# 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 ${\displaystyle f(x,y)={\frac {2xy}{x-y}}.}$

Foundations:
1) Which derivative rules do you have to use for this problem?
2) What is the partial derivative of ${\displaystyle xy}$, with respect to ${\displaystyle x}$?
1) You have to use the quotient rule and product rule. The quotient rule says that
${\displaystyle {\frac {\partial }{\partial x}}\left({\frac {f(x)}{g(x)}}\right)={\frac {f'(x)g(x)-g'(x)f(x)}{g(x)^{2}}},}$

so

${\displaystyle {\frac {\partial }{\partial x}}\left({\frac {x^{2}}{x+1}}\right)={\frac {2x(x+1)-x^{2}}{(x+1)^{2}}}.}$

The product rule says

${\displaystyle {\frac {\partial }{\partial x}}f(x)g(x)=f'(x)g(x)+g'(x)f(x).}$

This means

${\displaystyle {\frac {\partial }{\partial x}}[x(x+1)]=(x+1)+x.}$
2) The partial derivative is ${\displaystyle y}$, since we treat anything not involving ${\displaystyle x}$ as a constant and take the derivative with respect to ${\displaystyle x}$. In more detail, we have
${\displaystyle {\frac {\partial }{\partial x}}xy=y{\frac {\partial }{\partial x}}x=y.}$

Solution:

Step 1:
First, we start by finding the first partial derivatives. So we have to take the partial derivative of ${\displaystyle f(x,y)}$ with respect to ${\displaystyle x}$, and the partial derivative of ${\displaystyle f(x,y)}$ with respect to ${\displaystyle y}$. This gives us the following:
${\displaystyle {\begin{array}{rcl}\displaystyle {{\frac {\partial }{\partial x}}f(x,y)}&=&\displaystyle {{\frac {\partial }{\partial x}}\left({\frac {2xy}{x-y}}\right)}\\&&\\&=&\displaystyle {\frac {2y(x-y)-2xy}{(x-y)^{2}}}\\&&\\&=&\displaystyle {{\frac {-2y^{2}}{(x-y)^{2}}}.}\end{array}}}$
This gives us the derivative with respect to ${\displaystyle x}$. To find the derivative with respect to ${\displaystyle y}$, we do the following:
${\displaystyle {\begin{array}{rcl}\displaystyle {{\frac {\partial }{\partial y}}f(x,y)}&=&\displaystyle {{\frac {\partial }{\partial y}}\left({\frac {2xy}{x-y}}\right)}\\&&\\&=&\displaystyle {\frac {2x(x-y)+2xy}{(x-y)^{2}}}\\&&\\&=&\displaystyle {{\frac {2x^{2}}{(x-y)^{2}}}.}\end{array}}}$
Step 2:
Now we have to find the 4 second derivatives:

${\displaystyle {\begin{array}{rcl}\displaystyle {{\frac {\partial }{\partial x}}{\frac {\partial f(x,y)}{\partial x}}}&=&\displaystyle {{\frac {\partial }{\partial x}}\left({\frac {-2y^{2}}{(x-y)^{2}}}\right)}\\&&\\&=&\displaystyle {\frac {0-2(x-y)(-2y^{2})}{(x-y)^{4}}}\\&&\\&=&\displaystyle {\frac {4xy^{2}-4y^{3}}{(x-y)^{4}}}\end{array}}}$

${\displaystyle {\begin{array}{rcl}\displaystyle {{\frac {\partial }{\partial y}}{\frac {\partial f(x,y)}{\partial x}}}&=&\displaystyle {{\frac {\partial }{\partial y}}\left({\frac {-2y^{2}}{(x-y)^{2}}}\right)}\\&&\\&=&\displaystyle {\frac {-4y(x-y)^{2}-4y^{2}(x-y)}{(x-y)^{4}}}\\&&\\&=&\displaystyle {\frac {-4y(x^{2}-2xy+y^{2})-4xy^{2}+4y^{3}}{(x-y)^{4}}}\\&=&\displaystyle {\frac {4xy^{2}-4x^{2}y}{(x-y)^{4}}}\end{array}}}$

${\displaystyle {\begin{array}{rcl}\displaystyle {{\frac {\partial }{\partial x}}{\frac {\partial f(x,y)}{\partial y}}}&=&\displaystyle {{\frac {\partial }{\partial x}}\left({\frac {2x^{2}}{(x-y)^{2}}}\right)}\\&&\\&=&\displaystyle {\frac {4x(x-y)^{2}-2(x-y)2x^{2}}{(x-y)^{4}}}\\&&\\&=&\displaystyle {\frac {4x(x^{2}-2xy+y^{2})-4x^{3}+4x^{2}y}{(x-y)^{4}}}\\&&\\&=&\displaystyle {\frac {4xy^{2}-4x^{2}y}{(x-y)^{4}}}\end{array}}}$

${\displaystyle {\begin{array}{rcl}\displaystyle {{\frac {\partial }{\partial y}}{\frac {\partial f(x,y)}{\partial y}}}&=&\displaystyle {{\frac {\partial }{\partial y}}\left({\frac {2x^{2}}{(x-y)^{2}}}\right)}\\&&\\&=&\displaystyle {\frac {0+2(x-y)(2x^{2})}{(x-y)^{4}}}\\&&\\&=&\displaystyle {\frac {4x^{3}-4x^{2}y}{(x-y)^{4}}}\end{array}}}$

${\displaystyle \displaystyle {{\frac {\partial }{\partial x}}f(x,y)={\frac {-2y^{2}}{(x-y)^{2}}}\qquad {\frac {\partial }{\partial y}}f(x,y)={\frac {2x^{2}}{(x-y)^{2}}}\qquad }}$
${\displaystyle \displaystyle {{\frac {\partial }{\partial x}}{\frac {\partial f(x,y)}{\partial x}}={\frac {4xy^{2}-4y^{3}}{(x-y)^{4}}}\qquad {\frac {\partial }{\partial y}}{\frac {\partial f(x,y)}{\partial x}}={\frac {4xy^{2}-4x^{2}y}{(x-y)^{4}}}\qquad {\frac {\partial }{\partial x}}{\frac {\partial f(x,y)}{\partial y}}={\frac {4xy^{2}-4x^{2}y}{(x-y)^{4}}}\qquad {\frac {\partial }{\partial y}}{\frac {\partial f(x,y)}{\partial y}}={\frac {4x^{3}-4x^{2}y}{(x-y)^{4}}}}}$