009A Sample Midterm 1, Problem 4 - Revision history

Kayla Murray at 01:39, 7 November 2017

2017-11-07T01:39:35Z

2017-11-03T14:44:52Z

2017-11-03T14:37:00Z

2017-02-18T23:55:22Z

2017-02-18T21:59:35Z

2017-02-18T19:45:06Z

2017-02-16T21:00:24Z

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@@ Line 5: / Line 5: @@
 <span class="exam">(b) Find the equation of the tangent line to &nbsp;<math style="vertical-align: -5px">y=\sqrt{3x-5}</math>&nbsp; at &nbsp;<math style="vertical-align: -5px">(2,1).</math>
 <hr>
-(insert picture of handwritten solution)
+[[File:9ASM1P4.jpg|600px|thumb|center]]
 [[009A_Sample_Midterm_1|'''<u>Return to Sample Exam</u>''']]

@@ Line 8: / Line 8: @@
 (insert picture of handwritten solution)
-[[009A Sample Midterm 2, Problem 2 Detailed Solution|'''<u>Detailed Solution for this Problem</u>''']]
+[[009A Sample Midterm 1, Problem 4 Detailed Solution|'''<u>Detailed Solution for this Problem</u>''']]
 [[009A_Sample_Midterm_1|'''<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"> Let &nbsp;<math style="vertical-align: -5px">y=\sqrt{3x-5}.</math>
-<span class="exam">(a) &nbsp; <math style="vertical-align: -5px">f(x)=\sqrt{x}(x^2+2)</math>
+<span class="exam">(a) Use the definition of the derivative to compute &nbsp; <math>\frac{dy}{dx}</math> &nbsp; for &nbsp;<math style="vertical-align: -5px">y=\sqrt{3x-5}.</math>
-<span class="exam">(b) &nbsp; <math style="vertical-align: -17px">g(x)=\frac{x+3}{x^{\frac{3}{2}}+2}</math> where <math style="vertical-align: 0px">x>0</math>
+<span class="exam">(b) Find the equation of the tangent line to &nbsp;<math style="vertical-align: -5px">y=\sqrt{3x-5}</math>&nbsp; at &nbsp;<math style="vertical-align: -5px">(2,1).</math>
-<span class="exam">(c) &nbsp; <math style="vertical-align: -20px">h(x)=\frac{e^{-5x^3}}{\sqrt{x^2+1}}</math>
+(insert picture of handwritten solution)
 [[009A_Sample_Midterm_1|'''<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)=\sqrt{x}(x^2+2)</math>
+<span class="exam">(a) &nbsp; <math style="vertical-align: -5px">f(x)=\sqrt{x}(x^2+2)</math>
-<span class="exam">(b) &nbsp; <math>g(x)=\frac{x+3}{x^{\frac{3}{2}}+2}</math> where <math>x>0</math>
+<span class="exam">(b) &nbsp; <math style="vertical-align: -17px">g(x)=\frac{x+3}{x^{\frac{3}{2}}+2}</math> where <math style="vertical-align: 0px">x>0</math>
-<span class="exam">(c) &nbsp; <math>h(x)=\frac{e^{-5x^3}}{\sqrt{x^2+1}}</math>
+<span class="exam">(c) &nbsp; <math style="vertical-align: -20px">h(x)=\frac{e^{-5x^3}}{\sqrt{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)=\sqrt{x}(x^2+2)</math>
+<span class="exam">(a) &nbsp; <math>f(x)=\sqrt{x}(x^2+2)</math>
-::<span class="exam">b) <math>g(x)=\frac{x+3}{x^{\frac{3}{2}}+2}</math> where <math>x>0</math>
-::<span class="exam">c) <math>h(x)=\frac{e^{-5x^3}}{\sqrt{x^2+1}}</math>
+<span class="exam">(b) &nbsp; <math>g(x)=\frac{x+3}{x^{\frac{3}{2}}+2}</math> where <math>x>0</math>

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

@@ Line 15: / Line 15: @@
 |'''3.''' Chain Rule
 |}
 '''Solution:'''
@@ Line 82: / Line 83: @@
 \end{array}</math>
 |}
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"

@@ Line 64: / Line 64: @@
 !Step 1: &nbsp;
 |-
+|Using the Quotient Rule, we have
 |-
+|&nbsp; &nbsp; &nbsp; &nbsp; <math>h'(x)=\frac{\sqrt{x^2+1}(e^{-5x^3})'-e^{-5x^3}(\sqrt{x^2+1})'}{(\sqrt{x^2+1})^2}.</math>
 |}
@@ Line 76: / Line 72: @@
 !Step 2: &nbsp;
 |-
+|Now, using the Chain Rule, we have
 |-
+|&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
 |}
@@ Line 92: / Line 90: @@
 |&nbsp; &nbsp; '''(b)''' &nbsp; &nbsp; <math>\frac{(x^{\frac{3}{2}}+2)(1)-(x+3)(\frac{3}{2}x^{\frac{1}{2}})}{(x^{\frac{3}{2}}+2)^2}</math>
 |-
-|'''(c)'''
+|&nbsp; &nbsp; '''(c)''' &nbsp; &nbsp; <math>\frac{\sqrt{x^2+1}(e^{-5x^3})(-15x^2)-e^{-5x^3}\frac{1}{2}(x^2+1)^{\frac{-1}{2}}(2x)}{(\sqrt{x^2+1})^2}</math>
 |}
 [[009A_Sample_Midterm_1|'''<u>Return to Sample Exam</u>''']]

@@ Line 45: / Line 45: @@
 |Using the Quotient Rule, we have
 |-
-|&nbsp; &nbsp; &nbsp; &nbsp; <math>g'(x)=.</math>
+|&nbsp; &nbsp; &nbsp; &nbsp; <math>g'(x)=\frac{(x^{\frac{3}{2}}+2)(x+3)'-(x+3)(x^{\frac{3}{2}}+2)'}{(x^{\frac{3}{2}}+2)^2}.</math>
 |}
@@ Line 51: / Line 51: @@
 !Step 2: &nbsp;
 |-
+|Now, we have
 |-
+|&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
 |}
@@ Line 90: / Line 90: @@
 |&nbsp; &nbsp; '''(a)''' &nbsp; &nbsp; <math>\bigg(\frac{1}{2}x^{-\frac{1}{2}}\bigg)(x^2+2)+\sqrt{x}(2x)</math>
 |-
-|'''(b)'''
+|&nbsp; &nbsp; '''(b)''' &nbsp; &nbsp; <math>\frac{(x^{\frac{3}{2}}+2)(1)-(x+3)(\frac{3}{2}x^{\frac{1}{2}})}{(x^{\frac{3}{2}}+2)^2}</math>
 |-
 |'''(c)'''
 |}
 [[009A_Sample_Midterm_1|'''<u>Return to Sample Exam</u>''']]