022 Sample Final A, Problem 1 - Revision history

MathAdmin at 16:20, 7 June 2015

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MathAdmin at 02:59, 7 June 2015

2015-06-07T02:59:11Z

MathAdmin at 02:38, 7 June 2015

2015-06-07T02:38:18Z

MathAdmin at 21:11, 6 June 2015

2015-06-06T21:11:38Z

MathAdmin at 19:27, 6 June 2015

2015-06-06T19:27:55Z

MathAdmin at 17:57, 5 June 2015

2015-06-05T17:57:39Z

MathAdmin at 17:48, 5 June 2015

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MathAdmin at 17:29, 5 June 2015

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MathAdmin at 16:45, 5 June 2015

2015-06-05T16:45:13Z

MathAdmin at 16:30, 5 June 2015

2015-06-05T16:30:58Z

@@ Line 22: / Line 22: @@
 |-
 |2) The partial derivative is <math style="vertical-align: -4px">y</math>, since we treat anything not involving <math style="vertical-align: 0px">x</math> as a constant and take the derivative with respect to <math style="vertical-align: 0px">x</math>. In more detail, we have
-::<math style="vertical-align: 0px">\frac{\partial}{\partial x} xy = y\frac{\partial}{\partial x} x = y.</math>
+::<math style="vertical-align: 0px">\frac{\partial}{\partial x} xy \,=\, y\frac{\partial}{\partial x} x \,=\, y.</math>
 |}

