022 Exam 1 Sample A, Problem 2 - Revision history

MathAdmin at 21:48, 17 April 2017

2017-04-17T21:48:14Z

2015-04-02T18:37:53Z

2015-04-02T04:44:30Z

2015-04-01T05:50:52Z

2015-04-01T05:50:27Z

2015-04-01T05:49:58Z

2015-04-01T05:49:23Z

2015-04-01T05:46:51Z

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 |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>
 |}

@@ Line 29: / Line 29: @@
 |}
-'''Correct Answer:''' <math style="vertical-align:-24%">dy/dx=2.</math>
+{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
+!Final Answer: &nbsp;
 [[022_Exam_1_Sample_A|'''<u>Return to Sample Exam</u>''']]

@@ Line 28: / Line 28: @@
 |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>
 |}
 [[022_Exam_1_Sample_A|'''<u>Return to Sample Exam</u>''']]

@@ Line 5: / Line 5: @@
 !Foundations: &nbsp;
 |-
-|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: -23%">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
+|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>

@@ Line 15: / Line 15: @@
 !Step 1: &nbsp;
 |-
-|First, we differentiate each term separately with respect to x 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> to find that&thinsp; <math style="vertical-align: -18%">x^{3}-y^{3}-y=x</math> &thinsp;differentiates implicitly to
 |-
 |&nbsp;&nbsp;&nbsp;&nbsp; <math>3x^{2}-3y^{2}\cdot\frac{dy}{dx}-\frac{dy}{dx}=1</math>.