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 |
---|---|---|
![]() ![]() |
![]() |
use the quotient rule for differentiation and simplify |
![]() ![]() |
![]() |
use the quotient rule for differentiation and simplify |
![]() ![]() |
![]() |
plug in ![]() ![]() |
![]() ![]() |
![]() |
plug in ![]() ![]() |
![]() |
0 | start with ![]() |
![]() |
0 | start with ![]() |
![]() |
-1 | start with ![]() ![]() ![]() |
![]() |
1 | start with ![]() ![]() ![]() |