Some Pysics Insights

Partials Commute

Partial derivatives commute.  This is a basic fact about functions of several variables, and anyone who's studied them knows it and has seen it proved using a bit of algebra and some limits.  But why is it true?

Let z = f(x,y) be a function of two variables.  In the diagram, a small piece of the x,y plane is shown.  I've assumed f(0,0) is zero to keep the algebra simple.

Partials Commute -- Diagram 1

The proof is contained in the diagram, and can be understood just by looking at it.  The following discussion is intended to help with understanding the details of the diagram, but doesn't really add anything.

As we move to the right by δx, the value of f, shown on the red line, goes up by .  If we then move to the rear by δy, the value of f increases by approximately , and we arrive at the blue dot, labeled.

The difference between the blue dot and f(δx, δy) in that case is due to the difference between the actual partial of f with respect to y at (δx,0) and the value we used, which was the partial of f with respect to y at (0,0).  This difference is

On the other hand, if we go around the box the other way, then the function first increases by as we move to the rear, and then by approximately as we move to the right.  Again, we arrive at the blue dot:  .  And again, the actual value of f(δx,δy) will be different from the blue dot -- but this time the difference will be due to the change in the partial of f with respect to x rather than y.  The partial of f with respect to x at (0,δy) will be different from its value at (0,0), and the difference in the value of f will be .

But the value of f(δx,δy) must be the same no matter how we get there, so the second partials must be the same, no matter which order we evaluate them in.

As with most "proof-by-picture" demonstrations, the point of this is not rigor, but rather to show intuitively what's going on.  We've ignored a number of details here, such as the difference between the second derivatives at the rear corner and the front corner.  To produce a rigorous proof of this requires explicitly evaluating the limits used to define the derivatives to be sure that nothing we ignored made a difference, and the algebra gets a bit messy.  But any standard calculus text should include a full proof of this.

Back to Proof-by-Picture