# Difference between revisions of "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 xy, with respect to 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 y, since we treat anything not involving x as a constant and take the derivative with respect to x. So ${\displaystyle {\frac {\partial }{\partial y}}xy=x{\frac {\partial }{\partial y}}y=x.}$

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

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

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

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

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

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