009A Sample Midterm 3, Problem 3 - Revision history

Kayla Murray at 20:13, 11 November 2017

2017-11-11T20:13:55Z

Kayla Murray at 20:05, 11 November 2017

2017-11-11T20:05:00Z

Kayla Murray at 20:03, 11 November 2017

2017-11-11T20:03:55Z

Kayla Murray at 20:01, 11 November 2017

2017-11-11T20:01:01Z

Kayla Murray at 17:51, 27 March 2017

2017-03-27T17:51:27Z

Kayla Murray at 17:19, 13 March 2017

2017-03-13T17:19:57Z

Kayla Murray at 00:46, 27 February 2017

2017-02-27T00:46:08Z

Kayla Murray at 22:45, 18 February 2017

2017-02-18T22:45:45Z

Kayla Murray at 22:45, 18 February 2017

2017-02-18T22:45:06Z

Kayla Murray at 21:41, 18 February 2017

2017-02-18T21:41:22Z

← Older revision		Revision as of 20:13, 11 November 2017
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−	<span class="exam"> Let  <math style="vertical-align: -3px">y=\sqrt{2x+5},x\ge 0.</math>	+	<span class="exam"> Let  <math style="vertical-align: -3px">y=3\sqrt{2x+5},x\ge 0.</math>

	<span class="exam">(a) Use the definition of the derivative to compute   <math style="vertical-align: -13px">\frac{dy}{dx}.</math>		<span class="exam">(a) Use the definition of the derivative to compute   <math style="vertical-align: -13px">\frac{dy}{dx}.</math>

← Older revision		Revision as of 20:05, 11 November 2017
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−	~~[[009A_Sample_Midterm_3\|'''<u>Return to Sample Exam</u>''']]~~
−
−
−	~~{\| 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: -5px">f(x)=3\sqrt{-2x+5}.</math>~~
−	\|-
−	~~\|Using the limit definition of the 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{3\sqrt{-2(x+h)+5}-3\sqrt{-2x+5}}{h}}\\~~
−	~~&&\\~~
−	~~& = & \displaystyle{\lim_{h\rightarrow 0} \frac{3\sqrt{-2x+-2h+5}-3\sqrt{-2x+5}}{h}}\\~~
−	~~&&\\~~
−	~~& = & \displaystyle{3\lim_{h\rightarrow 0} \frac{\sqrt{-2x+-2h+5}-\sqrt{-2x+5}}{h}.}~~
−	~~\end{array}</math>~~
−	\|-
−	\|
−	\|}
−
−	~~{\| class="mw-collapsible mw-collapsed" style = "text-align:left;"~~
−	~~!Step 2:  ~~
−	\|-
−	~~\|Now, we multiply the numerator and denominator by the conjugate of the numerator.~~
−	\|-
−	~~\|Hence, we have~~
−	\|-
−	~~\|        <math>\begin{array}{rcl}~~
−	~~\displaystyle{f'(x)} & = & \displaystyle{3\lim_{h\rightarrow 0} \frac{(\sqrt{-2x+-2h+5}-\sqrt{-2x+5})}{h} \frac{(\sqrt{-2x+-2h+5}+\sqrt{-2x+5})}{(\sqrt{-2x+-2h+5}+\sqrt{-2x+5})}}\\~~
−	~~&&\\~~
−	~~& = & \displaystyle{3\lim_{h\rightarrow 0} \frac{(-2x+-2h+5)-(-2x+5)}{h(\sqrt{-2x+-2h+5}+\sqrt{-2x+5})}}\\~~
−	~~&&\\~~
−	~~& = & \displaystyle{3\lim_{h\rightarrow 0} \frac{-2h}{h(\sqrt{-2x+-2h+5}+\sqrt{-2x+5})}}\\~~
−	~~&&\\~~
−	~~& = & \displaystyle{3\lim_{h\rightarrow 0} \frac{-2}{\sqrt{-2x+-2h+5}+\sqrt{-2x+5}}}\\~~
−	~~&&\\~~
−	~~& = & \displaystyle{3\frac{-2}{\sqrt{-2x+5}+\sqrt{-2x+5}}}\\~~
−	~~&&\\~~
−	~~& = & \displaystyle{-\frac{3}{\sqrt{-2x+5}}.}~~
−	~~\end{array}</math>~~
−	\|-
−	\|
−	\|}
−
−
−	~~{\| class="mw-collapsible mw-collapsed" style = "text-align:left;"~~
−	~~!Final Answer:  ~~
−	\|-
−	~~\|        <math>-\frac{3}{\sqrt{-2x+5}}</math>~~
−	\|-
−	\|
−	\|}
	[[009A_Sample_Midterm_3\|'''<u>Return to Sample Exam</u>''']]		[[009A_Sample_Midterm_3\|'''<u>Return to Sample Exam</u>''']]

