# Failure of Clairaut's theorem where only one of the mixed partials is defined

From Calculus

## Statement

### For a function of two variables at a point

It is possible to have a function of two variables and a point in the domain of such that the second-order mixed partial derivative exists but the second-order mixed partial derivative does not exist.

## Proof

### Example

Consider a function defined on all of as:

We note that:

- Since depends only on , is identically the zero function.
- Thus, the second-order mixed partial derivative is identically the zero function.
- On the other hand, for and is undefined for .
- In particular, this means that is not defined on the line (i.e., the -axis).