><math>y'\,=\,\displaystyle{\frac{4}{x+1}-\frac{2}{2x-5}-\frac{1}{x+4}}.</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 21:22, 20 January 2017

Find the derivative of  

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

    
while
    
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 each term to find

Final Answer:  

Return to Sample Exam

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