Clairaut's Theorem

When mixed partials are continuous, the order of differentiation does not matter.

The idea

Theorem (Clairaut's theorem).

Let $f$ be defined on an open set containing the point $p$, suppose both mixed second partials $\partial_x\partial_y f$ and $\partial_y\partial_x f$ exist on that set, and suppose both are continuous at $p$. Then $\partial_x\partial_y f(p) = \partial_y\partial_x f(p).$

The theorem is also called Schwarz's theorem. Behind the equality is one picture: over a small rectangle with sides $h$ and $k$, both mixed partials measure the same signed combination of $f s values at the four corners — differentiating in $x$ then $y$, or in $y$ then $x$, is walking the two routes around that rectangle.

Under these hypotheses the order of differentiation makes no difference, so a mixed second partial may be computed in whichever order is easier.

The hypothesis concerns the second partials, not the first: continuity of $f$, or the existence of $f_x$ and $f_y$, is not enough. Nor can the continuity requirement be dropped. There are functions whose mixed partials both exist at a point and take different values there; in every such example the mixed partials fail to be continuous at that point, which is precisely the case the theorem excludes.

For a polynomial the hypothesis always holds: its partial derivatives of every order are again polynomials, hence continuous everywhere, so the two orders agree with nothing to check.

Ways to work on it

• Walkthrough. The statement of Clairaut's theorem, checked on a polynomial.
• Practice. Compute a mixed partial that the theorem lets you take in either order.
• Hardest. The counterexample where continuity fails and the orders disagree.

Not sure where to start? Take the ten-question placement test.