009A Sample Midterm 2, Problem 5 - Revision history

Kayla Murray at 22:40, 9 November 2017

2017-11-09T22:40:11Z

2017-11-07T01:43:47Z

2017-11-03T17:12:25Z

2017-02-18T23:21:26Z

2017-02-18T21:53:09Z

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@@ 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 8: / Line 8: @@
 <hr>
-(insert picture of handwritten solution)
+[[009A Sample Midterm 2, Problem 5 Detailed Solution|'''<u>Detailed Solution with Background Information</u>''']]
-[[009A Sample Midterm 2, Problem 5 Detailed Solution|'''<u>Detailed Solution for this Problem</u>''']]
+[[File:9ASM2P5.jpg|600px|thumb|center]]
 [[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;"
+[[009A Sample Midterm 2, Problem 5 Detailed Solution|'''<u>Detailed Solution for this Problem</u>''']]
 [[009A_Sample_Midterm_2|'''<u>Return to Sample Exam</u>''']]

@@ Line 1: / Line 1: @@
 <span class="exam">Find the derivatives of the following functions. Do not simplify.
-<span class="exam">(a) &nbsp; <math>f(x)=\tan^3(7x^2+5) </math>
+<span class="exam">(a) &nbsp; <math style="vertical-align: -5px">f(x)=\tan^3(7x^2+5) </math>
-<span class="exam">(b) &nbsp; <math>g(x)=\sin(\cos(e^x)) </math>
+<span class="exam">(b) &nbsp; <math style="vertical-align: -5px">g(x)=\sin(\cos(e^x)) </math>
-<span class="exam">(c) &nbsp; <math>h(x)=\frac{(5x^2+7x)^3}{\ln(x^2+1)} </math>
+<span class="exam">(c) &nbsp; <math style="vertical-align: -18px">h(x)=\frac{(5x^2+7x)^3}{\ln(x^2+1)} </math>

@@ Line 1: / Line 1: @@
 <span class="exam">Find the derivatives of the following functions. Do not simplify.
-::<span class="exam">a) <math>f(x)=\tan^3(7x^2+5) </math>
+<span class="exam">(a) &nbsp; <math>f(x)=\tan^3(7x^2+5) </math>
-::<span class="exam">b) <math>g(x)=\sin(\cos(e^x)) </math>
-::<span class="exam">c) <math>h(x)=\frac{(5x^2+7x)^3}{\ln(x^2+1)} </math>
+<span class="exam">(b) &nbsp; <math>g(x)=\sin(\cos(e^x)) </math>

@@ Line 9: / Line 9: @@
 !Foundations: &nbsp;
 |-
-|'''1.''' Chain Rule
+|'''1.''' '''Chain Rule'''
 |-
-|'''2.''' Derivatives of trig/ln
+|&nbsp; &nbsp; &nbsp; &nbsp; <math>\frac{d}{dx}(f(g(x)))=f'(g(x))g'(x)</math>
 |-
-|'''3.''' Quotient Rule
+|'''2.''' '''Trig Derivatives'''
 |}

@@ Line 71: / Line 71: @@
 !Step 1: &nbsp;
 |-
+|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>
 |}
@@ Line 83: / Line 79: @@
 !Step 2: &nbsp;
 |-
+|Now, we use the Chain Rule to get
 |-
+|&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
 |}
@@ Line 100: / Line 98: @@
 |&nbsp; &nbsp; '''(b)''' &nbsp; &nbsp; <math>\cos(\cos(e^x))(-\sin(e^x))(e^x)</math>
 |-
-|'''(c)'''
+|&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>
 |}
 [[009A_Sample_Midterm_2|'''<u>Return to Sample Exam</u>''']]

@@ Line 47: / Line 47: @@
 !Step 1: &nbsp;
 |-
+|First, we use the Chain Rule to get
 |-
+|&nbsp; &nbsp; &nbsp; &nbsp; <math>g'(x)=\cos(\cos(e^x))(\cos(e^x))'.</math>
 |}
@@ Line 59: / Line 55: @@
 !Step 2: &nbsp;
 |-
+|Now, we use the Chain Rule again to get
 |-
+|
 |}
@@ Line 99: / Line 98: @@
 |&nbsp; &nbsp; '''(a)''' &nbsp; &nbsp; <math>3\tan^2(7x^2+5)\sec^2(7x^2+5)(14x)</math>
 |-
-|'''(b)'''
+|&nbsp; &nbsp; '''(b)''' &nbsp; &nbsp; <math>\cos(\cos(e^x))(-\sin(e^x))(e^x)</math>
 |-
 |'''(c)'''
 |}
 [[009A_Sample_Midterm_2|'''<u>Return to Sample Exam</u>''']]

@@ Line 25: / Line 25: @@
 |First, we use the Chain Rule to get
 |-
-|&nbsp; &nbsp; &nbsp; &nbsp; <math>f'(x)=3\tan^2(
+|&nbsp; &nbsp; &nbsp; &nbsp; <math>f'(x)=3\tan^2(7x^2+5)(\tan(7x^2+5))'.</math>
 |}
@@ Line 31: / Line 31: @@
 !Step 2: &nbsp;
 |-
+|Now, we use the Chain Rule again to get
 |-
+|
 |}
@@ Line 90: / Line 97: @@
 !Final Answer: &nbsp;
 |-
-|'''(a)'''
+|&nbsp; &nbsp; '''(a)''' &nbsp; &nbsp; <math>3\tan^2(7x^2+5)\sec^2(7x^2+5)(14x)</math>
 |-
 |'''(b)'''