I have come across two theorems, which I think are essentially trying to say the same thing:

1) The mixed derivative theorem:

If $f(x, y)$ and its partial derivatives $f_x$, $f_y$, $f_{xy}$ and $f_{yx}$ are defined in a neighborhood of $(x_0, y_0)$ and all are continuous at $(x_0, y_0)$, then :

$$f_{xy}(x_0, y_0) = f_{yx}(x_0, y_0).$$

2) Clairaut's theorem:

Suppose that $f$ is defined on a disk $D$ that contains the point $(a,b)$. If the functions $f_{xy}$ and $f_{yx}$ are continuous on this disk then:

$$f_{xy}(a,b) = f_{yx}(a ,b).$$

Why are there two separate theorems for conveying the same thing? Is it that the second one is an improved version of the first one? It seems that the mixed derivative theorem lists some redundant conditions, because existence of the second order partial derivatives would imply the continuity of $f_x$ and $f_y$. Is it wrong to conclude this?

  • 1
    $\begingroup$ It happens quite often in mathematics that (1) there are multiple names for essentially the same thing; and (2) the same theorem is presented in a slightly different (but essentially equivalent) way by different authors. $\endgroup$
    – user7530
    Commented Aug 31, 2021 at 8:12

2 Answers 2


Yes, it's the same theorem. It's also known as Schwarz's theorem or Symmetry of second derivatives. This kind of situation happens a lot. And you are right when you claim that there is no need to assume that $\frac{\partial f}{\partial x}$ and $\frac{\partial f}{\partial y}$ are continuous.


It is actually incorrect to conclude that continuity and equality of the mixed partial derivatives implies continuity of the first-order partial derivatives. For example, let $g(x)=\begin{cases} x^2\sin(1/x), & x\ne 0 \\ 0, & x=0 \end{cases}$, as usual. Take $f(x,y)=g(x)+g(y)$. Then $f_{xy}=0=f_{yx}$ everywhere, and yet $f_x(x,y)=g'(x)$ and $f_y(x,y)=g'(y)$, so $f_x$ and $f_y$ are not continuous at the origin.


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