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?

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    $\begingroup$ I don't think this is, at least as stated, a good question for MO (and apparently I am not the only one, because I did not go as far as to downvote). Also, I would challenge some of your assumptions. The notation $\frac{df}{dx}$ is used very frequently in the books I'm reading, for instance. Also, does $\frac{d}{dx}$ denote a single operator? As Poincare remarked: "Mathematics is the art of giving the same name to different things". $\endgroup$ Dec 4, 2012 at 16:30
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    $\begingroup$ Mathematics is a human activity, hence it often benefits from some en.wikipedia.org/wiki/Abuse_of_notation $\endgroup$
    – Qfwfq
    Dec 4, 2012 at 17:14
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    $\begingroup$ Jason -- you could try asking this on math.stackexchange.com; there is a chance you'll get a detailed answer there. $\endgroup$
    – algori
    Dec 4, 2012 at 17:39
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    $\begingroup$ I find this question to be both deeper and more interesting than it appears by the comments that some others here do. So I have voted to reopen, and would look forward to reading a thoughtful answer that takes the issue seriously. $\endgroup$ Dec 4, 2012 at 18:10
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    $\begingroup$ I've always thought of $x$ as just a choice of coordinate on the $1$-manifold $\mathbb R$ (i.e. picking a diffeomorphism with a ``standard'' copy of $\mathbb R$). Without choosing coordinates, we have for each function $f:\mathbb R \to \mathbb R$ a linear operator $df$ on each tangent space. Picking a coordinate function $x$ allows us to express these as numbers, hence we get a function $df/dx$. $\endgroup$ Dec 4, 2012 at 23:41

7 Answers 7


(From the post on my blog:)

To my way of thinking, this is a serious question, and I am not really satisfied by the other answers and comments, which seem to answer a different question than the one that I find interesting here.

The problem is this. We want to regard $\frac{d}{dx}$ as an operator in the abstract senses mentioned by several of the other comments and answers. In the most elementary situation, it operates on a functions of a single real variable, returning another such function, the derivative. And the same for $\frac{d}{dt}$.

The problem is that, described this way, the operators $\frac{d}{dx}$ and $\frac{d}{dt}$ seem to be the same operator, namely, the operator that takes a function to its derivative, but nevertheless we cannot seem freely to substitute these symbols for one another in formal expressions. For example, if an instructor were to write $\frac{d}{dt}x^3=3x^2$, a student might object, "don't you mean $\frac{d}{dx}$?" and the instructor would likely reply, "Oh, yes, excuse me, I meant $\frac{d}{dx}x^3=3x^2$. The other expression would have a different meaning."

But if they are the same operator, why don't the two expressions have the same meaning? Why can't we freely substitute different names for this operator and get the same result? What is going on with the logic of reference here?

The situation is that the operator $\frac{d}{dx}$ seems to make sense only when applied to functions whose independent variable is described by the symbol "x". But this collides with the idea that what the function is at bottom has nothing to do with the way we represent it, with the particular symbols that we might use to express which function is meant. That is, the function is the abstract object (whether interpreted in set theory or category theory or whatever foundational theory), and is not connected in any intimate way with the symbol "$x$". Surely the functions $x\mapsto x^3$ and $t\mapsto t^3$, with the same domain and codomain, are simply different ways of describing exactly the same function. So why can't we seem to substitute them for one another in the formal expressions?

The answer is that the syntactic use of $\frac{d}{dx}$ in a formal expression involves a kind of binding of the variable $x$.

Consider the issue of collision of bound variables in first order logic: if $\varphi(x)$ is the assertion that $x$ is not maximal with respect to $\lt$, expressed by $\exists y\ x\lt y$, then $\varphi(y)$, the assertion that $y$ is not maximal, is not correctly described as the assertion $\exists y\ y\lt y$, which is what would be obtained by simply replacing the occurrence of $x$ in $\varphi(x)$ with the symbol $y$. For the intended meaning, we cannot simply syntactically replace the occurrence of $x$ with the symbol $y$, if that occurrence of $x$ falls under the scope of a quantifier.

