# Failure of Clairaut's theorem where both mixed partials are defined but not equal

From Calculus

## Contents

## 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 both the second-order mixed partial derivatives of exist at , i.e., both the numbers and exist, but they are not equal.

### For a function of two variables overall

It is possible to have a function of two variables such that both the second-order mixed partial derivatives and exist everywhere on but they are not equal as functions, i.e., there exists a point where the values of the second-order mixed partial derivatives are not equal.

## Related facts

- Failure of Clairaut's theorem where only one of the mixed partials is defined
- Separately continuous not implies continuous
- Continuous in every linear direction not implies continuous
- Existence of partial derivatives not implies differentiable
- Existence of directional derivatives in every direction not implies differentiable

## Proof

### Example

Consider the function:

We do some computations:

Item | Value | Explanation |
---|---|---|

for | use the quotient rule for differentiation and simplify | |

for | use the quotient rule for differentiation and simplify | |

for | plug in in the general expression for and simplify. | |

for | plug in in the general expression for and simplify. | |

0 | start with and simplify, noting that the numerator is identically zero | |

0 | start with and simplify, noting that the numerator is identically zero | |

-1 | start with and simplify using the expressions obtained above for and | |

1 | start with and simplify using the expressions obtained above for and |