__Unequal Mixed Partials:__

__By Joe Schinco__

**According to the text, in order for mixed partial derivatives to be equal, f _{xy} = f_{yx} the mixed partials must be continuos. Are there functions where the mixed partials are not equal?**

__Partial Derivative:__

In countless areas of mathematics, physics, and chemistry there are equations that contain many variables. When taking the derivative of these equations it is often necessary to keep all variables constant, except for the one being studied. When every variable in a given equation are held constant during differentiation, other than the variable of specific interest, the process is called a partial derivative.

An example of an equation containing multiple variables is that of the volume of a cone:

**V= (r ^ 2 * h * Pi) / 3**

An example of a partial derivative is taking the derivative of V with respect to r, the radius:

In order to go forward in the explanation of unequal mixed partials it is necessary to note the standard definition of a partial derivative, which can be written as follows:

__Equal Mixed Partials:__

In order for the mixed partial derivatives of a certain function to be equal Fxy needs to be continuous throughout a given limit. This will result in Fyx existing being equal to Fxy in that given limit.

__Unequal Mixed Partials:__

There

**is**a standard example that proves that there is a specific case where the mixed partials are not equal. “Functions whose second partials are discontinuous need not have their mixed partials equal.” http://www.math.tamu.edu/~tom.vogel/gallery/node18.html

An example of a function that satisfies the unequal mixed partials idea is

**f (x,y) = (xy^3 – x^3y) / (x^2 + y^2) , (x,y) (0,0)**

, (x,y) = (0,0)

**f x (x,y) = (3x^2 – y^3)/(x^2 + y^2) – (2x(x^3y – xy^3)) / (x^2 + y^2)^2**

, (x,y)

, (x,y) = (0,0)

The value of f x (0,0) is found through the use of a limit:

**fx (0,0) = lim f(h,0) – f(0,0) / h**

h—>0

Just as f x (x,y) was found so is f y (x,y) found in a very similar manner:

**f y (0,0) = (x^3 – 3xy^2) / (x^2 + y^2) – (2y(x^3y – xy^3)) / (x^2 + y^2)^2**

, (x,y) (0,0)

**, (x,y) = (0,0)**

This finally results in the following for fxy (0,0):

**fxy (0,0) = lim fx(o,h) – fx (0,0) / h lim -h – 0 / h = -1**

h—>0 = **h—>0**

The same idea is applied to find the answer to f yx (0,0):

**f yx (0,0) = lim fy(h,0) – fy (0,0) / h lim h – 0 / h = 1**

h—>0 = **h—>0**

Because of the fact that -1 is clearly not equal to 1 it can be stated that

**f xy (0,0) f yx (0,0)**

The above mentioned function is clear proof that there is a function whose mixed partials are not equal to each other.

__References:__

1. Vogel, Tom Dr. “A Function Whose Mixed Partials are Unequal,” 5 May, 1997.

<http://www.math.tamu.edu/~tom.vogel/gallery/node18.html>

2. Wagon, S. New York: W. H. Freeman, pp. 83-85, 1991.

3. Weisstein, Eric W. “Partial Derivative.” From Mathworld–A Wolfram Web Resource.

<http://mathworld.wolfram.com/partialderivative.html>

4. “Partial Derivative.” Wikipedia, The Free Encyclopedia. 11 September 2006. Wikipedia Foundation, Inc.

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