009A Sample Final 2, Problem 3 - Revision history

Kayla Murray at 17:10, 1 December 2017

2017-12-01T17:10:55Z

Kayla Murray at 17:10, 1 December 2017

2017-12-01T17:10:31Z

Kayla Murray at 00:40, 21 May 2017

2017-05-21T00:40:48Z

Kayla Murray at 01:36, 8 March 2017

2017-03-08T01:36:38Z

Kayla Murray at 01:33, 8 March 2017

2017-03-08T01:33:45Z

Kayla Murray at 01:22, 8 March 2017

2017-03-08T01:22:49Z

Kayla Murray at 01:16, 8 March 2017

2017-03-08T01:16:37Z

Kayla Murray at 00:06, 8 March 2017

2017-03-08T00:06:14Z

Kayla Murray at 21:45, 6 March 2017

2017-03-06T21:45:28Z

Kayla Murray at 01:54, 19 February 2017

2017-02-19T01:54:22Z

@@ Line 8: / Line 8: @@
 <hr>
-[[009A Sample Midterm 2, Problem 3 Solution|'''<u>Solution</u>''']]
+[[009A Sample Final 2, Problem 3 Solution|'''<u>Solution</u>''']]
-[[009A Sample Midterm 2, Problem 3 Detailed Solution|'''<u>Detailed Solution</u>''']]
+[[009A Sample Final 2, Problem 3 Detailed Solution|'''<u>Detailed Solution</u>''']]
-[[009A_Sample_Midterm_2|'''<u>Return to Sample Exam</u>''']]
+[[009A_Sample_Final_2|'''<u>Return to Sample Exam</u>''']]

@@ Line 7: / Line 7: @@
 <span class="exam">(c) &nbsp;<math style="vertical-align: -5px">y=\sin^{-1} x</math>
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
+<hr>
-!Foundations: &nbsp;
+[[009A Sample Midterm 2, Problem 3 Solution|'''<u>Solution</u>''']]
-'''Solution:'''
+[[009A Sample Midterm 2, Problem 3 Detailed Solution|'''<u>Detailed Solution</u>''']]
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
+[[009A_Sample_Midterm_2|'''<u>Return to Sample Exam</u>''']]

@@ Line 88: / Line 88: @@
 !Step 1: &nbsp;
 |-
-|Let &nbsp;<math style="vertical-align: -5px">f(x)=\sin(x)</math>&nbsp; and &nbsp;<math style="vertical-align: -5px">g(x)=\sin^{-1}x.</math>
+|Let &nbsp;<math style="vertical-align: -5px">y=\sin^{-1}(x).</math>&nbsp; Then,
 |-
-|These functions are inverses of each other since
+|&nbsp; &nbsp; &nbsp; &nbsp;<math>\sin(y)=x</math>
 |-
-|&nbsp; &nbsp; &nbsp; &nbsp;<math style="vertical-align: -5px">f(g(x))=x</math>&nbsp; and &nbsp; <math style="vertical-align: -5px">g(f(x))=x.</math>
+|for &nbsp;<math>y</math>&nbsp; in the interval &nbsp;<math>\bigg[-\frac{\pi}{2},\frac{\pi}{2}\bigg].</math>
 |-
-|Therefore,
+|Using implicit differentiation, we have
 |-
-|&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
+|&nbsp; &nbsp; &nbsp; &nbsp;<math>\cos(y) \frac{dy}{dx}=1.</math>
 |-
-|Now, let &nbsp;<math style="vertical-align: -5px">y=\sin^{-1}(x).</math>&nbsp; Then, &nbsp;<math style="vertical-align: -5px">x=\sin(y).</math>
+|Solving for &nbsp;<math style="vertical-align: -15px">\frac{dy}{dx},</math>&nbsp; we get
 |-
-|So, &nbsp;<math style="vertical-align: -5px">\cos(\sin^{-1} x)=\cos(y).</math>
+|&nbsp; &nbsp; &nbsp; &nbsp; <math>\frac{dy}{dx}=\frac{1}{\cos(y)}.</math>
 |}
@@ Line 114: / Line 106: @@
 !Step 2: &nbsp;
 |-
-|Now, since
+|Now, since &nbsp;<math>\sin(y)=x,</math>&nbsp; we have the following diagram.
 |-
-|&nbsp; &nbsp; &nbsp; &nbsp;<math>\cos^2 y+\sin^2 y =1,</math>
+|(Insert diagram)
 |-
-|we have
+|Therefore,
 |-
 |&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
-\displaystyle{\cos(y)} & = & \displaystyle{\sqrt{1-\sin^2 y}}\\
+\displaystyle{\frac{dy}{dx}} & = & \displaystyle{\frac{1}{\cos(y)}}\\
 &&\\
-& = & \displaystyle{\sqrt{1-x^2}.}
+& = & \displaystyle{\frac{1}{\sqrt{1-x^2}}.}
 \end{array}</math>
 |}

