User:Egm6321.f12.team5.hou

Given:
Theorem: If f(x,y) is twice continuously differentiable, then the mixed partial derivatives are equal; that is $$\begin{align} \frac{\partial f^2(x,y)}{\partial x\partial y}=\frac{\partial f^2(x,y)}{\partial y\partial x} \\\end{align}$$

The mean value theorem: given an arc between two endpoints, there is at least one point at which the tangent to the arc is parallel to the secant through its endpoints. If a function f(x) is continuous on the closed interval [a, b], where a < b, and differentiable on the open interval (a, b), then there exists a point c in (a, b) such that $$\begin{align} f'(c)=\frac{f(b)-f(a)}{b-a} \\\end{align}$$

http://en.wikipedia.org/wiki/Mean_value_theorem Image there are four infinite closed points in the x-y coordinate system, as shown above. The distance between 2 point shown as m and n are infinite small. $$\begin{align} F(m,n):=\frac{1}{mn}[f(x+m,y+n)-f(x+m,y)-f(x,y+n)+f(x,y)] \\\end{align}$$

Set F (m,n) is a smooth function, which means F(m,n) can be infinitely differentiated. $$\begin{align} F(m,n)=\frac{1}{mn}[f(x+m,y+n)-f(x+m,y)]-\frac{1}{mn}[f(x,y+n)-f(x,y)] \\\end{align}$$

$$\begin{align} F(m,n)=\frac{1}{m}[\frac{f(x+m,y+n)-f(x+m,y)}{n}]-\frac{1}{n}[\frac{f(x,y+n)-f(x,y)}{m}] \\\end{align}$$

$$\begin{align}F(m,n)=\frac{1}{m}[\frac{\partial f(x+m,y)}{\partial y}]-\frac{1}{m}[\frac{\partial f(x,y)}{\partial y}] \\\end{align}$$

$$\begin{align} F(m,n)=\frac{\frac{\partial f(x+m,y)}{\partial y}-\frac{\partial f(x,y)}{\partial y}}{m} \\\end{align}$$

$$\begin{align} F(m,n)=\frac{\partial f^2(x,y)}{\partial y\partial x} \\\end{align}$$

On the other hand $$\begin{align}F(m,n) =\frac{1}{mn}[f(x+m,y+n)-f(x+m,y)-f(x,y+n)+f(x,y)] \\\end{align}$$

$$\begin{align} F(m,n)=\frac{1}{mn}[f(x+m,y+n)-f(x,y+n)]-\frac{1}{mn}[f(x+m,y)-f(x,y)] \\\end{align}$$

$$\begin{align} F(m,n)=\frac{1}{n}[\frac{f(x+m,y+n)-f(x,y+n)}{m}]-\frac{1}{n}[\frac{f(x+m,y)-f(x,y)}{m}]\\\end{align}$$

$$\begin{align} F(m,n)=\frac{1}{n}[\frac{\partial f(x,y+n)}{\partial x}]-\frac{1}{n}[\frac{\partial f(x,y)}{\partial x}] \\\end{align}$$

$$\begin{align} F(m,n)=\frac{[\frac{\partial f(x,y+n)}{\partial x}-\frac{\partial f(x,y)}{\partial x}]}{n} \\\end{align}$$

$$\begin{align} F(m,n)=\frac{\partial f^2(x,y)}{\partial x\partial y} \\\end{align}$$

As $$\begin{align} F(m,n)=F(m,n) \\\end{align}$$ , $$\begin{align} \frac{\partial f^2(x,y)}{\partial x\partial y}=\frac{\partial f^2(x,y)}{\partial y\partial x} \\\end{align}$$