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