@@ Line 1: / Line 1: @@
-<span class="exam"> Let &nbsp;<math>y=\sqrt{2x+5},x\ge 0.</math>
+<span class="exam"> Let &nbsp;<math style="vertical-align: -3px">y=\sqrt{2x+5},x\ge 0.</math>
-<span class="exam">(a) Use the definition of the derivative to compute &nbsp; <math>\frac{dy}{dx}.</math>
+<span class="exam">(a) Use the definition of the derivative to compute &nbsp; <math style="vertical-align: -13px">\frac{dy}{dx}.</math>
-<span class="exam">(b) Find the equation of the tangent line to &nbsp;<math>y=3\sqrt{2x+5}</math>&nbsp; at &nbsp;<math>(2,9).</math>
+<span class="exam">(b) Find the equation of the tangent line to &nbsp;<math style="vertical-align: -3px">y=3\sqrt{2x+5}</math>&nbsp; at &nbsp;<math style="vertical-align: -3px">(2,9).</math>
 <hr>

← Older revision		Revision as of 20:01, 11 November 2017
Line 1:		Line 1:
−	<span class="exam"> Use the definition of the derivative to compute   <math>\frac{dy}{dx}</math> ~~  for~~  <math ~~style="vertical-align: -4px"~~>y=3\sqrt{-2x+5}.</math>	+	<span class="exam"> Let  <math>y=\sqrt{2x+5},x\ge 0.</math>
		+
		+	<span class="exam">(a) Use the definition of the derivative to compute   <math>\frac{dy}{dx}.</math>
		+
		+	<span class="exam">(b) Find the equation of the tangent line to  <math>y=3\sqrt{2x+5}</math>  at  <math>(2,9).</math>
		+	<hr>
		+
		+	[[009A Sample Midterm 3, Problem 3 Solution\|'''<u>Solution</u>''']]
		+
		+
		+	[[009A Sample Midterm 3, Problem 3 Detailed Solution\|'''<u>Detailed Solution</u>''']]
		+
		+
		+	[[009A_Sample_Midterm_3\|'''<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 50: / Line 50: @@
 & = & \displaystyle{3\frac{-2}{\sqrt{-2x+5}+\sqrt{-2x+5}}}\\
 &&\\
-& = & \displaystyle{\frac{-3}{\sqrt{-2x+5}}.}
+& = & \displaystyle{-\frac{3}{\sqrt{-2x+5}}.}
 \end{array}</math>
 |-
@@ Line 60: / Line 60: @@
 !Final Answer: &nbsp;
 |-
-|&nbsp; &nbsp; &nbsp; &nbsp; <math>\frac{-3}{\sqrt{-2x+5}}</math>
+|&nbsp; &nbsp; &nbsp; &nbsp; <math>-\frac{3}{\sqrt{-2x+5}}</math>
 |-
+|
 |}
 [[009A_Sample_Midterm_3|'''<u>Return to Sample Exam</u>''']]

@@ Line 1: / Line 1: @@
-<span class="exam"> Use the definition of the derivative to compute &nbsp; <math>\frac{dy}{dx}</math> &nbsp; for <math style="vertical-align: -4px">y=3\sqrt{-2x+5}.</math>
+<span class="exam"> Use the definition of the derivative to compute &nbsp; <math>\frac{dy}{dx}</math> &nbsp; for &nbsp;<math style="vertical-align: -4px">y=3\sqrt{-2x+5}.</math>
@@ 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: -5px">f(x)=3\sqrt{-2x+5}.</math>
+|Let &nbsp;<math style="vertical-align: -5px">f(x)=3\sqrt{-2x+5}.</math>
 |-
 |Using the limit definition of the derivative, we have

← Older revision		Revision as of 22:45, 18 February 2017
Line 1:		Line 1:
−	<span class="exam"> Use the definition of the derivative to compute   <math>\frac{dy}{dx}</math>   for <math>y=3\sqrt{-2x+5}.</math>	+	<span class="exam"> Use the definition of the derivative to compute   <math>\frac{dy}{dx}</math>   for <math style="vertical-align: -4px">y=3\sqrt{-2x+5}.</math>

← Older revision		Revision as of 21:41, 18 February 2017
Line 1:		Line 1:
−	<span class="exam"> Use the definition of the derivative to compute <math>\frac{dy}{dx}</math> for <math>y=3\sqrt{-2x+5}.</math>	+	<span class="exam"> Use the definition of the derivative to compute   <math>\frac{dy}{dx}</math>   for <math>y=3\sqrt{-2x+5}.</math>