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

Revision as of 18:59, 6 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 $f(x,y)={\frac {2xy}{x-y}}.$

Foundations:
1) Which derivative rules do you have to use for this problem?
2) What is the partial derivative of $xy$ , with respect to $x$ ?
Answers:
1) You have to use the quotient rule and product rule. The quotient rule says that ${\frac {\partial }{\partial x}}\left({\frac {f(x)}{g(x)}}\right)={\frac {f'(x)g(x)-g'(x)f(x)}{g(x)^{2}}},$ so ${\frac {\partial }{\partial x}}\left({\frac {x^{2}}{x+1}}\right)={\frac {2x(x+1)-x^{2}}{(x+1)^{2}}}.$ The product rule says ${\frac {\partial }{\partial x}}f(x)g(x)=f'(x)g(x)+g'(x)f(x).$ This means ${\frac {\partial }{\partial x}}[x(x+1)]=(x+1)+x.$
2) The partial derivative is $y$ , since we treat anything not involving $x$ as a constant and take the derivative with respect to $x$ . In more detail, we have ${\frac {\partial }{\partial x}}xy=y{\frac {\partial }{\partial x}}x=y.$

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

@@ 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{\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{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{\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}}
+& = & \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{\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}}\\
@@ 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
+\frac{\partial}{\partial y} f(x, y) = \frac{2x^2}{(x - y)^2}.}</math>
 |-
-|
+|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>
-\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>''']]

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

Revision as of 18:59, 6 June 2015

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