009A Sample Midterm 2, Problem 3 - Revision history

MathAdmin at 22:21, 9 November 2017

2017-11-09T22:21:12Z

MathAdmin at 17:08, 7 November 2017

2017-11-07T17:08:53Z

MathAdmin at 02:50, 14 April 2017

2017-04-14T02:50:26Z

MathAdmin: Created page with "Use the definition of the derivative to find $\frac{dy}{dx}$ for the function $y=\frac{1+x..."$

2017-04-09T18:52:12Z

Created page with "<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..."

New page

<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>

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Foundations:  
|-
|Recall
|-
|       <math style="vertical-align: -13px">f'(x)=\lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}</math>
|}

'''Solution:'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|Let  <math style="vertical-align: -14px">f(x)=\frac{1+x}{3x}.</math>
|-
|Using the limit definition of derivative, we have
|-
|        <math>\begin{array}{rcl}
\displaystyle{f'(x)} & = & \displaystyle{\lim_{h\rightarrow 0} \frac{f(x+h)-f(x)}{h}}\\
&&\\
& = & \displaystyle{\lim_{h\rightarrow 0} \frac{(\frac{1+(x+h)}{3(x+h)})-(\frac{1+x}{3x})}{h}}\\
&&\\
& = & \displaystyle{\lim_{h\rightarrow 0} \frac{(\frac{1+x+h}{3x+3h})-(\frac{1+x}{h})}{h}.}
\end{array}</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Now, we get a common denominator for the fractions in the numerator.
|-
|Hence, we have
|-
|        <math>\begin{array}{rcl}
\displaystyle{f'(x)} & = & \displaystyle{\lim_{h\rightarrow 0}\frac{\frac{(1+x+h)3x}{(3x+3h)(3x)}-\frac{(1+x)(3x+3h)}{(3x+3h)(3x)}}{h}}\\
&&\\
& = & \displaystyle{\lim_{h\rightarrow 0}\frac{\frac{3x+3x^2+3xh-(3x+3h+3x^2+3hx)}{(3x+3h)(3x)}}{h}}\\
&&\\
& = & \displaystyle{\lim_{h\rightarrow 0} \frac{-3h}{h(3x+3h)(3x)}}\\
&&\\
& = & \displaystyle{\lim_{h\rightarrow 0} \frac{-3}{(3x+3h)(3x)}}\\
&&\\
& = & \displaystyle{\frac{-3}{(3x)(3x)}}\\
& = & \displaystyle{-\frac{1}{3x^2}.}
\end{array}</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
|-
|        <math>-\frac{1}{3x^2}</math>
|-
|
|}
[[009A_Sample_Midterm_2|'''<u>Return to Sample Exam</u>''']]

@@ Line 2: / Line 2: @@
 <hr>
-[[009A Sample Midterm 2, Problem 3 Detailed Solution|'''<u>Detailed Solution with Background Information</u>''']]
+[[009A Sample Midterm 2, Problem 3 Solution|'''<u>Solution</u>''']]
 [[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;"
+[[File:9ASM2P3.jpg|600px|thumb|center]]
 [[009A_Sample_Midterm_2|'''<u>Return to Sample Exam</u>''']]

@@ Line 54: / Line 54: @@
 !Final Answer: &nbsp;
 |-
-|&nbsp; &nbsp; &nbsp; &nbsp; <math>-\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>''']]

009A Sample Midterm 2, Problem 3 - Revision history

MathAdmin at 22:21, 9 November 2017

MathAdmin at 17:08, 7 November 2017

MathAdmin at 02:50, 14 April 2017

MathAdmin: Created page with "Use the definition of the derivative to find \frac{dy}{dx} for the function y=\frac{1+x..."

MathAdmin: Created page with "Use the definition of the derivative to find $\frac{dy}{dx}$ for the function $y=\frac{1+x..."$