@@ Line 88: / Line 88: @@
 !Step 1: &nbsp;
 |-
-|Let &nbsp;<math>f(x)=\sin(x)</math>&nbsp; and &nbsp;<math>g(x)=\sin^{-1}x.</math>
+|Let &nbsp;<math style="vertical-align: -5px">f(x)=\sin(x)</math>&nbsp; and &nbsp;<math style="vertical-align: -5px">g(x)=\sin^{-1}x.</math>
 |-
 |These functions are inverses of each other since
 |-
-|&nbsp; &nbsp; &nbsp; &nbsp;<math>f(g(x))=x</math>&nbsp; and &nbsp; <math>g(f(x))=x.</math>
+|&nbsp; &nbsp; &nbsp; &nbsp;<math style="vertical-align: -5px">f(g(x))=x</math>&nbsp; and &nbsp; <math style="vertical-align: -5px">g(f(x))=x.</math>
 |-
 |Therefore,
@@ Line 102: / Line 102: @@
 \end{array}</math>
 |-
-|Now, let &nbsp;<math>y=\sin^{-1}(x).</math>&nbsp; Then, &nbsp;<math>x=\sin(y).</math>
+|Now, let &nbsp;<math style="vertical-align: -5px">y=\sin^{-1}(x).</math>&nbsp; Then, &nbsp;<math style="vertical-align: -5px">x=\sin(y).</math>
 |-
-|So, &nbsp;<math>\cos(\sin^{-1} x)=\cos(y).</math>
+|So, &nbsp;<math style="vertical-align: -5px">\cos(\sin^{-1} x)=\cos(y).</math>
 |-
 |Therefore,
@@ Line 117: / Line 117: @@
 |-
 |&nbsp; &nbsp; &nbsp; &nbsp;<math>\cos^2 y+\sin^2 y =1,</math>
 |-
 |&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
@@ Line 137: / Line 139: @@
 |&nbsp; &nbsp;'''(b)'''&nbsp; &nbsp; <math>\frac{-x\sin(\sqrt{x+1})}{2\sqrt{x+1}}+\cos(\sqrt{x+1})</math>
 |-
-|&nbsp; &nbsp;'''(c)'''&nbsp; &nbsp; <math>g'(x)=\frac{1}{\sqrt{1-x^2}}</math>
+|&nbsp; &nbsp;'''(c)'''&nbsp; &nbsp; <math>\frac{1}{\sqrt{1-x^2}}</math>
 |}
 [[009A_Sample_Final_2|'''<u>Return to Sample Exam</u>''']]