Similarly, although the functions $x\mapsto x^3$ and $t\mapsto t^3$ are equal as functions of a real variable, we cannot simply syntactically substitute the expression $x^3$ for $t^3$ in $\frac{d}{dt}t^3$ to get $\frac{d}{dt}x^3$. One might even take the latter as a kind of ill-formed expression, without further explanation of how $x^3$ is to be taken as a function of $t$.

So the expression $\frac{d}{dx}$ causes a binding of the variable $x$, much like a quantifier might, and this prevents free substitution in just the way that collision does. But the case here is not quite the same as the way $x$ is a bound variable in $\int_0^1 x^3\ dx$, since $x$ remains free in $\frac{d}{dx}x^3$, but we would say that $\int_0^1 x^3\ dx$ has the same meaning as $\int_0^1 y^3\ dy$.

Of course, the issue evaporates if one uses a notation, such as the $\lambda$-calculus, which insists that one be completely explicit about which syntactic variables are to be regarded as the independent variables of a functional term, as in $\lambda x.x^3$, which means the function of the variable $x$ with value $x^3$. And this is how I take several of the other answers to the question, namely, that the use of the operator $\frac{d}{dx}$ indicates that one has previously indicated which of the arguments of the given function is to be regarded as $x$, and it is with respect to this argument that one is differentiating. In practice, this is almost always clear without much remark. For example, our use of $\frac{\partial}{\partial x}$ and $\frac{\partial}{\partial y}$ seems to manage very well in complex situations, sometimes with dozens of variables running around, without adopting the onerous formalism of the $\lambda$-calculus, even if that formalism is what these solutions are essentially really about.

Meanwhile, it is easy to make examples where one must be very specific about which variables are the independent variable and which are not, as Todd mentions in his comment to David's answer. For example, cases like

$$\frac{d}{dx}\int_0^x(t^2+x^3)dt\qquad \frac{d}{dt}\int_t^x(t^2+x^3)dt$$

are surely clarified for students by a discussion of the usage of variables in formal expressions and more specifically the issue of bound and free variables.

  • $\begingroup$ I assume you still define integeration as an operation on a function, as opposed to a purely syntactic process (which for calculus would mostly be possible). So, if you write $\int_{0}^x x^3 dt$ and a student ask: please explain precisely which function is meant (with domain of definition and all that) by $x^3$ what do you reply? Same question for $7$ in $\int_{0}^x 7 dt$. $\endgroup$
    – user9072
    Dec 6, 2012 at 15:31
  • $\begingroup$ Well, of course I would give the answer explaining precisely which function I had meant; the notation is ambiguous without doing so, and this is the point of the question and my answer. One can imagine a variety of perfectly reasonable answers that would correspond to different intended meanings here. The discussion of whether $x$ in the integrand was meant to depend on $t$ or be an independent variable from $t$ and so on was the discussion I alluded to at the end of my answer. The part of the question interesting me is the precise nature of this particular ambiguity. $\endgroup$ Dec 6, 2012 at 15:48
  • $\begingroup$ I did not mean anything fancy. The situation I envision is just $x^3$ does not depend on $t$ for a result of $x^4$ and $7x$ in the latter case. But what is the precise nature of the finction in the integeral. In particular, please make it so that the $d/dx$ application you give makes still sense. $\endgroup$
    – user9072
    Dec 6, 2012 at 16:07
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    $\begingroup$ I am grateful to Joel for his support of the question, including this interesting answer. Certainly $\frac{d}{dx}$ is similar to a quantifier: It "shields" occurrences of the variable $x$ in its scope from direct substitution. It is defined in terms of the limit, which also binds a variable, as a quantifier could. It is a very strange quantifier, though, as $x$ once again occurs free in the ("bound"?) expression $\frac{d}{dx} x^3$ since $\frac{d}{dx} x^3 = 3x^2$. $\endgroup$ Dec 6, 2012 at 19:35
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    $\begingroup$ I agree with this answer. While professional mathematician differentiate functions, in basic calculus we differentiate expressions. $\endgroup$ Dec 7, 2012 at 14:58

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.

