This was a shower thought I had today. Consider, for example, the function of two variables f(x, y) = x^2 + 3y. Then compute

    \[ \frac{\partial f}{\partial x}(3, 5). \]

Now, you might not know how to do partial differentiation, but my point is that the expression above is not even well-defined.

The reason for this is that even though we defined f as f(x, y), the x and y are just placeholders, so we might as well write f(3, 5) (when evaluating at specific numbers), f(a, b) if putting in other variables, or even f(y, x). In the latter case, f(y, x) = y^2 + 3x. So, how would you compute \dfrac{\partial f}{\partial x}(y, x)?

Strictly speaking, it is not a property of the function what we call the arguments in its definition. It is exactly the same thing to define f(x, y) = x^2 + 3y and f(a, b) = a^2 + 3b. Thus, when writing \dfrac{\partial f}{\partial x}, we can’t really know which argument is meant to differentiate by.

But obviously we mean the first argument if f was defined using the symbols f(x, y) (and the second if it was f(t, x) or something). Mathematics is not programming where you need to communicate to an entirely stringent computer that cannot fill in gaps by its own. We don’t build everything from the axioms all the time. We use shortcuts, known results, and let the reader do some work, and ambiguities are nothing new. Here, the reader can reasonably assume that it’s the first variable if it says \dfrac{\partial f}{\partial x}.

Anyway, that was my shower thought.

</rant>

2 thoughts on “Why the Partial Derivative Notation is not Well-Defined (but everyone knows what you mean)”
    1. Well, it seems hard to get around the fact that we may insert concrete values such as f(3, 5) without redefining what a function even is. But it would be interesting to investigate further. So far, we will just have to live with the convention, as far as I am aware.

Leave a Reply

Your email address will not be published. Required fields are marked *