Is this acceptable notation? (Partial Derivatives)

Question

I have a question about the notation which I have decided to use. I'm writing lecture notes for my own Calculus course and I've introduced notation such as $[R_x-Q_y]_x$ and so on. It's supposed to mean the following

Find the first partial derivative of the argument in the brackets w.r.t the variable that appears after the brackets

Here is the text that I've written (below)

As always, let us define $\vec{F}(x,y,z)$ as $$\vec{F}(x,y,z)=P(x,y,z)\, \vec{i}+Q(x,y,z)\, \vec{j}+R(x,y,z)\, \vec{k}$$ Computing the curl, we get \begin{align*} \nabla \times \vec{F}&=\begin{vmatrix} \vec{i}& \vec{j}& \vec{k} \\ \partial_x & \partial_y & \partial_z \\ P&Q&R \end{vmatrix}\\ &=(R_y-Q_z)\vec{i}-(R_x-P_z)\vec{j}+(Q_x-P_y)\vec{k}\end{align*} Computing the divergence, \begin{align*} \nabla\cdot(\nabla\times\vec{F})&=[R_y-Q_z]_x- [R_x-P_z]_y+[Q_x-P_y]_z\\ &=R_{yx}-Q_{zx}-R_{xy}+P_{zy}+Q_{xz}-P_{yz} \\ &=(R_{yz}-R_{yx})+(Q_{xz}-Q_{xz})+(P_{zy}-P_{zy}) \qquad \text{(by Clairaut's theorem)} \\ &=0 \end{align*} (QED)

Is this a good notation to use? Thanks

Answer 1

$$(P_x-Q_x)_y\equiv\dfrac{\partial}{\partial y}(P_x-Q_x)$$ seems a natural convention.

Answer 2

In general, the use of subscripts to denote derivatives is pretty common, so you're breaking no new ground here. I see two possible problems.

1. It might be tempting (to students!) to keep the subscripts in alphabetical order for "neatness", i.e., write $R_{xy}$ instead of $R_{yx}$. If you start out by saying "we're going to assume a level of continuity that lets us know that mixed partials are all equal," then this is OK ... indeed, it's probably good! But if you're not doing that, then you should probably make a note that order matters, and that $[R_x]_y$ becomes $R_{xy}$, i.e., that the order in which partials are computed is the left-to-right-order within the subscript.

2. I personally both use and hate notations that depend on names. Suppose that we have two functions $(x, y) \mapsto f(x, y)$ and $(x, y) \mapsto g(x, y)$. Further suppose that $f(x, y) = g(y, x)$, and that both are differentiable. How is the derivative of $f$ with respect to its first argument related to the derivatives of $g$? Well, it looks as if $$\frac{\partial f}{\partial x} = \frac{\partial g}{\partial y}, $$ ...sort of. Perhaps if we add arguments, it'll be clearer: $$\frac{\partial f}{\partial x}(x, y) = \frac{\partial g}{\partial y}(y, x). $$ So ... when we compute that second thing, since $y$ is the first argument, did we differentiate with respect to the first argument or the second? You and I both know what's meant here (probably), but a beginning student is left just baffled. Far better, perhaps, is to say something like $$ (\partial_1 f)(x, y) = (\partial_2 g)(x, y) $$ where the "1" and "2" name the positions of the arguments with respect to which we are differentiating. Of course, this loses us lots of the mnemonic beauty of the chain rule, but it provides a gain in clarity as a trade-off.

Given problem 2, I'd advise using this notation for the course of this one proof, and perhaps a couple of other div/grad/etc proofs, and then abandoning it.