022 Sample Final A, Problem 4 - Revision history

MathAdmin at 01:33, 7 June 2015

2015-06-07T01:33:26Z

MathAdmin at 23:15, 4 June 2015

2015-06-04T23:15:31Z

MathAdmin at 23:13, 4 June 2015

2015-06-04T23:13:46Z

MathAdmin: Created page with " Use implicit differentiation to find $\frac{dy}{dx}: \qquad x+y = x^3y^3$ {| class="mw-collapsible mw-collapsed" style = "text-align:left;" !..."

2015-06-04T23:13:03Z

Created page with " <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;" !..."

New page

<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;"
!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}{dx}.</math>
|-
|For this problem we also need to use the product rule.
|}

'''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> and apply the product rule on the right hand side to find that  <math style="vertical-align: -18%">x + y = x^3y^3</math>  differentiates implicitly to
|-
|
::<math>1 + \frac{dy}{dx} = 3x^2y^3 + 3x^3y^2 \cdot \frac{dy}{dx}</math>.
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|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>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
|-
|<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 23:15, 4 June 2015
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−	<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;"

@@ 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}{dx}.</math>
 |-
-|For this problem we also need to use the product rule.
+|For this problem, we also need to use the product rule.
 |}
@@ Line 18: / Line 19: @@
 !Step 1: &nbsp;
 |-
-|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
+|First, we differentiate each term separately with respect to <math style="vertical-align: 0px">x</math> and apply the product rule on the right hand side to find that&thinsp; <math style="vertical-align: -4px">x + y = x^3y^3</math> &thinsp;differentiates implicitly to
 |-
+|
-::<math>1 + \frac{dy}{dx} = 3x^2y^3 + 3x^3y^2 \cdot \frac{dy}{dx}</math>.
+::<math>1 + \frac{dy}{dx} \,=\, 3x^2y^3 + 3x^3y^2 \cdot \frac{dy}{dx}</math>.
 |}
@@ Line 27: / Line 28: @@
 !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 &thinsp;<math style="vertical-align: -14px">\frac{dy}{dx}</math>&thinsp;, and doing so we find that &nbsp;<math style="vertical-align: -18px">\frac{dy}{dx} \,=\, \frac{3x^2y^3 - 1}{1 - 3x^3y^2}</math>.
 |}
@@ Line 33: / Line 34: @@
 !Final Answer: &nbsp;
 |-
 |<math>\frac{dy}{dx} = \frac{3x^2y^3 - 1}{1 - 3x^3y^2}</math>
 |}
 [[022_Sample Final A|'''<u>Return to Sample Exam</u>''']]

022 Sample Final A, Problem 4 - Revision history

MathAdmin at 01:33, 7 June 2015

MathAdmin at 23:15, 4 June 2015

MathAdmin at 23:13, 4 June 2015

MathAdmin: Created page with " Use implicit differentiation to find \frac{dy}{dx}: \qquad x+y = x^3y^3 {| class="mw-collapsible mw-collapsed" style = "text-align:left;" !..."

MathAdmin: Created page with " Use implicit differentiation to find $\frac{dy}{dx}: \qquad x+y = x^3y^3$ {| class="mw-collapsible mw-collapsed" style = "text-align:left;" !..."