Difference between revisions of "022 Sample Final A, Problem 1"

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<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>
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<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>
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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!Foundations: &nbsp;
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|-
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|1) Which derivative rules do you have to use for this problem?
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|-
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|2) What is the partial derivative of <math style="vertical-align: -4px">xy</math>, with respect to <math style="vertical-align: 0px">x</math>?
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|-
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|-
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|Answers:
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|-
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|1) You have to use the quotient rule and product rule. The quotient rule says that
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::<math style="vertical-align: 0px">\frac{\partial}{\partial x} \left(\frac{f(x)}{g(x)}\right) = \frac{f'(x)g(x) - g'(x)f(x)}{g(x)^2},</math>
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so
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::<math style="vertical-align: 0px">\frac{\partial}{\partial x} \left(\frac{x^2}{x + 1}\right) = \frac{2x(x + 1) - x^2}{(x + 1)^2}.</math>
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The product rule says
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::<math style="vertical-align: 0px">\frac{\partial}{\partial x} f(x)g(x) = f'(x)g(x) + g'(x)f(x). </math>
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This means
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::<math style="vertical-align: 0px">\frac{\partial}{\partial x} [x(x + 1)] = (x + 1) + x. </math>
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|-
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|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
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::<math style="vertical-align: 0px">\frac{\partial}{\partial x} xy \,=\, y\frac{\partial}{\partial x} x \,=\, y.</math>
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|}
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'''Solution:'''
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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!Step 1: &nbsp;
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|-
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|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:
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|-
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|
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::<math>\begin{array}{rcl}
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\displaystyle{\frac{\partial}{\partial x} f(x, y)} & = & \displaystyle{\frac{\partial}{\partial x} \left( \frac{2xy}{x - y}\right)}\\
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&&\\
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& = & \displaystyle{\frac{2y(x - y) -2xy}{(x - y)^2}}\\
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&&\\
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& = & \displaystyle{\frac{-2y^2}{(x - y)^2}.}
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\end{array}</math>
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|-
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|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:
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|-
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|
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::<math>\begin{array}{rcl}
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\displaystyle{\frac{\partial}{\partial y}f(x, y)} & = & \displaystyle{\frac{\partial}{\partial y}\left(\frac{2xy}{x - y}\right)}\\
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&&\\
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& = & \displaystyle{\frac{2x(x - y) +2xy}{(x - y)^2}}\\
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&&\\
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& = & \displaystyle{\frac{2x^2}{(x - y)^2}.}
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\end{array}</math>
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|}
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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!Step 2: &nbsp;
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|-
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|Now we have to find the 4 second derivatives,  We have
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<br>
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::<math>\begin{array}{rcl}
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\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)}\\
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&&\\
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& = & \displaystyle{\frac{0 - 2(x - y)(-2y^2)}{(x - y)^4}}\\
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&&\\
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& = & \displaystyle{\frac{4xy^2 - 4y^3}{(x - y)^4}.}
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\end{array}</math>
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|-
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|Also,
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<br>
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::<math>\begin{array}{rcl}
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\displaystyle{\frac{\partial^2f(x,y)}{\partial y\partial x}\,=\,\frac{\partial}{\partial y} \left(\frac{\partial f(x, y)}{\partial x}\right)} & = & \displaystyle{\frac{\partial}{\partial y}\left(\frac{-2y^2}{(x - y)^2}\right)}\\
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&&\\
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& = & \displaystyle{\frac{-4y(x - y)^2 -4y^2(x - y)}{(x - y)^4}}\\
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&&\\
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& = & \displaystyle{\frac{-4y(x^2 - 2xy + y^2) - 4xy^2 + 4y^3}{(x - y)^4}}\\
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& = & \displaystyle{\frac{4xy^2 - 4x^2y}{(x - y)^4}.}
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\end{array}</math>
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|-
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|Showing the equality of mixed partial derivatives,
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<br>
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::<math>\begin{array}{rcl}
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\displaystyle{\frac{\partial^2f(x,y)}{\partial x\partial y}\,=\,\frac{\partial}{\partial x} \left(\frac{\partial f(x, y)}{\partial y}\right)} & = & \displaystyle{\frac{\partial}{\partial x}\left(\frac{2x^2}{(x - y)^2}\right)}\\
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&&\\
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& = & \displaystyle{\frac{4x(x - y)^2 -2(x - y)2x^2}{(x - y)^4}}\\
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&&\\
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& = & \displaystyle{\frac{4x(x^2 - 2xy + y^2) -4x^3+ 4x^2y}{(x - y)^4}}\\
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&&\\
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& = & \displaystyle{\frac{4xy^2 -4x^2y}{(x - y)^4}.}
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\end{array}</math>
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|-
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|Finally,
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<br>
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::<math>\begin{array}{rcl}
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\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)}\\
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&&\\
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& = & \displaystyle{\frac{0 + 2(x - y)(2x^2)}{(x - y)^4}}\\
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&&\\
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& = & \displaystyle{\frac{4x^3 - 4x^2y}{(x - y)^4}.}
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\end{array}</math>
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|}
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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!Final Answer: &nbsp;
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|-
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|The first partial derivatives are:
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::<math>\displaystyle{\frac{\partial}{\partial x} f(x, y) = \frac{-2y^2}{(x - y)^2}, \qquad
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\frac{\partial}{\partial y} f(x, y) = \frac{2x^2}{(x - y)^2}.}</math>
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|-
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|The second partial derivatives are:
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::<math>\frac{\partial^{2}f(x,y)}{\partial x^{2}}\,=\,\frac{4xy^{2}-4y^{3}}{(x-y)^{4}},
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\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>
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|}
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[[022_Sample_Final_A|'''<u>Return to Sample Exam</u>''']]

Latest revision as of 09:20, 7 June 2015

Find all first and second partial derivatives of the following function, and demostrate that the mixed second partials are equal for the function


Foundations:  
1) Which derivative rules do you have to use for this problem?
2) What is the partial derivative of , with respect to ?
Answers:
1) You have to use the quotient rule and product rule. The quotient rule says that

so

The product rule says

This means

2) The partial derivative is , since we treat anything not involving as a constant and take the derivative with respect to . In more detail, we have

Solution:

Step 1:  
First, we start by finding the first partial derivatives. So we have to take the partial derivative of with respect to , and the partial derivative of with respect to . This gives us the following:
This gives us the derivative with respect to . To find the derivative with respect to , we do the following:
Step 2:  
Now we have to find the 4 second derivatives, We have


Also,


Showing the equality of mixed partial derivatives,


Finally,


Final Answer:  
The first partial derivatives are:
The second partial derivatives are:

Return to Sample Exam