009A Sample Midterm 2, Problem 3 - Revision history

Kayla Murray at 22:37, 9 November 2017

2017-11-09T22:37:02Z

2017-11-07T01:42:48Z

2017-11-03T17:07:16Z

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@@ Line 1: / Line 1: @@
 <span class="exam">Use the definition of the derivative to find &nbsp; <math>\frac{dy}{dx}</math> &nbsp; for the function &nbsp;<math style="vertical-align: -12px">y=\frac{1+x}{3x}.</math>
 <hr>
-[[File:9ASM2P3.jpg|600px|thumb|center]]
+[[009A Sample Midterm 2, Problem 3 Detailed Solution|'''<u>Detailed Solution</u>''']]
 [[009A_Sample_Midterm_2|'''<u>Return to Sample Exam</u>''']]

@@ Line 2: / Line 2: @@
 <hr>
-(insert picture of handwritten solution)
+[[009A Sample Midterm 2, Problem 3 Detailed Solution|'''<u>Detailed Solution with Background Information</u>''']]
-[[009A Sample Midterm 2, Problem 3 Detailed Solution|'''<u>Detailed Solution for this Problem</u>''']]
+[[File:9ASM2P3.jpg|600px|thumb|center]]
 [[009A_Sample_Midterm_2|'''<u>Return to Sample Exam</u>''']]

@@ Line 1: / Line 1: @@
 <span class="exam">Use the definition of the derivative to find &nbsp; <math>\frac{dy}{dx}</math> &nbsp; for the function &nbsp;<math style="vertical-align: -12px">y=\frac{1+x}{3x}.</math>
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
+[[009A Sample Midterm 2, Problem 3 Detailed Solution|'''<u>Detailed Solution for this Problem</u>''']]
 [[009A_Sample_Midterm_2|'''<u>Return to Sample Exam</u>''']]

@@ Line 5: / Line 5: @@
 !Foundations: &nbsp;
 |-
-|<math style="vertical-align: -13px">f'(x)=\lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}</math>
+|Recall
 |}

@@ Line 52: / Line 52: @@
 !Final Answer: &nbsp;
 |-
-|&nbsp; &nbsp; &nbsp; &nbsp; <math>\frac{dy}{dx}=-\frac{1}{3x^2}</math>
+|&nbsp; &nbsp; &nbsp; &nbsp; <math>-\frac{1}{3x^2}</math>
 |-
+|
 |}
 [[009A_Sample_Midterm_2|'''<u>Return to Sample Exam</u>''']]

@@ Line 44: / Line 44: @@
 &&\\
 & = & \displaystyle{\frac{-3}{(3x)(3x)}}\\
-& = & \displaystyle{\frac{-1}{3x^2}.}
+& = & \displaystyle{-\frac{1}{3x^2}.}
 \end{array}</math>
 |}
@@ Line 52: / Line 52: @@
 !Final Answer: &nbsp;
 |-
-|&nbsp; &nbsp; &nbsp; &nbsp; <math>\frac{dy}{dx}=\frac{-1}{3x^2}</math>
+|&nbsp; &nbsp; &nbsp; &nbsp; <math>\frac{dy}{dx}=-\frac{1}{3x^2}</math>
 |-
+|
 |}
 [[009A_Sample_Midterm_2|'''<u>Return to Sample Exam</u>''']]

@@ Line 5: / Line 5: @@
 !Foundations: &nbsp;
 |-
-|'''Limit Definition of Derivative'''
+|<math style="vertical-align: -13px">f'(x)=\lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}</math>
 |}
@@ Line 16: / Line 14: @@
 !Step 1: &nbsp;
 |-
-|Let <math style="vertical-align: -14px">f(x)=\frac{1+x}{3x}.</math>
+|Let &nbsp;<math style="vertical-align: -14px">f(x)=\frac{1+x}{3x}.</math>
 |-
 |Using the limit definition of derivative, we have

← Older revision		Revision as of 23:55, 26 February 2017
Line 1:		Line 1:
−	<span class="exam">Use the definition of the derivative to find   <math>\frac{dy}{dx}</math>   for the function <math style="vertical-align: -12px">y=\frac{1+x}{3x}.</math>	+	<span class="exam">Use the definition of the derivative to find   <math>\frac{dy}{dx}</math>   for the function  <math style="vertical-align: -12px">y=\frac{1+x}{3x}.</math>

← Older revision		Revision as of 23:16, 18 February 2017
Line 1:		Line 1:
−	<span class="exam">Use the definition of the derivative to find   <math>\frac{dy}{dx}</math>   for the function <math>y=\frac{1+x}{3x}.</math>	+	<span class="exam">Use the definition of the derivative to find   <math>\frac{dy}{dx}</math>   for the function <math style="vertical-align: -12px">y=\frac{1+x}{3x}.</math>