TITLE: If d/dx is an operator, on what does it operate?
QUESTION [21 upvotes]: If $\frac{d}{dx}$ is a differential operator, what are its inputs?  If the answer is "(differentiable) functions" (i.e., variable-agnostic sets of ordered pairs), we have difficulty distinguishing between $\frac{d}{dx}$ and $\frac{d}{dt}$, which in practice have different meanings.  If the answer is "(differentiable) functions of $x$", what does that mean?  It sounds like a peculiar hybrid of mathematical object (function) with mathematical notation (variable $x$). 
Does $\frac{d}{dx}$ have an interpretation as an operator, distinct from $\frac{d}{dt}$, and consistent with its use in first-year Calculus?

REPLY [20 votes]: Not sure why this question is back on the front page, but I just wanted to add that the situation seems to be clarified by temporarily generalising to higher dimensions and to curved spaces, i.e., by taking a differential geometry perspective.
Firstly, a quick reminder of the concept of a dual basis in linear algebra: if one has an $n$-dimensional vector space $V$ (let's say over the reals ${\bf R}$ for sake of discussion), and one has a basis $e^1,\dots,e^n$ of it, then there is a unique dual basis $e_1,\dots,e_n$ of the dual space $V^* = \mathrm{Hom}(V,{\bf R})$, such that $e_i(e^j) = \delta_i^j$ for all $i,j=1,\dots,n$ ($\delta_i^j$ being the Kronecker delta, and where I am trying to choose subscripts and superscripts in accordance with Einstein notation).  It is worth pointing out that while each dual basis element $e_i$ is "dual" to its counterpart $e^i$ in the sense that $e_i(e^i) = 1$, $e_i$ is not determined purely by $e^i$ (except in the one-dimensional case $n=1$); one must also know all the other vectors in the basis besides $e^i$ in order to calculate $e_i$.
In a similar spirit, whenever one has an $n$-dimensional smooth manifold $M$, and (locally) one has $n$ smooth coordinate functions $x^1,\dots,x^n: M \to {\bf R}$ on this manifold, whose differentials $dx^1,\dots,dx^n$ form a basis of the cotangent space at every point $p$ of the manifold $M$, then (locally at least) there is a unique "dual basis" of derivations $\partial_1,\dots,\partial_n$ on $C^\infty(M)$ with the property $\partial_i x^j = \delta_i^j$ for $i,j=1,\dots,n$.  (By the way, proving this claim is an excellent exercise for someone who really wants to understand the modern foundations of differential geometry.)
Now, traditionally, the derivation $\partial_i$ is instead denoted $\frac{\partial}{\partial x^i}$.  But the notation is a bit misleading as it suggests that $\frac{\partial}{\partial x^i}$ only depends on the $i^{th}$ coordinate function $x^i$, when in fact it depends on the entire basis $x^1,\dots,x^n$ of coordinate functions.  One can fix this by using more complicated notation, e.g., $\frac{\partial}{\partial x^i}|_{x^1,\dots,x^{i-1},x^{i+1},\dots,x^n}$, which informally means "differentiate with respect to $x^i$ while holding the other coordinates $x^1,\dots,x^{i-1},\dots,x^{i+1},\dots,x^n$ fixed".  One sees this sort of notation for instance in thermodynamics.  Of course, things are much simpler in the one-dimensional setting $n=1$; here, any coordinate function $x$ (with differential $dx$ nowhere vanishing) gives rise to a unique derivation $\frac{d}{dx}$ such that $\frac{d}{dx} x = 1$.
With this perspective, we can finally answer the original question.  The symbol $x$ refers to a coordinate function $x: M \to {\bf R}$ on the one-dimensional domain $M$ that one is working on.  Usually, one "simplifies" things by identifying $M$ with ${\bf R}$ (or maybe a subset thereof, such as an interval $[a,b]$) and setting $x$ to be the identity function $x(p) = p$, but here we will adopt instead a more differential geometric perspective and refuse to make this identification.  The inputs to $\frac{d}{dx}$ are smooth (or at least differentiable) functions $f$ on the one-dimensional domain $M$.  Again, one usually "simplifies" things by thinking of $f$ as functions of the coordinate function $x$, but really they are functions of the position variable $p$; this distinction between $x$ and $p$ is usually obscured due to the above-mentioned "simplification" $x(p)=p$, which is convenient for calculation but causes conceptual confusion by conflating the map with the territory.
Thus, for instance, the identity
$$ \frac{d}{dx} x^2 = 2x$$
should actually be interpreted as
$$ \frac{d}{dx} (p \mapsto x(p)^2) = (p \mapsto 2x(p)),$$
where $p \mapsto x(p)^2$ denotes the function that takes the position variable $p$ to the quantity $x(p)^2$, and similarly for $p \mapsto 2x(p)$.
If one also had another coordinate $t: M \to {\bf R}$ on the same domain $M$, then one would have another differential $\frac{d}{dt}$ on $M$, which is related to the original differential $\frac{d}{dx}$ by the usual chain rule
$$ \frac{d}{dt} f = \left(\frac{d}{dt} x\right) \left(\frac{d}{dx} f\right).$$
Again, for conceptual clarity, $t, x, f: M \to {\bf R}$ should all be viewed here as functions of a position variable $p \in M$, rather than being viewed as functions of each other.