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| | <span class="exam">(c) <math style="vertical-align: -18px">h(x)=\frac{(5x^2+7x)^3}{\ln(x^2+1)} </math> | | <span class="exam">(c) <math style="vertical-align: -18px">h(x)=\frac{(5x^2+7x)^3}{\ln(x^2+1)} </math> |
| | + | <hr> |
| | + | [[009A Sample Midterm 2, Problem 5 Solution|'''<u>Solution</u>''']] |
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
| + | [[009A Sample Midterm 2, Problem 5 Detailed Solution|'''<u>Detailed Solution</u>''']] |
| − | !Foundations:
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| − | |-
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| − | |'''1.''' '''Chain Rule'''
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| − | |-
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| − | | <math>\frac{d}{dx}(f(g(x)))=f'(g(x))g'(x)</math>
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| − | |-
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| − | |'''2.''' '''Trig Derivatives'''
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| − | |-
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| − | | <math>\frac{d}{dx}(\sin x)=\cos x,\quad\frac{d}{dx}(\cos x)=-\sin x</math>
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| − | |-
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| − | |'''3.''' '''Quotient Rule''' | |
| − | |-
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| − | | <math>\frac{d}{dx}\bigg(\frac{f(x)}{g(x)}\bigg)=\frac{g(x)f'(x)-f(x)g'(x)}{(g(x))^2}</math>
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| − | |-
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| − | |'''4.''' '''Derivative of natural logarithm
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| − | |-
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| − | | <math>\frac{d}{dx}(\ln x)=\frac{1}{x}</math>
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| − | |}
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| − | '''Solution:'''
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| − |
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| − | '''(a)'''
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Step 1:
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| − | |-
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| − | |First, we use the Chain Rule to get
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| − | |-
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| − | | <math>f'(x)=3\tan^2(7x^2+5)(\tan(7x^2+5))'.</math>
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| − | |}
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| − |
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Step 2:
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| − | |-
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| − | |Now, we use the Chain Rule again to get
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| − | |-
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| − | |
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| − | <math>\begin{array}{rcl}
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| − | \displaystyle{f'(x)} & = & \displaystyle{3\tan^2(7x^2+5)(\tan(7x^2+5))'}\\
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| − | &&\\
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| − | & = & \displaystyle{3\tan^2(7x^2+5)\sec^2(7x^2+5)(7x^2+5)'}\\
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| − | &&\\
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| − | & = & \displaystyle{3\tan^2(7x^2+5)\sec^2(7x^2+5)(14x).}
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| − | \end{array}</math>
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| − | |}
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| − |
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| − | '''(b)'''
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Step 1:
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| − | |-
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| − | |First, we use the Chain Rule to get
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| − | |-
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| − | | <math>g'(x)=\cos(\cos(e^x))(\cos(e^x))'.</math>
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| − | |}
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| − |
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Step 2:
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| − | |-
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| − | |Now, we use the Chain Rule again to get
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| − | |-
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| − | |
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| − | <math>\begin{array}{rcl}
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| − | \displaystyle{g'(x)} & = & \displaystyle{\cos(\cos(e^x))(\cos(e^x))'}\\
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| − | &&\\
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| − | & = & \displaystyle{\cos(\cos(e^x))(-\sin(e^x))(e^x)'}\\
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| − | &&\\
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| − | & = & \displaystyle{\cos(\cos(e^x))(-\sin(e^x))(e^x).}
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| − | \end{array}</math>
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| − | |}
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| − |
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| − | '''(c)'''
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Step 1:
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| − | |-
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| − | |First, we use the Quotient Rule to get
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| − | |-
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| − | | <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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| − | |}
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| − |
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Step 2:
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| − | |-
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| − | |Now, we use the Chain Rule to get
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| − | |-
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| − | | <math>\begin{array}{rcl}
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| − | \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}}\\
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| − | &&\\
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| − | & = & \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}}\\
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| − | &&\\
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| − | & = & \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}.}
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| − | \end{array}</math>
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| − | |}
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| − |
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| − |
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Final Answer:
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| − | |-
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| − | | '''(a)''' <math>f'(x)=3\tan^2(7x^2+5)\sec^2(7x^2+5)(14x)</math>
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| − | |-
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| − | | '''(b)''' <math>g'(x)=\cos(\cos(e^x))(-\sin(e^x))(e^x)</math>
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| − | |-
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| − | | '''(c)''' <math>h'(x)=\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>
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| − | |}
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| | [[009A_Sample_Midterm_2|'''<u>Return to Sample Exam</u>''']] | | [[009A_Sample_Midterm_2|'''<u>Return to Sample Exam</u>''']] |