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    $\begingroup$ The answer to why this thread is on the front page now is that Michael Bächtold has been reactivating interest in these issues recently: see his answer here which links to some related threads he's interested in. Some of the people participating in those threads are also recently participating here (e.g., Mike Shulman in comments). $\endgroup$
    – Todd Trimble
    Aug 21, 2018 at 2:50
  • $\begingroup$ Ah, thanks for clearing that up. And now I see that Michael's answer in fact has a lot of overlap with mine. $\endgroup$
    – Terry Tao
    Aug 21, 2018 at 3:00
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    $\begingroup$ I noticed that you switched from $\partial/\partial{x^1}$ to $d/dx$ when $n=1$ without comment. We might think of it as a traditional variation in notation, but it can also be interpreted as taking the differential and then dividing by $dx$. Dividing by a vector (in the cotangent space) is unusual, but it makes perfect sense in $1$ dimension, and then it's a theorem that the two interpretations of $d/dx$ are equivalent. (Even in $1$ dimension, you can't divide by the zero vector, but $dx$ cannot be zero anywhere if $x$ is a coordinate, so that's OK.) $\endgroup$ Aug 23, 2018 at 15:42

The accepted answer is good in that it draws attention to the subtleties involved, but as far as I can tell it doesn't really settle the matter.

Joel is careful to speak of a kind of binding of $x$ by $\frac{d}{dx}$, but at the same time he mentions that $x$ remains free in $\frac{d}{dx}x^3$. So is it free or bound?

It cannot be bound in the traditional sense (and Joel says that), otherwise we'd be allowed to rename bound variables ($\alpha$-convert) and write $$ \frac{d}{dx}x^2 = \frac{d}{dt}t^2, $$ which everyone since Leibniz would simplify to $$ 2x=2t. $$ It's probably a bad idea to have a mechanism wich allows us to conlcude that any two free variables are equal.

On the other hand $x$ cannot be free in the traditional sense, since if we substitute say $5$ for $x$ we'd get $$ \frac{d}{d5}5^2. $$ Most people would consider this meaningless. Even if we don't consider it meaningless, I fail to see how one could arrive from there to the expected result of $10$. (Certainly if you allow substituting $5$ for $x$ in $\frac{d}{dx}x^2$ you would also allow substituting $25$ for $5^2$ in $\frac{d}{d5}5^2$ to rewrite it as $\frac{d}{d5}25.$ But the same expression results if we substitute $5$ for $x$ in $\frac{d}{dx}(20+x)$, with the expected result now being 1.)

So we conclude that $x$ it is neither bound nor free in $\frac{d}{dx}x^2$. But which kind of binding is it then?

From a modern perspective it's tempting to say that $\frac{d}{dx}x^2$ is 'syntactic sugar' for $(\lambda x.x^2)' (x)$, where $f'$ denotes the derivative of a map $f:\mathbb{R}\to \mathbb{R}$ and $\lambda x.x^2$ is lambda calculus notation for the map $x\mapsto x^2$. But the expression $(\lambda x.x^2)' (x)$ has both a free $x$ (in the second parenthesis) and a bound $x$ (inside the $\lambda x.x^2$), while it's not clear which $x$ in $\frac{d x^2}{dx}$ is free/bound. So if we really want to interpret $\frac{d x^2}{dx}$ as syntactic sugar for $(\lambda x.x^2)' (x)$, there seems to be a proof missing that this notation is correct (which reminds me of Mike Shulman's question). We might also conclude what Andrej Bauer suggested elsewhere, that maybe $\frac{d f(x)}{dx}$ is broken notation that we should stop teaching.

Instead I'll argue that there is a consistent way of making sense of the notation $\frac{dy}{dx}$. It was already suggested in your question: interpret $\frac{d}{dx}$ as acting on "functions of $x$". You rightly asks what functions of $x$ are. Here's one way to answer that: interpret the variables $x$, $y$ of calculus as differentiable maps from a manifold $M$ (the state space) to $\mathbb{R}$. Call one such variable $y$ a function of $x$, if there exists $f:\mathbb{R}\to\mathbb{R}$ such that $y=f\circ x$. One can easily prove that if $y$ is a function of $x$ in this sense, then there is a unique $z:M\to \mathbb{R}$ such that $dy=z\cdot dx$ where $dx,dy$ are differential forms in the sense of modern differential geometry. (Indeed $z=f'\circ x$ and used to be called the differential coefficient of $dy$ wrt $dx$). Denote this unique $z$ with $\frac{dy}{dx}$.

