Difference between revisions of "022 Exam 2 Sample A, Problem 1"
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!Foundations: | !Foundations: | ||
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− | |This problem | + | |This problem is best approached through properties of logarithms. Remember that |
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− | | | + | |<br> <math>\ln (xy) = \ln x + \ln y,</math> |
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− | + | |and | |
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− | |<br> | + | |<br> <math>\ln \left( \frac{x}{y}\right) = \ln x - \ln y,</math> |
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− | | | + | |You will also need to apply |
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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> <math>(f\circ g)'(x) = f'(g(x))\cdot g'(x).</math> | ||
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− | + | |Finally, recall that the derivative of natural log is | |
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!Step 1: | !Step 1: | ||
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− | |We | + | |We can use the log rules to rewrite our function as |
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− | ::<math> | + | ::<math>y\,=\,\ln (x+5)+\ln(x-1)-\ln(x).</math> |
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!Step 2: | !Step 2: | ||
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− | |We can | + | |We can differentiate term-by-term, applying the chain rule to the first two to find |
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|<br> | |<br> | ||
::<math>\begin{array}{rcl} | ::<math>\begin{array}{rcl} | ||
− | + | 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{ | + | & = & \displaystyle{\frac{1}{x+5}+\frac{1}{x-1}+\frac{1}{x}}. |
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\end{array}</math> | \end{array}</math> | ||
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|} | |} | ||
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!Final Answer: | !Final Answer: | ||
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− | |<math>y'\,=\,\displaystyle{\frac{x | + | |<math>y'\,=\,\displaystyle{\frac{1}{x+5}+\frac{1}{x-1}+\frac{1}{x}}.</math> |
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[[022_Exam_2_Sample_A|'''<u>Return to Sample Exam</u>''']] | [[022_Exam_2_Sample_A|'''<u>Return to Sample Exam</u>''']] |
Revision as of 21:40, 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 to find | |
Final Answer: |
---|