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

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|<br>&nbsp;&nbsp;&nbsp;&nbsp; <math>\ln \left( \frac{x}{y}\right) = \ln x - \ln y,</math>
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|&nbsp;&nbsp;&nbsp;&nbsp; <math>\ln \left( \frac{x}{y}\right) = \ln x - \ln y,</math>
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|You will also need to apply
 
|You will also need to apply
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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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|&nbsp;&nbsp;&nbsp;&nbsp; <math>(f\circ g)'(x) = f'(g(x))\cdot g'(x).</math>
 
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!Step 2: &nbsp;
 
!Step 2: &nbsp;
 
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|We can differentiate term-by-term, applying the chain rule to the first two to find
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|We can differentiate term-by-term, applying the chain rule to the first two terms to find
 
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!Final Answer: &nbsp;
 
!Final Answer: &nbsp;
 
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|<math>y'\,=\,\displaystyle{\frac{1}{x+5}+\frac{1}{x-1}+\frac{1}{x}}.</math>
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|<br><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>''']]

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

     Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \ln(xy)=\ln x+\ln y,}
and
     Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ln \left( \frac{x}{y}\right) = \ln x - \ln y,}
You will also need to apply
The Chain Rule: If Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f} and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle g} are differentiable functions, then
     Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (f\circ g)'(x) = f'(g(x))\cdot g'(x).}
Finally, recall that the derivative of natural log is
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \left(\ln x\right)'\,=\,\frac{1}{x}.}

 Solution:

Step 1:  
We can use the log rules to rewrite our function as
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle y\,=\,\ln (x+5)+\ln(x-1)-\ln(x).}
Step 2:  
We can differentiate term-by-term, applying the chain rule to the first two terms to find

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{array}{rcl} y' & = & \displaystyle{\frac{1}{x+5}\cdot(x+5)'+\frac{1}{x-1}\cdot(x-1)'+\frac{1}{x}}\\ \\ & = & \displaystyle{\frac{1}{x+5}+\frac{1}{x-1}+\frac{1}{x}}. \end{array}}
Final Answer:  

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle y'\,=\,\displaystyle{\frac{1}{x+5}+\frac{1}{x-1}+\frac{1}{x}}.}

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