022 Sample Final A, Problem 4 - Revision history

Matt Lee at 20:16, 30 May 2015

2015-05-30T20:16:46Z

Matt Lee at 20:16, 30 May 2015

2015-05-30T20:16:16Z

Matt Lee at 20:06, 30 May 2015

2015-05-30T20:06:19Z

Matt Lee at 20:05, 30 May 2015

2015-05-30T20:05:50Z

Matt Lee: Created page with " 4. Use implicit differentiation to find $\frac{dy}{dx}: \qquad x+y = x^3y^3$ {| class="mw..."

2015-05-30T20:05:20Z

Created page with " 4. Use implicit differentiation to find <math>\frac{dy}{dx}: \qquad x+y = x^3y^3</math> {| class="mw..."

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4. Use implicit differentiation to find <math>\frac{dy}{dx}: \qquad x+y = x^3y^3</math>

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Foundations:  
|-
|When we use implicit differentiation, we combine the chain rule with the fact that <math style="vertical-align: -18%">y</math> is a function of <math style="vertical-align: 0%">x</math>, and could really be written as <math style="vertical-align: -21%">y(x).</math> Because of this, the derivative of <math style="vertical-align:-17%">y^3</math> with respect to <math style="vertical-align: 0%">x</math> requires the chain rule, so
|-
|     <math>\frac{d}{dx}\left(y^{3}\right)=3y^{2}\cdot\frac{dy}{dt}.</math>
|}

 '''Solution:'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|First, we differentiate each term separately with respect to <math style="vertical-align: 0%">x</math> to find that  <math style="vertical-align: -18%">x^{3}-y^{3}-y=x</math>  differentiates implicitly to
|-
|     <math>3x^{2}-3y^{2}\cdot\frac{dy}{dx}-\frac{dy}{dx}=1</math>.
|}
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Since they don't ask for a general expression of <math style="vertical-align:-24%">dy/dx</math>, but rather a particular value at a particular point, we can plug in the values <math style="vertical-align:-5%">x=1</math> and <math style="vertical-align:-18%">y=0</math>  to find
|-
|     <math>3(1)^{2}-3(0)^{2}\cdot\frac{dy}{dx}-\frac{dy}{dx}=1,</math>
|-
|which is equivalent to <math style="vertical-align:-60%">3-\frac{dy}{dx}=1</math>. This solves to <math style="vertical-align:-24%">dy/dx=2.</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
|-
|     <math style="vertical-align:-24%">dy/dx=2.</math>
|}
[[022_Exam_1_Sample_A|'''Return to Sample Exam''']]

@@ Line 27: / Line 27: @@
 !Step 2: &nbsp;
 |-
-|Now we need to solve for <math>\frac{dy}{dx}</math> and doing so we find that <math>\frac{dy}dx} = \frac{3x^2y^3 - 1}{1 - 3x^3y^2}</math>
+|Now we need to solve for <math>\frac{dy}{dx}</math> and doing so we find that <math>\frac{dy}{dx} = \frac{3x^2y^3 - 1}{1 - 3x^3y^2}</math>
 |}
@@ Line 33: / Line 33: @@
 !Final Answer: &nbsp;
 |-
-|<math>\frac{dy}dx} = \frac{3x^2y^3 - 1}{1 - 3x^3y^2}</math>
+|<math>\frac{dy}{dx} = \frac{3x^2y^3 - 1}{1 - 3x^3y^2}</math>
 |}
 [[022_Exam_1_Sample_A|'''<u>Return to Sample Exam</u>''']]

← Older revision		Revision as of 20:06, 30 May 2015
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−	~~<span style="font-size:135%"><font face=Times Roman>~~<span class="exam"> Use implicit differentiation to find <math>\frac{dy}{dx}: \qquad x+y = x^3y^3</math>	+	<span class="exam"> Use implicit differentiation to find <math>\frac{dy}{dx}: \qquad x+y = x^3y^3</math>