@@ Line 88: / Line 88: @@
 !Step 1: &nbsp;
 |-
+|Let &nbsp;<math>f(x)=\sin(x)</math>&nbsp; and &nbsp;<math>g(x)=\sin^{-1}x.</math>
 |-
+|These functions are inverses of each other since
 |-
+|&nbsp; &nbsp; &nbsp; &nbsp;<math>f(g(x))=x</math>&nbsp; and &nbsp; <math>g(f(x))=x.</math>
 |-
+|Therefore,
 |-
+|&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
 |}
@@ Line 102: / Line 114: @@
 !Step 2: &nbsp;
 |-
+|Now, since
 |-
+|&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
 |-
+|Hence,
 |-
+|&nbsp; &nbsp; &nbsp; &nbsp;<math>g'(x)=\frac{1}{\sqrt{1-x^2}}.</math>
 |}
@@ Line 119: / Line 137: @@
 |&nbsp; &nbsp;'''(b)'''&nbsp; &nbsp; <math>\frac{-x\sin(\sqrt{x+1})}{2\sqrt{x+1}}+\cos(\sqrt{x+1})</math>
 |-
-|'''(c)'''
+|&nbsp; &nbsp;'''(c)'''&nbsp; &nbsp; <math>g'(x)=\frac{1}{\sqrt{1-x^2}}</math>
 |}
 [[009A_Sample_Final_2|'''<u>Return to Sample Exam</u>''']]

@@ Line 60: / Line 60: @@
 !Step 1: &nbsp;
 |-
+|Using the Product Rule, we have
 |-
+|
@@ Line 68: / Line 70: @@
 !Step 2: &nbsp;
 |-
+|Now, using the Chain Rule, we get
 |}
@@ Line 105: / Line 117: @@
 |&nbsp; &nbsp;'''(a)'''&nbsp; &nbsp; <math>\frac{3(x^2+3)^2(-8x)}{(x^2-1)^4}</math>
 |-
-|'''(b)'''
+|&nbsp; &nbsp;'''(b)'''&nbsp; &nbsp; <math>\frac{-x\sin(\sqrt{x+1})}{2\sqrt{x+1}}+\cos(\sqrt{x+1})</math>
 |-
 |'''(c)'''
 |}
 [[009A_Sample_Final_2|'''<u>Return to Sample Exam</u>''']]

@@ Line 31: / Line 31: @@
 !Step 1: &nbsp;
 |-
+|Using the Chain Rule, we have
 |-
+|
 |}
@@ Line 43: / Line 40: @@
 !Step 2: &nbsp;
 |-
+|Now, using the Quotient Rule, we have
 |-
+|&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
 |}
@@ Line 96: / Line 103: @@
 !Final Answer: &nbsp;
 |-
-|'''(a)'''
+|&nbsp; &nbsp;'''(a)'''&nbsp; &nbsp; <math>\frac{3(x^2+3)^2(-8x)}{(x^2-1)^4}</math>
 |-
 |'''(b)'''

@@ Line 10: / Line 10: @@
 !Foundations: &nbsp;
 |-
+|'''1.''' '''Product Rule'''
 |-
+|&nbsp; &nbsp; &nbsp; &nbsp; <math>\frac{d}{dx}(f(x)g(x))=f(x)g'(x)+f'(x)g(x)</math>
 |-
+|'''2.''' '''Quotient Rule'''
 |-
+|&nbsp; &nbsp; &nbsp; &nbsp; <math>\frac{d}{dx}\bigg(\frac{f(x)}{g(x)}\bigg)=\frac{g(x)f'(x)-f(x)g'(x)}{(g(x))^2}</math>
 |}

@@ Line 1: / Line 1: @@
 <span class="exam">Compute &nbsp; <math>\frac{dy}{dx}.</math>
-<span class="exam">(a) <math style="vertical-align: -14px">y=\bigg(\frac{x^2+3}{x^2-1}\bigg)^3</math>
+<span class="exam">(a) &nbsp;<math style="vertical-align: -15px">y=\bigg(\frac{x^2+3}{x^2-1}\bigg)^3</math>
-<span class="exam">(b) <math style="vertical-align: -3px">y=x\cos(\sqrt{x+1})</math>
+<span class="exam">(b) &nbsp;<math style="vertical-align: -4px">y=x\cos(\sqrt{x+1})</math>
-<span class="exam">(c) <math>y=\sin^{-1} x</math>
+<span class="exam">(c) &nbsp;<math style="vertical-align: -5px">y=\sin^{-1} x</math>
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"