Difference between revisions of "009A Sample Midterm 2, Problem 5"
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!Step 1: | !Step 1: | ||
|- | |- | ||
| − | | | + | |First, we use the Quotient Rule to get |
|- | |- | ||
| − | | | + | | <math>h'(x)=\frac{\ln(x^2+1)((5x^2+7x)^2)'-(5x^2+7x)^2(\ln(x^2+1))'}{(\ln(x^2+1))^2}.</math> |
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| − | |||
| − | |||
|} | |} | ||
| Line 83: | Line 79: | ||
!Step 2: | !Step 2: | ||
|- | |- | ||
| − | | | + | |Now, we use the Chain Rule to get |
| − | |||
| − | |||
| − | |||
| − | |||
|- | |- | ||
| − | | | + | | <math>\begin{array}{rcl} |
| + | \displaystyle{h'(x)} & = & \displaystyle{\frac{\ln(x^2+1)((5x^2+7x)^2)'-(5x^2+7x)^2(\ln(x^2+1))'}{(\ln(x^2+1))^2}}\\ | ||
| + | &&\\ | ||
| + | & = & \displaystyle{\frac{\ln(x^2+1)2(5x^2+7x)(5x^2+7x)'-(5x^2+7x)^2\frac{1}{x^2+1}(x^2+1)'}{(\ln(x^2+1))^2}}\\ | ||
| + | &&\\ | ||
| + | & = & \displaystyle{\frac{\ln(x^2+1)2(5x^2+7x)(10x+7)-(5x^2+7x)^2\frac{1}{x^2+1}(2x)}{(\ln(x^2+1))^2}.} | ||
| + | \end{array}</math> | ||
|} | |} | ||
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| '''(b)''' <math>\cos(\cos(e^x))(-\sin(e^x))(e^x)</math> | | '''(b)''' <math>\cos(\cos(e^x))(-\sin(e^x))(e^x)</math> | ||
|- | |- | ||
| − | |'''(c)''' | + | | '''(c)''' <math>\frac{\ln(x^2+1)2(5x^2+7x)(10x+7)-(5x^2+7x)^2\frac{1}{x^2+1}(2x)}{(\ln(x^2+1))^2}</math> |
|} | |} | ||
[[009A_Sample_Midterm_2|'''<u>Return to Sample Exam</u>''']] | [[009A_Sample_Midterm_2|'''<u>Return to Sample Exam</u>''']] | ||
Revision as of 12:45, 17 February 2017
Find the derivatives of the following functions. Do not simplify.
- a) 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(x)=\tan^3(7x^2+5) }
- b) 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(x)=\sin(\cos(e^x)) }
- c) 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 h(x)=\frac{(5x^2+7x)^3}{\ln(x^2+1)} }
| Foundations: |
|---|
| 1. Chain Rule |
| 2. Derivatives of trig/ln |
| 3. Quotient Rule |
Solution:
(a)
| Step 1: |
|---|
| First, we use the Chain Rule to get |
| 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'(x)=3\tan^2(7x^2+5)(\tan(7x^2+5))'.} |
| Step 2: |
|---|
| Now, we use the Chain Rule again to get |
|
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} \displaystyle{f'(x)} & = & \displaystyle{3\tan^2(7x^2+5)(\tan(7x^2+5))'}\\ &&\\ & = & \displaystyle{3\tan^2(7x^2+5)\sec^2(7x^2+5)(7x^2+5)'}\\ &&\\ & = & \displaystyle{3\tan^2(7x^2+5)\sec^2(7x^2+5)(14x).} \end{array}} |
(b)
| Step 1: |
|---|
| First, we use the Chain Rule to get |
| 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'(x)=\cos(\cos(e^x))(\cos(e^x))'.} |
| Step 2: |
|---|
| Now, we use the Chain Rule again to get |
|
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} \displaystyle{g'(x)} & = & \displaystyle{\cos(\cos(e^x))(\cos(e^x))'}\\ &&\\ & = & \displaystyle{\cos(\cos(e^x))(-\sin(e^x))(e^x)'}\\ &&\\ & = & \displaystyle{\cos(\cos(e^x))(-\sin(e^x))(e^x).} \end{array}} |
(c)
| Step 1: |
|---|
| First, we use the Quotient Rule to get |
| 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 h'(x)=\frac{\ln(x^2+1)((5x^2+7x)^2)'-(5x^2+7x)^2(\ln(x^2+1))'}{(\ln(x^2+1))^2}.} |
| Step 2: |
|---|
| Now, we use the Chain Rule to get |
| 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} \displaystyle{h'(x)} & = & \displaystyle{\frac{\ln(x^2+1)((5x^2+7x)^2)'-(5x^2+7x)^2(\ln(x^2+1))'}{(\ln(x^2+1))^2}}\\ &&\\ & = & \displaystyle{\frac{\ln(x^2+1)2(5x^2+7x)(5x^2+7x)'-(5x^2+7x)^2\frac{1}{x^2+1}(x^2+1)'}{(\ln(x^2+1))^2}}\\ &&\\ & = & \displaystyle{\frac{\ln(x^2+1)2(5x^2+7x)(10x+7)-(5x^2+7x)^2\frac{1}{x^2+1}(2x)}{(\ln(x^2+1))^2}.} \end{array}} |
| Final Answer: |
|---|
| (a) 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 3\tan^2(7x^2+5)\sec^2(7x^2+5)(14x)} |
| (b) 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 \cos(\cos(e^x))(-\sin(e^x))(e^x)} |
| (c) |