	{\| class="mw-collapsible mw-collapsed" style = "text-align:left;"		{\| class="mw-collapsible mw-collapsed" style = "text-align:left;"

← Older revision		Revision as of 20:05, 30 May 2015
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−	<span style="font-size:135%"><font face=Times Roman>4. <span class="exam"> Use implicit differentiation to find <math>\frac{dy}{dx}: \qquad x+y = x^3y^3</math>	+	<span style="font-size:135%"><font face=Times Roman><span class="exam"> Use implicit differentiation to find <math>\frac{dy}{dx}: \qquad x+y = x^3y^3</math>

	{\| class="mw-collapsible mw-collapsed" style = "text-align:left;"		{\| class="mw-collapsible mw-collapsed" style = "text-align:left;"

@@ Line 8: / Line 8: @@
 |When we use implicit differentiation, we combine the chain rule with the fact that <math style="vertical-align: -18%">y</math> is a function of <math style="vertical-align: 0%">x</math>, and could really be written as <math style="vertical-align: -21%">y(x).</math> Because of this, the derivative of <math style="vertical-align:-17%">y^3</math> with respect to <math style="vertical-align: 0%">x</math> requires the chain rule, so
 |-
-|&nbsp;&nbsp;&nbsp;&nbsp; <math>\frac{d}{dx}\left(y^{3}\right)=3y^{2}\cdot\frac{dy}{dt}.</math>
+|&nbsp;&nbsp;&nbsp;&nbsp; <math>\frac{d}{dx}\left(y^{3}\right)=3y^{2}\cdot\frac{dy}{dx}.</math>
 |}
-&nbsp;'''Solution:'''
+'''Solution:'''
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 !Step 1: &nbsp;
 |-
-|First, we differentiate each term separately with respect to <math style="vertical-align: 0%">x</math> to find that&thinsp; <math style="vertical-align: -18%">x^{3}-y^{3}-y=x</math> &thinsp;differentiates implicitly to
+|First, we differentiate each term separately with respect to <math style="vertical-align: 0%">x</math> and apply the product rule on the right hand side to find that&thinsp; <math style="vertical-align: -18%">x + y = x^3y^3</math> &thinsp;differentiates implicitly to
 |-
-|&nbsp;&nbsp;&nbsp;&nbsp; <math>3x^{2}-3y^{2}\cdot\frac{dy}{dx}-\frac{dy}{dx}=1</math>.
 |}
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 !Step 2: &nbsp;
 |-
-|Since they don't ask for a general expression of <math style="vertical-align:-24%">dy/dx</math>, but rather a particular value at a particular point, we can plug in the values <math style="vertical-align:-5%">x=1</math> and <math style="vertical-align:-18%">y=0</math> &thinsp;to find
+|Now we need to solve for <math>\frac{dy}{dx}</math> and doing so we find that <math>\frac{dy}dx} = \frac{3x^2y^3 - 1}{1 - 3x^3y^2}</math>
 |}
@@ Line 33: / Line 33: @@
 !Final Answer: &nbsp;
 |-
-|&nbsp;&nbsp;&nbsp;&nbsp; <math style="vertical-align:-24%">dy/dx=2.</math>
+|<math>\frac{dy}dx} = \frac{3x^2y^3 - 1}{1 - 3x^3y^2}</math>
 |}
 [[022_Exam_1_Sample_A|'''<u>Return to Sample Exam</u>''']]

022 Sample Final A, Problem 4 - Revision history

Matt Lee at 20:16, 30 May 2015

Matt Lee at 20:16, 30 May 2015

Matt Lee at 20:06, 30 May 2015

Matt Lee at 20:05, 30 May 2015

Matt Lee: Created page with " 4. Use implicit differentiation to find \frac{dy}{dx}: \qquad x+y = x^3y^3 {| class="mw..."

Matt Lee: Created page with " 4. Use implicit differentiation to find $\frac{dy}{dx}: \qquad x+y = x^3y^3$ {| class="mw..."