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, ${\begin{array}{rcl}\displaystyle {{\frac {\partial ^{2}f(x,y)}{\partial y\partial x}}\,=\,{\frac {\partial }{\partial y}}\left({\frac {\partial f(x,y)}{\partial x}}\right)}&=&\displaystyle {{\frac {\partial }{\partial y}}\left({\frac {-2y^{2}}{(x-y)^{2}}}\right)}\\&&\\&=&\displaystyle {\frac {-4y(x-y)^{2}-4y^{2}(x-y)}{(x-y)^{4}}}\\&&\\&=&\displaystyle {\frac {-4y(x^{2}-2xy+y^{2})-4xy^{2}+4y^{3}}{(x-y)^{4}}}\\&=&\displaystyle {{\frac {4xy^{2}-4x^{2}y}{(x-y)^{4}}}.}\end{array}}$
Showing the equality of mixed partial derivatives, ${\begin{array}{rcl}\displaystyle {{\frac {\partial ^{2}f(x,y)}{\partial x\partial y}}\,=\,{\frac {\partial }{\partial x}}\left({\frac {\partial f(x,y)}{\partial y}}\right)}&=&\displaystyle {{\frac {\partial }{\partial x}}\left({\frac {2x^{2}}{(x-y)^{2}}}\right)}\\&&\\&=&\displaystyle {\frac {4x(x-y)^{2}-2(x-y)2x^{2}}{(x-y)^{4}}}\\&&\\&=&\displaystyle {\frac {4x(x^{2}-2xy+y^{2})-4x^{3}+4x^{2}y}{(x-y)^{4}}}\\&&\\&=&\displaystyle {{\frac {4xy^{2}-4x^{2}y}{(x-y)^{4}}}.}\end{array}}$
Finally, ${\begin{array}{rcl}\displaystyle {{\frac {\partial ^{2}f(x,y)}{\partial y^{2}}}\,=\,{\frac {\partial }{\partial y}}\left({\frac {\partial f(x,y)}{\partial y}}\right)}&=&\displaystyle {{\frac {\partial }{\partial y}}\left({\frac {2x^{2}}{(x-y)^{2}}}\right)}\\&&\\&=&\displaystyle {\frac {0+2(x-y)(2x^{2})}{(x-y)^{4}}}\\&&\\&=&\displaystyle {{\frac {4x^{3}-4x^{2}y}{(x-y)^{4}}}.}\end{array}}$

Final Answer:
The first partial derivatives are: $\displaystyle {{\frac {\partial }{\partial x}}f(x,y)={\frac {-2y^{2}}{(x-y)^{2}}},\qquad {\frac {\partial }{\partial y}}f(x,y)={\frac {2x^{2}}{(x-y)^{2}}}.}$
The second partial derivatives are: ${\frac {\partial ^{2}f(x,y)}{\partial x^{2}}}\,=\,{\frac {4xy^{2}-4y^{3}}{(x-y)^{4}}},\qquad {\frac {\partial ^{2}f(x,y)}{\partial x\partial y}}\,=\,{\frac {\partial ^{2}f(x,y)}{\partial y\partial x}}\,=\,{\frac {4xy^{2}-4x^{2}y}{(x-y)^{4}}},\qquad {\frac {\partial ^{2}f(x,y)}{\partial y^{2}}}\,=\,{\frac {4x^{3}-4x^{2y}}{(x-y)^{4}}}.$

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