You might not be very happy with the manifold $M$ appearing here, since it never appeared explicitly in the old calculus. I am not very happy with it either, which is why I asked this question, and only found now that you had already asked a very similar question several years earlier. (The answers you received there unfortunately don't satisfy me.)

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    $\begingroup$ Before seeing this answer, I just added essentially this proposal as an answer to the "very similar question" you linked to. (-: I don't think it's necessary to restrict this notation to act only on "functions of $x$", however. For instance, if $z = x^2+y^2$ for another "independent" variable $y$, then ${\rm d}z = 2 x \,{\rm d}x + 2 y\,{\rm d}y$, so that $\frac{{\rm d}z}{{\rm d}x} = 2 x + 2 y \frac{{\rm d}y}{{\rm d}x}$, which makes perfect sense. $\endgroup$ Aug 13, 2018 at 8:37
  • $\begingroup$ @MikeShulman Hmm, I'm not so sure if one can run into trouble with allowing arbitrary applications of d/dx. Consider the following example: take the equation $x=1$ and derive both sides with respect to $x$ to arrive at $0=1$. $\endgroup$ Aug 13, 2018 at 8:43
  • $\begingroup$ In that case the problem is that ${\rm d}x = 0$, so you can't divide by it. The state-space perspective is that there is no distinguished operator to call "d/dx", it really is literally taking the differential $\rm d$ followed by dividing pointwise by the differential of $x$, ${\rm d}x$, so there is nothing to "allow" except to worry about whether ${\rm d}x = 0$. $\endgroup$ Aug 13, 2018 at 8:59
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    $\begingroup$ Possibly a more faithful way to deal with the worry about whether ${\rm d}x=0$ is to consider all functions to be partial (as one generally does in calculus anyway). Then $\frac{{\rm d}y}{{\rm d}x}$ just has its domain restricted to the points of the tangent bundle where ${\rm d}x \neq 0$. Your calculation of $0=1$ is then perfectly valid as long as you keep track of domains, which are empty in this case -- the two functions $\emptyset \to \mathbb{R}$ constant at 0 and 1 are in fact equal! $\endgroup$ Aug 13, 2018 at 9:02
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    $\begingroup$ @MikeShulman: I just re-read you're answer on the other question and have a better picture of what's going on now. From your perspective $dy/dx$ will always be a partial function on $TX$, but only when $y$ is (locally) a function of $x$ will $dy/dx$ be the pullback of a function on $X$. So only then will $dy/dx$ be an observable quantity on the same state space as $y$ and $x$. $\endgroup$ Aug 13, 2018 at 19:14

I repeat (a variant of) my comment, even though I agree that it is shallow and has low entertainment value.

As long as we are only looking at functions in one variable, there is only one differential operator $D$, which may be called $\frac d{dx}$ or $\frac{d}{dt}$ depending on the context.

If you look at a composite function $f \circ g$, you may introduce the notation/abbreviation $x=g(t)$, $y=f(x)$, then

  • $\frac {d}{dx} f$ or $\frac d {dx} y$ is just $D(f)$,
  • and by $\frac{d}{dt} f$ or $\frac d{dt} y$ you mean $D(f\circ g)$.

So here both $\frac {d}{dx} $ and $\frac {d}{dt} $ have a meaning, and the meaning is different.

When we look at functions in, say, two variables (do they appear in first year calculus?), we implicitly introduce an (arbitrary) order of variables, say x is the first and t the second, and $\frac{\partial}{\partial x}$ is the partial derivative with respect to the first variable. This makes sense even if you treat functions as "variable-agnostic" sets of ordered pairs. (Which I do all the time, and do not find peculiar at all. Tastes differ.)

