022 Sample Final A, Problem 1

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 $f(x,y)$ with respect to $x$, and the partial derivative of $f(x,y)$ with respect to $y$. This gives us the following:
${\begin{array}{rcl}\displaystyle {{\frac {\partial }{\partial x}}f(x,y)}&=&\displaystyle {{\frac {\partial }{\partial x}}\left({\frac {2xy}{x-y}}\right)}\\&&\\&=&\displaystyle {\frac {2y(x-y)-2xy}{(x-y)^{2}}}\\&&\\&=&\displaystyle {{\frac {-2y^{2}}{(x-y)^{2}}}.}\end{array}}$
This gives us the derivative with respect to $x$. To find the derivative with respect to $y$, we do the following:
${\begin{array}{rcl}\displaystyle {{\frac {\partial }{\partial y}}f(x,y)}&=&\displaystyle {{\frac {\partial }{\partial y}}\left({\frac {2xy}{x-y}}\right)}\\&&\\&=&\displaystyle {\frac {2x(x-y)+2xy}{(x-y)^{2}}}\\&&\\&=&\displaystyle {{\frac {2x^{2}}{(x-y)^{2}}}.}\end{array}}$

Step 2:
Now we have to find the 4 second derivatives, We have ${\begin{array}{rcl}\displaystyle {{\frac {\partial ^{2}f(x,y)}{\partial x^{2}}}\,=\,{\frac {\partial }{\partial x}}\left({\frac {\partial f(x,y)}{\partial x}}\right)}&=&\displaystyle {{\frac {\partial }{\partial x}}\left({\frac {-2y^{2}}{(x-y)^{2}}}\right)}\\&&\\&=&\displaystyle {\frac {0-2(x-y)(-2y^{2})}{(x-y)^{4}}}\\&&\\&=&\displaystyle {{\frac {4xy^{2}-4y^{3}}{(x-y)^{4}}}.}\end{array}}$
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}}.}

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