009A Sample Final 3, Problem 2 - Revision history

Kayla Murray at 23:34, 2 December 2017

2017-12-02T23:34:39Z

2017-03-06T18:05:37Z

2017-03-06T18:02:18Z

2017-03-06T17:54:01Z

2017-03-06T17:41:52Z

2017-02-19T02:01:11Z

2017-02-18T02:32:09Z

@@ Line 5: / Line 5: @@
 <span class="exam">(b) &nbsp;<math style="vertical-align: -5px">y=\cos(3\pi)+\tan^{-1}(\sqrt{x})</math>
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
+<hr>
-!Foundations: &nbsp;
+[[009A Sample Final 3, Problem 2 Solution|'''<u>Solution</u>''']]
-'''Solution:'''
+[[009A Sample Final 3, Problem 2 Detailed Solution|'''<u>Detailed Solution</u>''']]
 [[009A_Sample_Final_3|'''<u>Return to Sample Exam</u>''']]

@@ Line 1: / Line 1: @@
 <span class="exam"> Find the derivative of the following functions:
-<span class="exam">(a) &nbsp;<math>g(\theta)=\frac{\pi^2}{(\sec\theta -\sin 2\theta)^2}</math>
+<span class="exam">(a) &nbsp;<math style="vertical-align: -18px">g(\theta)=\frac{\pi^2}{(\sec\theta -\sin 2\theta)^2}</math>
-<span class="exam">(b) &nbsp;<math>y=\cos(3\pi)+\tan^{-1}(\sqrt{x})</math>
+<span class="exam">(b) &nbsp;<math style="vertical-align: -5px">y=\cos(3\pi)+\tan^{-1}(\sqrt{x})</math>
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
@@ Line 64: / Line 64: @@
 |&nbsp; &nbsp; &nbsp; &nbsp; <math>y'=(\cos(3\pi))'+(\tan^{-1}(\sqrt{x}))'.</math>
 |-
-|Since &nbsp;<math style="vertical-align: 0px">\cos(3\pi)</math>&nbsp; is a constant,
+|Since &nbsp;<math style="vertical-align: -5px">\cos(3\pi)</math>&nbsp; is a constant,
 |-
 |we have

@@ Line 60: / Line 60: @@
 !Step 1: &nbsp;
 |-
+|First, we have
 |-
+|&nbsp; &nbsp; &nbsp; &nbsp; <math>y'=(\cos(3\pi))'+(\tan^{-1}(\sqrt{x}))'.</math>
 |}
@@ Line 68: / Line 78: @@
 !Step 2: &nbsp;
 |-
+|Now, using the Chain Rule, we have
 |}
@@ Line 77: / Line 95: @@
 |&nbsp; &nbsp;'''(a)'''&nbsp; &nbsp; <math>g'(\theta)=\frac{-2\pi^2(\sec\theta\tan\theta -2\cos (2\theta))}{(\sec\theta -\sin 2\theta)^{3}}</math>
 |-
-|&nbsp; &nbsp;'''(b)'''&nbsp; &nbsp;
+|&nbsp; &nbsp;'''(b)'''&nbsp; &nbsp; <math>y'=\bigg(\frac{1}{1+x}\bigg)\frac{1}{2\sqrt{x}}</math>
 |}
 [[009A_Sample_Final_3|'''<u>Return to Sample Exam</u>''']]

@@ Line 1: / Line 1: @@
 <span class="exam"> Find the derivative of the following functions:
-<span class="exam">(a) <math>g(\theta)=\frac{\pi^2}{(\sec\theta -\sin 2\theta)^2}</math>
+<span class="exam">(a) &nbsp;<math>g(\theta)=\frac{\pi^2}{(\sec\theta -\sin 2\theta)^2}</math>
-<span class="exam">(b) <math>y=\cos(3\pi)+\tan^{-1}(\sqrt{x})</math>
+<span class="exam">(b) &nbsp;<math>y=\cos(3\pi)+\tan^{-1}(\sqrt{x})</math>
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
@@ Line 30: / Line 30: @@
 !Step 1: &nbsp;
 |-
+|First, we write
 |-
+|&nbsp; &nbsp; &nbsp; &nbsp;<math>g(\theta)=\pi^2(\sec\theta -\sin 2\theta)^{-2}.</math>
 |-
+|Now, using the Chain Rule, we have
 |-
+|&nbsp; &nbsp; &nbsp; &nbsp;<math>g'(\theta)=(-2)\pi^2(\sec\theta -\sin 2\theta)^{-3}(\sec\theta -\sin 2\theta)'.</math>
 |}
@@ Line 42: / Line 42: @@
 !Step 2: &nbsp;
 |-
+|Now, using the Chain Rule a second time, we get
 |-
+|&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
 |}
@@ Line 59: / Line 67: @@
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 !Step 2: &nbsp;
 |-
+|
@@ Line 95: / Line 75: @@
 !Final Answer: &nbsp;
 |-
-|'''(a)'''
+|&nbsp; &nbsp;'''(a)'''&nbsp; &nbsp; <math>g'(\theta)=\frac{-2\pi^2(\sec\theta\tan\theta -2\cos (2\theta))}{(\sec\theta -\sin 2\theta)^{3}}</math>
 |-
-|'''(c)'''
+|&nbsp; &nbsp;'''(b)'''&nbsp; &nbsp;
 |}
 [[009A_Sample_Final_3|'''<u>Return to Sample Exam</u>''']]

@@ Line 1: / Line 1: @@
 <span class="exam"> Find the derivative of the following functions:
-<span class="exam">a) <math>g(\theta)=\frac{\pi^2}{(\sec\theta -\sin 2\theta)^2}</math>
+<span class="exam">(a) <math>g(\theta)=\frac{\pi^2}{(\sec\theta -\sin 2\theta)^2}</math>
-<span class="exam">b) <math>y=\cos(3\pi)+\tan^{-1}(\sqrt{x})</math>
+<span class="exam">(b) <math>y=\cos(3\pi)+\tan^{-1}(\sqrt{x})</math>
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
@@ Line 16: / Line 16: @@
+|
 |}
 '''Solution:'''
@@ Line 84: / Line 85: @@
+|
 |}
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"

@@ Line 8: / Line 8: @@
 !Foundations: &nbsp;
 |-
+|'''1.''' '''Chain Rule'''
 |-
+|&nbsp; &nbsp; &nbsp; &nbsp; <math>\frac{d}{dx}(f(g(x)))=f'(g(x))g'(x)</math>
 |-
+|'''2.''' '''Trig Derivatives'''
 |-
+|
 |}