@@ Line 55: / Line 55: @@
 !Step 2: &nbsp;
 |-
-|Now we have to find the 4 second derivatives:
+|Now we have to find the 4 second derivatives,  We have
-|-
+<br>
+::<math>\begin{array}{rcl}
-<math>\begin{array}{rcl}
+\displaystyle{\frac{\partial^2f(x,y)}{\partial x^2}\,=\,\frac{\partial}{\partial x} \left(\frac{\partial f(x, y)}{\partial x}\right)} & = & \displaystyle{\frac{\partial}{\partial x}\left(\frac{-2y^2}{(x - y)^2}\right)}\\
 &&\\
 & = & \displaystyle{\frac{0 - 2(x - y)(-2y^2)}{(x - y)^4}}\\
 &&\\
-& = & \displaystyle{\frac{4xy^2 - 4y^3}{(x - y)^4}}
+& = & \displaystyle{\frac{4xy^2 - 4y^3}{(x - y)^4}.}
 \end{array}</math>
 |-
+|Also,
-<math>\begin{array}{rcl}
+<br>
-\displaystyle{\frac{\partial}{\partial y} \frac{\partial f(x, y)}{\partial x}} & = & \displaystyle{\frac{\partial}{\partial y}\left(\frac{-2y^2}{(x - y)^2}\right)}\\
+::<math>\begin{array}{rcl}
 &&\\
 & = & \displaystyle{\frac{-4y(x - y)^2 -4y^2(x - y)}{(x - y)^4}}\\
 &&\\
 & = & \displaystyle{\frac{-4y(x^2 - 2xy + y^2) - 4xy^2 + 4y^3}{(x - y)^4}}\\
-& = & \displaystyle{\frac{4xy^2 - 4x^2y}{(x - y)^4}}
+& = & \displaystyle{\frac{4xy^2 - 4x^2y}{(x - y)^4}.}
 \end{array}</math>
 |-
+|Showing the equality of mixed partial derivatives,
-<math>\begin{array}{rcl}
+<br>
-\displaystyle{\frac{\partial}{\partial x} \frac{\partial f(x, y)}{\partial y}} & = & \displaystyle{\frac{\partial}{\partial x}\left(\frac{2x^2}{(x - y)^2}\right)}\\
+::<math>\begin{array}{rcl}
 &&\\
 & = & \displaystyle{\frac{4x(x - y)^2 -2(x - y)2x^2}{(x - y)^4}}\\
@@ Line 84: / Line 85: @@
 & = & \displaystyle{\frac{4x(x^2 - 2xy + y^2) -4x^3+ 4x^2y}{(x - y)^4}}\\
 &&\\
-& = & \displaystyle{\frac{4xy^2 -4x^2y}{(x - y)^4}}
+& = & \displaystyle{\frac{4xy^2 -4x^2y}{(x - y)^4}.}
 \end{array}</math>
 |-
+|Finally,
+<br>
-<math>\begin{array}{rcl}
+::<math>\begin{array}{rcl}
-\displaystyle{\frac{\partial}{\partial y} \frac{\partial f(x, y)}{\partial y}} & = & \displaystyle{\frac{\partial}{\partial y}\left(\frac{2x^2}{(x - y)^2}\right)}\\
+\displaystyle{\frac{\partial^2f(x,y)}{\partial y^2}\,=\,\frac{\partial}{\partial y} \left(\frac{\partial f(x, y)}{\partial y}\right)}  & = & \displaystyle{\frac{\partial}{\partial y}\left(\frac{2x^2}{(x - y)^2}\right)}\\
 &&\\
 & = & \displaystyle{\frac{0 + 2(x - y)(2x^2)}{(x - y)^4}}\\
 &&\\
-& = & \displaystyle{\frac{4x^3 - 4x^2y}{(x - y)^4}}
+& = & \displaystyle{\frac{4x^3 - 4x^2y}{(x - y)^4}.}
 \end{array}</math>
 |}
@@ Line 101: / Line 102: @@
 !Final Answer: &nbsp;
 |-
-|<math>\displaystyle{\frac{\partial}{\partial x} f(x, y) = \frac{-2y^2}{(x - y)^2} \qquad
+|The first partial derivatives are:
-\frac{\partial}{\partial y} f(x, y) = \frac{2x^2}{(x - y)^2} \qquad}</math>
+::<math>\displaystyle{\frac{\partial}{\partial x} f(x, y) = \frac{-2y^2}{(x - y)^2}, \qquad
 |-
+|The second partial derivatives are:
-<math>\displaystyle{\frac{\partial}{\partial x} \frac{\partial f(x, y)}{\partial x} = \frac{4xy^2 - 4y^3}{(x - y)^4} \qquad
+::<math>\frac{\partial^{2}f(x,y)}{\partial x^{2}}\,=\,\frac{4xy^{2}-4y^{3}}{(x-y)^{4}},
-\frac{\partial}{\partial y}\frac{\partial f(x, y)}{\partial x} = \frac{4xy^2 - 4x^2y}{(x - y)^4} \qquad
+\qquad\frac{\partial^{2}f(x,y)}{\partial x\partial y}\,=\,\frac{\partial^{2}f(x,y)}{\partial y\partial x}\,=\,\frac{4xy^{2}-4x^{2}y}{(x-y)^{4}},\qquad\frac{\partial^{2}f(x,y)}{\partial y^{2}}\,=\,\frac{4x^{3}-4x^{2y}}{(x-y)^{4}}.</math>
 |}
 [[022_Sample_Final_A|'''<u>Return to Sample Exam</u>''']]

