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

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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
 
|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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::<math style="vertical-align: 0px">\frac{\partial}{\partial x} xy \,=\, y\frac{\partial}{\partial x} x \,=\, y.</math>
 
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!Step 2: &nbsp;
 
!Step 2: &nbsp;
 
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|-
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|Now we have to find the 4 second derivatives:
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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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<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{\partial}{\partial x} \frac{\partial f(x, y)}{\partial x}} & = & \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{0 - 2(x - y)(-2y^2)}{(x - y)^4}}\\
 
&&\\
 
&&\\
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& = & \displaystyle{\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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|-
−
|
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|Also,
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<math>\begin{array}{rcl}
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<br>
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\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{\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)}\\
 
&&\\
 
&&\\
 
& = & \displaystyle{\frac{-4y(x - y)^2 -4y^2(x - y)}{(x - y)^4}}\\
 
& = & \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{-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>
 
\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{\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)}\\
 
&&\\
 
&&\\
 
& = & \displaystyle{\frac{4x(x - y)^2 -2(x - y)2x^2}{(x - y)^4}}\\
 
& = & \displaystyle{\frac{4x(x - y)^2 -2(x - y)2x^2}{(x - y)^4}}\\
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& = & \displaystyle{\frac{4x(x^2 - 2xy + y^2) -4x^3+ 4x^2y}{(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}}
+
& = & \displaystyle{\frac{4xy^2 -4x^2y}{(x - y)^4}.}
 
\end{array}</math>
 
\end{array}</math>
 
|-
 
|-
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|
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|Finally,
−
 
+
<br>
−
<math>\begin{array}{rcl}
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::<math>\begin{array}{rcl}
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\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{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>
 
\end{array}</math>
 
|}
 
|}
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!Final Answer: &nbsp;  
 
!Final Answer: &nbsp;  
 
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|-
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|<math>\displaystyle{\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:
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<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
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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>
−
\frac{\partial}{\partial x}\frac{\partial f(x, y)}{\partial y} = \frac{4xy^2 -4x^2y}{(x - y)^4} \qquad
 
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\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>''']]
 
[[022_Sample_Final_A|'''<u>Return to Sample Exam</u>''']]

Latest revision as of 08:20, 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 Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f(x, y) = \frac{2xy}{x-y}.}


Foundations:  
1) Which derivative rules do you have to use for this problem?
2) What is the partial derivative of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle xy} , with respect to Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x} ?
Answers:
1) You have to use the quotient rule and product rule. The quotient rule says that
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{\partial}{\partial x} \left(\frac{f(x)}{g(x)}\right) = \frac{f'(x)g(x) - g'(x)f(x)}{g(x)^2},}

so

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{\partial}{\partial x} \left(\frac{x^2}{x + 1}\right) = \frac{2x(x + 1) - x^2}{(x + 1)^2}.}

The product rule says

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{\partial}{\partial x} f(x)g(x) = f'(x)g(x) + g'(x)f(x). }

This means

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{\partial}{\partial x} [x(x + 1)] = (x + 1) + x. }
2) The partial derivative is Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle y} , since we treat anything not involving Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x} as a constant and take the derivative with respect to Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x} . In more detail, we have
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{\partial}{\partial x} xy \,=\, y\frac{\partial}{\partial x} x \,=\, y.}

Solution:

Step 1:  
First, we start by finding the first partial derivatives. So we have to take the partial derivative of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f(x, y)} with respect to Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x} , and the partial derivative of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f(x, y)} with respect to Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle y} . This gives us the following:
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{array}{rcl} \displaystyle{\frac{\partial}{\partial x} f(x, y)} & = & \displaystyle{\frac{\partial}{\partial x} \left( \frac{2xy}{x - y}\right)}\\ &&\\ & = & \displaystyle{\frac{2y(x - y) -2xy}{(x - y)^2}}\\ &&\\ & = & \displaystyle{\frac{-2y^2}{(x - y)^2}.} \end{array}}
This gives us the derivative with respect to Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x} . To find the derivative with respect to Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle y} , we do the following:
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{array}{rcl} \displaystyle{\frac{\partial}{\partial y}f(x, y)} & = & \displaystyle{\frac{\partial}{\partial y}\left(\frac{2xy}{x - y}\right)}\\ &&\\ & = & \displaystyle{\frac{2x(x - y) +2xy}{(x - y)^2}}\\ &&\\ & = & \displaystyle{\frac{2x^2}{(x - y)^2}.} \end{array}}
Step 2:  
Now we have to find the 4 second derivatives, We have


Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \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}.} \end{array}}
Also,


Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{array}{rcl} \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)}\\ &&\\ & = & \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}}
Showing the equality of mixed partial derivatives,


Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{array}{rcl} \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)}\\ &&\\ & = & \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}}
Finally,


Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{array}{rcl} \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}.} \end{array}}
Final Answer:  
The first partial derivatives are:
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \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}.}}
The second partial derivatives are:
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{\partial^{2}f(x,y)}{\partial x^{2}}\,=\,\frac{4xy^{2}-4y^{3}}{(x-y)^{4}}, \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}}.}

Return to Sample Exam

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