Clairaut's theorem on equality of mixed partials

Statement

Statement for second-order mixed partial of function of two variables

Suppose $f$ is a real-valued function of two variables $x,y$ and $f(x,y)$ is defined on an open subset $U$ of $\mathbb {R} ^{2}$. Suppose further that both the second-order mixed partial derivatives $f_{xy}(x,y)$ and $f_{yx}(x,y)$ exist and are continuous on $U$. Then, we have:

$\!f_{xy}=f_{yx}$

on all of $U$.

General statement

The statement can be generalized in two ways:

• We can generalize it to higher-order partial derivatives.
• We can generalize it to functions of more than two variables.

The general version states the following. Suppose $f$ is a function of $n$ variables defined on an open subset $U$ of $\mathbb {R} ^{n}$. Suppose all mixed partials with a certain number of differentiations in each input variable exist and are continuous on $U$. Then, all the mixed partials are continuous.

Some examples are given below:

• Suppose $f$ is a function of two variables $x$ and $y$, and the three mixed partials $f_{xxy},f_{xyx},f_{yxx}$ exist and are continuous on an open subset $U$ of $\mathbb {R} ^{3}$. Then, all three of them are equal on $U$. (Note that these mixed partials all involve differentiating twice with respect to $x$ and once with respect to $y$).
• Suppose $f$ is a function of three variables $x,y,z$, and the six mixed partials $f_{xyz},f_{xzy},f_{yxz},f_{yzx},f_{zxy},f_{zyx}$ exist and are continuous on an open subset $U$ of $\mathbb {R} ^{3}$. Then, all six of them are equal on $U$.

Particular cases

Clairaut's theorem can be verified in a number of special cases through direct computations. Some of these are illustrated below.

Additively separable functions

For further information, refer: Additively separable function

Suppose $F(x,y)$ is an additively separable function of two variables, i.e., we can write:

$\!F(x,y):=f(x)+g(y)$

where $f,g$ are both functions of one variable. Assume that both $f$ and $g$ are both differentiable. Then, we have:

$\!F_{x}(x,y)=f'(x),F_{y}(x,y)=g'(y)$

Differentiating each of these a second time with respect to the other variable, we obtain that:

$\!F_{xy}(x,y)=0,F_{yx}(x,y)=0$

Thus, we see that both second-order mixed partial derivatives are the zero function. In particular, they are equal to each other.

Multiplicatively separable functions

For further information, refer: Multiplicatively separable function

Suppose $G(x,y)$ is a multiplicatively separable function of two variables, i.e., we can write:

$\!G(x,y):=f(x)g(y)$

where $f,g$ are both functions of one variable. Assume that both $f$ and $g$ are both differentiable. Then, we have:

$\!G_{x}(x,y)=f'(x)g(y),G_{y}(x,y)=f(x)g'(y)$

Differentiating each of these a second time with respect to the other variable, we obtain that:

$\!G_{xy}(x,y)=f'(x)g'(y),G_{yx}(x,y)=f'(x)g'(y)$

We thus see that both the second-order mixed partial derivatives are equal.

Note that the proof for multiplicatively separable functions can be easily generalized to functions that are expressible as sums of multiplicatively separable functions. This, therefore, includes all polynomial functions.

Related facts

• Failure of Clairaut's theorem where only one of the mixed partials is defined
• Failure of Clairaut's theorem where both mixed partials are defined but not equal (this happens because one or both of them is not continuous)