@@ Line 1: / Line 1: @@
-<span class="exam">Find all first and second partial derivatives of the following function, and demostrate that the mixed second partials are equal for the function <math> f(x, y) = \frac{2xy}{x-y}.</math>
+<span class="exam">Find all first and second partial derivatives of the following function, and demostrate that the mixed second partials are equal for the function <math style="vertical-align: -17px"> f(x, y) = \frac{2xy}{x-y}.</math>
@@ Line 5: / Line 5: @@
 !Foundations: &nbsp;
 |-
-|1)Which derivative rules do you have to use for this problem?
+|1) Which derivative rules do you have to use for this problem?
 |-
-|2)What is the partial derivative of xy, with respect to x?
+|2) What is the partial derivative of <math style="vertical-align: -4px">xy</math>, with respect to <math style="vertical-align: 0px">x</math>?
 |-
 |-
-|2) The partial derivative is y, since we treat anything not involving x as a constant and take the derivative with respect to x. So <math>\frac{\partial}{\partial y} xy = x\frac{\partial}{\partial y} y = x.</math>
+|Answers:
 |}
@@ Line 18: / Line 29: @@
 !Step 1: &nbsp;
 |-
-|First, we start by finding the first partial derivatives. So we have to take the partial derivative of f(x, y)with respect to x, and the partial derivative of f(x, y)with respect to y. This gives us the following:
+|First, we start by finding the first partial derivatives. So we have to take the partial derivative of <math style="vertical-align: -4px">f(x, y)</math> with respect to <math style="vertical-align: 0px">x</math>, and the partial derivative of <math style="vertical-align: -4px">f(x, y)</math> with respect to <math style="vertical-align: -4px">y</math>. This gives us the following:
 |-
+|
@@ Line 26: / Line 37: @@
 & = & \displaystyle{\frac{2y(x - y) -2xy}{(x - y)^2}}\\
 &&\\
-& = & \displaystyle{\frac{-2y^2}{(x - y)^2}}
+& = & \displaystyle{\frac{-2y^2}{(x - y)^2}.}
 \end{array}</math>
 |-
-|This gives us the derivative with respect to x. To find the derivative with respect to y, we do the following:
+|This gives us the derivative with respect to <math style="vertical-align: 0px">x</math>. To find the derivative with respect to <math style="vertical-align: -4px">y</math>, we do the following:
 |-
+|
@@ Line 37: / Line 48: @@
 & = & \displaystyle{\frac{2x(x - y) +2xy}{(x - y)^2}}\\
 &&\\
-& = & \displaystyle{\frac{2x^2}{(x - y)^2}}
+& = & \displaystyle{\frac{2x^2}{(x - y)^2}.}
 \end{array}</math>
 |}

