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I'm constantly seeing questions here where people are confused about the notation $\frac {\partial f}{\partial x}$ or $\frac {\partial f}{\partial x} (x,y)$ or $\frac {\partial f(x,y)}{\partial x}$. Is there some better notation which exists for partial derivatives? If not, can anyone suggest one?

Problems with this notation:

  • Is the derivative evaluated at the point $(x,y)$ or is that part of the definition of the function? If it's evaluated at the point, then should we really use $x$ as the name of the first parameter of the function? If it's just a part of the definition of the function, how do we specify where the function is evaluated? And if we leave off the $(x,y)$, how do we even know that $x$ is supposed to be the first parameter as opposed to anything else?
  • The same problem that $\frac {df}{dx}$ shares, it looks like division. But of course $\frac {\partial f}{\partial x}$ is not as simple as division: it's a shorthand for $\lim_{h\to 0} \frac {f(x+h,y)-f(x,y)}{h}$. This leads students to think that $\frac {\partial f}{\partial g}\frac {\partial g}{\partial x} = \frac {\partial f}{\partial x}$ and other such "cancellations" hold. However, an operator which looks like division isn't entirely a bad thing, in that you get the correct units when doing a physical problem -- obviously this is because the definition does include a division, but only as part of a limit.
  • Kind of a nitpick, but $\frac {\partial^n f}{\partial x^n}$ looks awfully weird to me. And worse, it could be confusing to some who see the "denominator" as $\partial(x^2)$ instead of $(\partial x)^2$.

Other notations include $f_x$, $\partial_x f$, and $D_x f$. I don't like $f_x$ because that looks like an indexed function to me, not like $f$ is being operated on by a differential operator. $\partial_x f$ and $D_x f$ both look like operators acting on $f$ but don't tell you the units and we get back to "can we really use $x$ to refer to the first parameter of $f$ outside of the definition of $f$"?

Does anyone know of a better notation? Can anyone come up with one?

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    $\begingroup$ Instead of using indices like $x$ and $y$, one can use the numbers 1 and 2 referring to the first and second arguments of the function respectively. This will give $\partial_1 f$, $\partial_2 f$, etc. $\endgroup$
    – Ulrik
    Jan 4, 2015 at 21:44

1 Answer 1

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In case of doubt use $(\partial_{1}f)(x,y)$, a function of $(x,y)$ whose value at $(a,b)$ is $(\partial_{1}f)(a,b)$.
The penalty for not doing so is having your readers scratch their heads over$$\frac {\partial f}{\partial x } (y,x) \quad ??$$ You can then iterate: define $\partial_{12}f=\partial_{1}(\partial_{2}f)$, etc.
A pleasant result is that for a twice continuously differentiable function we have $\partial_{12}f=\partial_{21}f$ so that the convention underlying the order in which we take partial derivatives doesn't matter: some authors define $\partial_{12}f=\partial_{2}(\partial_{1}f)$ .

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    $\begingroup$ Perhaps Georges should have said $\partial_{1, 2}$, but just as we write $a_{13}$ for the first-row, third-col entry of a matrix $A$, we can probably forgive $\partial_{12}$. I suppose we sometimes encounter functions of 12 variables, but ... $\endgroup$ Jan 4, 2015 at 22:26
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    $\begingroup$ I believe that his $\partial_{12}$ means partial with respect to the first and then the second slot. For the derivative with respect to the first, twice, you'd use $\partial_{11}$. There might be a case for $\partial^2_{12}$, but the superscript isn't really necessary, since you can count the subscripts --esp. if there are commas. To write the thing you asked about, I'd write $\partial_{2, 1}$. And $\partial_2 \partial_1$ also works. $\endgroup$ Jan 4, 2015 at 22:38

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