Suppose \(f\) is defined on a disk \(D\) that contains the point \((a,b)\), and suppose the functions \(f_{xy}\) and \(f_{yx}\) are both continuous on \(D\). The parial derivatives are defiend as:
$$\begin{align} f_x(x,y) &= \frac{\partial}{\partial x}f(x,y) = \lim_{h \to 0} \frac{f(x+h,y) - f(x,y)}{h} \\ f_y(x,y) &= \frac{\partial}{\partial y}f(x,y) = \lim_{h \to 0} \frac{f(x,y+h) - f(x,y)}{h} \end{align}$$
The second partial derivative is defined as:
$$\begin{align} f_{xx} &= \frac{\partial}{\partial x} \left( \frac{\partial f}{\partial x} \right) \qquad f_{xy} = \frac{\partial}{\partial y} \left( \frac{\partial f}{\partial x} \right) \\ f_{yx} &= \frac{\partial}{\partial x} \left( \frac{\partial f}{\partial y} \right) \qquad f_{yy} = \frac{\partial}{\partial y} \left( \frac{\partial f}{\partial y} \right) \end{align}$$
For small positive values of \(h\), consider the difference:
$$\Delta h = [f(a+h,b+h) - f(a+h,b)] - [f(a,b+h)- f(a,b)]$$
Let \(g(x) = f(x, b+h) - f(x, b)\):
$$\Delta h = g(a+h) - g(a)$$
According to the Mean Value Theorem, there is a number \(c\) between \(a\) and \(a+h\) such that:
$$ g(a+h) - g(a) = g'(c)h $$
Since \(g(x) = f(x, b+h) - f(x, b)\):
$$h \left[ \left. \frac{d}{dx} \right|_{x=c}g(x) \right] = h[f_x(c,b+h) - f_x(c,b)] $$
By the Mean Value Theorem to \(f_x\), we get a number \(d\) between \(b\) and \(b+h\) such that:
$$[f_x(c,b+h) - f_x(c,b)] = f_{xy}(c,d)h$$
Combining these equations:
$$ \Delta h = h^2 f_{xy}(c,d)$$
If \(h \to 0\), then \((c,d) \to (a,b)\):
$$ \lim_{h \to 0} \frac{\Delta h}{h^2} = \lim_{(c,d) \to (a,b)} f_{xy}(c,d) = f_{xy}(a,b)$$
Similarly, by writing:
$$\Delta h = [f(a+h,b+h) - f(a+h,b)] - [f(a,b+h)- f(a,b)]$$
By using the Mean Value Theorem twice and the continuity of \(f_{yx}\) at \((a,b)\), we obtain:
$$ \lim_{h \to 0} \frac{\Delta h}{h^2} = f_{yx}(a,b)$$
It follows that \(f_{yx}(a,b)=f_{xy}(a,b)\).
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