@@ Line 22: / Line 22: @@
+|
 ::<math>\begin{array}{rcl}
-\frac{\partial}{\partial x} f(x, y) & = & \frac{\partial}{\partial x} \left( \frac{2xy}{x - y}\right)\\
+\displaystyle{\frac{\partial}{\partial x} f(x, y)} & = & \displaystyle{\frac{\partial}{\partial x} \left( \frac{2xy}{x - y}\right)}\\
-& = & \frac{2y(x - y) -2xy}{(x - y)^2}\\
+&&\\
-& = & \frac{-2y^2}{(x - y)^2}
+& = & \displaystyle{\frac{2y(x - y) -2xy}{(x - y)^2}}\\
 \end{array}</math>
 |-
@@ Line 31: / Line 33: @@
+|
 ::<math>\begin{array}{rcl}
-\frac{\partial}{\partial y}f(x, y) & = & \frac{\partial}{\partial y}\left(\frac{2xy}{x - y}\right)\\
+\displaystyle{\frac{\partial}{\partial y}f(x, y)} & = & \displaystyle{\frac{\partial}{\partial y}\left(\frac{2xy}{x - y}\right)}\\
-& = & \frac{2x(x - y) +2xy}{(x - y)^2}\\
+&&\\
-& = & \frac{2x^2}{(x - y)^2}
+& = & \displaystyle{\frac{2x(x - y) +2xy}{(x - y)^2}}\\
 \end{array}</math>
 |}
@@ Line 44: / Line 48: @@
+|
 <math>\begin{array}{rcl}
-\frac{\partial}{\partial x} \frac{\partial f(x, y)}{\partial x} & = & \frac{\partial}{\partial x}\left(\frac{-2y^2}{(x - y)^2}\right)\\
+\displaystyle{\frac{\partial}{\partial x} \frac{\partial f(x, y)}{\partial x}} & = & \displaystyle{\frac{\partial}{\partial x}\left(\frac{-2y^2}{(x - y)^2}\right)}\\
-& = & \frac{0 - 2(x - y)(-2y^2)}{(x - y)^4}\\
+&&\\
-& = & \frac{4xy^2 - 4y^3}{(x - y)^4}
+& = & \displaystyle{\frac{0 - 2(x - y)(-2y^2)}{(x - y)^4}}\\
 \end{array}</math>
 |-
+|
 <math>\begin{array}{rcl}
-\frac{\partial}{\partial y} \frac{\partial f(x, y)}{\partial x} & = & \frac{\partial}{\partial y}\left(\frac{-2y^2}{(x - y)^2}\right)\\
+\displaystyle{\frac{\partial}{\partial y} \frac{\partial f(x, y)}{\partial x}} & = & \displaystyle{\frac{\partial}{\partial y}\left(\frac{-2y^2}{(x - y)^2}\right)}\\
-& = & \frac{-4y(x - y)^2 -4y^2(x - y)}{(x - y)^4}\\
+&&\\
-& = & \frac{-4y(x^2 - 2xy + y^2) - 4xy^2 + 4y^3}{(x - y)^4}\\
+& = & \displaystyle{\frac{-4y(x - y)^2 -4y^2(x - y)}{(x - y)^4}}\\
-& = & \frac{4xy^2 - 4x^2y}{(x - y)^4}
+&&\\
 \end{array}</math>
 |-
+|
 <math>\begin{array}{rcl}
-\frac{\partial}{\partial x} \frac{\partial f(x, y)}{\partial y} & = & \frac{\partial}{\partial x}\left(\frac{2x^2}{(x - y)^2}\right)\\
+\displaystyle{\frac{\partial}{\partial x} \frac{\partial f(x, y)}{\partial y}} & = & \displaystyle{\frac{\partial}{\partial x}\left(\frac{2x^2}{(x - y)^2}\right)}\\
-& = & \frac{4x(x - y)^2 -2(x - y)2x^2}{(x - y)^4}\\
+&&\\
-& = & \frac{4x(x^2 - 2xy + y^2) -4x^3+ 4x^2y}{(x - y)^4}\\
+& = & \displaystyle{\frac{4x(x - y)^2 -2(x - y)2x^2}{(x - y)^4}}\\
-& = & \frac{4xy^2 -4x^2y}{(x - y)^4}
+&&\\
 \end{array}</math>
 |-
@@ Line 68: / Line 79: @@
 <math>\begin{array}{rcl}
-\frac{\partial}{\partial y} \frac{\partial f(x, y)}{\partial y} & = & \frac{\partial}{\partial y}\left(\frac{2x^2}{(x - y)^2}\right)\\
+\displaystyle{\frac{\partial}{\partial y} \frac{\partial f(x, y)}{\partial y}} & = & \displaystyle{\frac{\partial}{\partial y}\left(\frac{2x^2}{(x - y)^2}\right)}\\
-& = & \frac{0 + 2(x - y)(2x^2)}{(x - y)^4}\\
+&&\\
-& = & \frac{4x^3 - 4x^2y}{(x - y)^4}
+& = & \displaystyle{\frac{0 + 2(x - y)(2x^2)}{(x - y)^4}}\\
 \end{array}</math>
 |}
@@ Line 77: / Line 90: @@
 !Final Answer: &nbsp;
 |-
-|<math>\frac{\partial}{\partial x} f(x, y) = \frac{-2y^2}{(x - y)^2} \qquad
+|<math>\displaystyle{\frac{\partial}{\partial x} f(x, y) = \frac{-2y^2}{(x - y)^2} \qquad
-\frac{\partial}{\partial y} f(x, y) = \frac{2x^2}{(x - y)^2} \qquad</math>
+\frac{\partial}{\partial y} f(x, y) = \frac{2x^2}{(x - y)^2} \qquad}</math>
 |-
+|
-<math>\frac{\partial}{\partial x} \frac{\partial f(x, y)}{\partial x} = \frac{4xy^2 - 4y^3}{(x - y)^4} \qquad
+<math>\displaystyle{\frac{\partial}{\partial x} \frac{\partial f(x, y)}{\partial x} = \frac{4xy^2 - 4y^3}{(x - y)^4} \qquad
 \frac{\partial}{\partial y}\frac{\partial f(x, y)}{\partial x} = \frac{4xy^2 - 4x^2y}{(x - y)^4} \qquad
 \frac{\partial}{\partial x}\frac{\partial f(x, y)}{\partial y} = \frac{4xy^2 -4x^2y}{(x - y)^4} \qquad
-\frac{\partial}{\partial y}\frac{\partial f(x, y)}{\partial y} = \frac{4x^3 - 4x^2y}{(x - y)^4}
+\frac{\partial}{\partial y}\frac{\partial f(x, y)}{\partial y} = \frac{4x^3 - 4x^2y}{(x - y)^4}}
 </math>
 |}
 [[022_Sample_Final_A|'''<u>Return to Sample Exam</u>''']]

