009A Sample Midterm 2, Problem 5 - Revision history

MathAdmin at 22:22, 9 November 2017

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MathAdmin: Created page with "Find the derivatives of the following functions. Do not simplify. (a) $f(x)=\tan^3(7x^2+5)$
2017-04-09T18:53:16Z

Created page with "Find the derivatives of the following functions. Do not simplify. (a) <math style="vertical-align: -5px">f(x)=\tan^3(7x^2+5) </m..."

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Find the derivatives of the following functions. Do not simplify.

(a)   <math style="vertical-align: -5px">f(x)=\tan^3(7x^2+5) </math>

(b)   <math style="vertical-align: -5px">g(x)=\sin(\cos(e^x)) </math>

(c)   <math style="vertical-align: -18px">h(x)=\frac{(5x^2+7x)^3}{\ln(x^2+1)} </math>

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Foundations:  
|-
|'''1.''' '''Chain Rule'''
|-
|        <math>\frac{d}{dx}(f(g(x)))=f'(g(x))g'(x)</math>
|-
|'''2.''' '''Trig Derivatives'''
|-
|        <math>\frac{d}{dx}(\sin x)=\cos x,\quad\frac{d}{dx}(\cos x)=-\sin x</math>
|-
|'''3.''' '''Quotient Rule'''
|-
|        <math>\frac{d}{dx}\bigg(\frac{f(x)}{g(x)}\bigg)=\frac{g(x)f'(x)-f(x)g'(x)}{(g(x))^2}</math>
|-
|'''4.''' '''Derivative of natural logarithm
|-
|        <math>\frac{d}{dx}(\ln x)=\frac{1}{x}</math>
|}

'''Solution:'''

'''(a)'''
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|First, we use the Chain Rule to get
|-
|        <math>f'(x)=3\tan^2(7x^2+5)(\tan(7x^2+5))'.</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Now, we use the Chain Rule again to get
|-
|
        <math>\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}</math>
|}

'''(b)'''
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|First, we use the Chain Rule to get
|-
|        <math>g'(x)=\cos(\cos(e^x))(\cos(e^x))'.</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Now, we use the Chain Rule again to get
|-
|
        <math>\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}</math>
|}

'''(c)'''
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!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>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!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>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
|-
|    '''(a)'''     <math>3\tan^2(7x^2+5)\sec^2(7x^2+5)(14x)</math>
|-
|    '''(b)'''     <math>\cos(\cos(e^x))(-\sin(e^x))(e^x)</math>
|-
|    '''(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|'''Return to Sample Exam''']]

@@ Line 7: / Line 7: @@
 <span class="exam">(c) &nbsp; <math style="vertical-align: -18px">h(x)=\frac{(5x^2+7x)^3}{\ln(x^2+1)} </math>
 <hr>
-[[File:9ASM2P5.jpg|600px|thumb|center]]
+[[009A Sample Midterm 2, Problem 5 Detailed Solution|'''<u>Detailed Solution</u>''']]
 [[009A_Sample_Midterm_2|'''<u>Return to Sample Exam</u>''']]

@@ Line 6: / Line 6: @@
 <span class="exam">(c) &nbsp; <math style="vertical-align: -18px">h(x)=\frac{(5x^2+7x)^3}{\ln(x^2+1)} </math>
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
+[[File:9ASM2P5.jpg|600px|thumb|center]]
 [[009A_Sample_Midterm_2|'''<u>Return to Sample Exam</u>''']]

@@ Line 85: / Line 85: @@
 |First, we use the Quotient Rule to get
 |-
-|&nbsp; &nbsp; &nbsp; &nbsp; <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>
+|&nbsp; &nbsp; &nbsp; &nbsp; <math>h'(x)=\frac{\ln(x^2+1)((5x^2+7x)^3)'-(5x^2+7x)^3(\ln(x^2+1))'}{(\ln(x^2+1))^2}.</math>
 |}
@@ Line 94: / Line 94: @@
 |-
 |&nbsp; &nbsp; &nbsp; &nbsp; <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{h'(x)} & = & \displaystyle{\frac{\ln(x^2+1)((5x^2+7x)^3)'-(5x^2+7x)^3(\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)3(5x^2+7x)^2(5x^2+7x)'-(5x^2+7x)^3\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}.}
+& = & \displaystyle{\frac{\ln(x^2+1)3(5x^2+7x)^2(10x+7)-(5x^2+7x)^3\frac{1}{x^2+1}(2x)}{(\ln(x^2+1))^2}.}
 \end{array}</math>
 |}
@@ Line 110: / Line 110: @@
 |&nbsp; &nbsp; '''(b)''' &nbsp; &nbsp; <math>g'(x)=\cos(\cos(e^x))(-\sin(e^x))(e^x)</math>
 |-
-|&nbsp; &nbsp; '''(c)''' &nbsp; &nbsp; <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>
+|&nbsp; &nbsp; '''(c)''' &nbsp; &nbsp; <math>h'(x)=\frac{\ln(x^2+1)3(5x^2+7x)^2(10x+7)-(5x^2+7x)^3\frac{1}{x^2+1}(2x)}{(\ln(x^2+1))^2}</math>
 |}
 [[009A_Sample_Midterm_2|'''<u>Return to Sample Exam</u>''']]

@@ Line 106: / Line 106: @@
 !Final Answer: &nbsp;
 |-
-|&nbsp; &nbsp; '''(a)''' &nbsp; &nbsp; <math>3\tan^2(7x^2+5)\sec^2(7x^2+5)(14x)</math>
+|&nbsp; &nbsp; '''(a)''' &nbsp; &nbsp; <math>f'(x)=3\tan^2(7x^2+5)\sec^2(7x^2+5)(14x)</math>
 |-
-|&nbsp; &nbsp; '''(b)''' &nbsp; &nbsp; <math>\cos(\cos(e^x))(-\sin(e^x))(e^x)</math>
+|&nbsp; &nbsp; '''(b)''' &nbsp; &nbsp; <math>g'(x)=\cos(\cos(e^x))(-\sin(e^x))(e^x)</math>
 |-
-|&nbsp; &nbsp; '''(c)''' &nbsp; &nbsp; <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>
+|&nbsp; &nbsp; '''(c)''' &nbsp; &nbsp; <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>
 |}
 [[009A_Sample_Midterm_2|'''<u>Return to Sample Exam</u>''']]