← Older revision		Revision as of 19:27, 6 June 2015
Line 87:		Line 87:
	</math>		</math>
	\|}		\|}
		+	[[022_Sample_Final_A\|'''<u>Return to Sample Exam</u>''']]

@@ Line 16: / Line 16: @@
 '''Solution:'''
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Foundations: &nbsp;
+!Step 1: &nbsp;
 |-
-|The word 'marginal' should make you immediately think of a derivative.  In this case, the marginal is just the partial derivative with respect to a particular variable.
+|First, we start by finding the first partial derivatives. So we have to take the partial derivative of f(x, y)with respect to x, and the partial derivative of f(x, y)with respect to y. This gives us the following:
 |-
-|The teacher has also added the additional restriction that you should not leave your answer with negative exponents.
 |}

← Older revision		Revision as of 16:45, 5 June 2015
Line 1:		Line 1:
	<span class="exam">Find all first and second partial derivatives of the following function, and demostrate that the mixed second partials are equal for the function <math> f(x, y) = \frac{2xy}{x-y}.</math>		<span class="exam">Find all first and second partial derivatives of the following function, and demostrate that the mixed second partials are equal for the function <math> f(x, y) = \frac{2xy}{x-y}.</math>
		+
		+
		+	{\| class="mw-collapsible mw-collapsed" style = "text-align:left;"
		+	!Foundations:
		+	\|-
		+	\|1)Which derivative rules do you have to use for this problem?
		+	\|-
		+	\|2)What is the partial derivative of xy, with respect to x?
		+	\|-
		+	\|1)You have to use the quotient rule, and product rule. The quotient rule says that <math>\frac{\partial}{\partial x} \left(\frac{f(x)}{g(x)}\right) = \frac{f'(x)g(x) - g'(x)f(x)}{g(x)^2}</math>,   so <math>\frac{\partial}{\partial x} \left(\frac{x^2}{x + 1}\right) = \frac{2x(x + 1) - x^2}{(x + 1)^2}</math>. The product rule says <math>\frac{\partial}{\partial x} f(x)g(x) = f'(x)g(x) + g'(x)f(x) </math>.   This means <math>\frac{\partial}{\partial x} x(x + 1) = (x + 1) + x </math>
		+	\|-
		+	\|2) The partial derivative is y, since we treat anything not involving x as a constant and take the derivative with respect to x. So <math>\frac{\partial}{\partial y} xy = x\frac{\partial}{\partial y} y = x.</math>
		+	\|}
		+
		+	'''Solution:'''
		+	{\| class="mw-collapsible mw-collapsed" style = "text-align:left;"
		+	!Foundations:
		+	\|-
		+	\|The word 'marginal' should make you immediately think of a derivative. In this case, the marginal is just the partial derivative with respect to a particular variable.
		+	\|-
		+	\|The teacher has also added the additional restriction that you should not leave your answer with negative exponents.
		+	\|}

← Older revision		Revision as of 16:30, 5 June 2015
Line 1:		Line 1:
−	Find all first and second partial derivatives of the following function, and demostrate that the mixed second partials are equal for the function <math> f(x, y) = \frac{2xy}{x-y}.</math>	+	<span class="exam">Find all first and second partial derivatives of the following function, and demostrate that the mixed second partials are equal for the function <math> f(x, y) = \frac{2xy}{x-y}.</math>