Of course, the intended meaning always depends on the context. If $f$ is a binary function, $\frac d {dt} f$ may be a variant notation for $\frac{\partial}{\partial t}f$, or it may be understood that we are really looking at a unary function $\hat f$ obtained by composing $f$ with some function $t \mapsto (x(t), y(t))$.


I am a little late to the question but wanted to add a low-tech answer which somehow complements JDH's answer:

The operators $\frac{d}{dx}$ and $\frac{d}{dt}$ are as distinguishable as $f(x)$ and $f(t)$.

Probably this formulation is a little too vague but it should just reflect that writing $d/dx$ says how one has named the free variables. As already illustrated, one gets into notational ambiguities in cases as $\frac{d}{dt} f(t,x(t))$...


Two answers: (1) Distribution theory. On the space $\mathcal D'(\mathbb R)$ of continuous linear forms on $\mathcal D(\mathbb R)=C_c^\infty(\mathbb R)$ it is easy to define the first derivative: $$ \langle\frac{du}{dx},\phi\rangle_{\mathcal D'(\mathbb R),\mathcal D(\mathbb R)}= -\langle u,\frac{d\phi}{dx}\rangle_{\mathcal D'(\mathbb R),\mathcal D(\mathbb R)}. $$ You get the ordinary derivative of a differentiable function, also $H'=\delta$ ($H$ is the Heaviside function, characteristic function of $\mathbb R_+$, $\delta$ the Dirac mass), $$ \frac{d}{dx}(\ln \vert x\vert)=\text{pv}\frac{1}{x} $$ and many other classical formulas. In particular, you can define the derivative of any $L^1_{loc} $ function, of course not pointwise but as above.

(2) Operator theory. In $L^2(\mathbb R)$, you consider the subspace $H^1(\mathbb R)=${$u\in L^2(\mathbb R), u'\in L^2(\mathbb R)$}, where the derivative is taken in the distribution sense. Then the operator $d/dx$ is an unbounded operator with domain $H^1(\mathbb R)$. It is even possible to prove that the operator $\frac{d}{idx}$ is selfadjoint.

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    $\begingroup$ I do not think this answers the question in any way, however it highlights a shortcoming of it. So thanks for the long comment. $\endgroup$
    – user9072
    Dec 5, 2012 at 11:22
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    $\begingroup$ @quid: How is "d/dx can be viewed as an operator on distributions/Sobolev spaces" not an answer in any way to "on what does d/dx operate/what are its inputs"? What are the particular shortcomings of the question you have in mind? $\endgroup$
    – Martin
    Dec 5, 2012 at 12:41
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    $\begingroup$ @user49437: Is now $d/dt$ also an operator on this space? If so, is it the same as $d/dx$ or perhaps the zero-operator or still something else. While what you quote is in the title the actual question was definitely not about giving spaces where a derivative is 'nicely' definable. One shortcoming of the question is that it does not make precise what $d/dx$ should even mean at all. There are, any number of ways to define some map somewhere one might reasonaby call '$d/dx$' consistent with calculus; specifically, is the context strictly single variable or not (cf Goldstern) $\endgroup$
    – user9072
    Dec 5, 2012 at 14:09
  • $\begingroup$ Thanks for the clarification. Well, I still think this does provide an answer to the the title + the first and last sentences without "distinct from d/dt". The rest of the post is apparently homotopic to an interesting question that eludes me so far, so I'm curious to see what answer Joel David Hamkins has prepared... $\endgroup$
    – Martin
    Dec 5, 2012 at 17:09
  • $\begingroup$ @user49437: You are welcome. And, it is true what you say, but it is my understanding the point of the question is precisely and only the 'distinct from d/dt' thing. But I agree it is not quite clear what OP wants. $\endgroup$
    – user9072
    Dec 5, 2012 at 17:24

Edit: obviously some people didn't realise this answer was tongue-in-cheek. Also, I read the question differently to others, given its ambiguity, and didn't bother with the last (possibly most crucial) part of the question.

The gist of my answer was an expansion of Sam Gunningham's comment, namely that the operator "$\frac{\mathrm{d}}{\mathrm{d}x}$" is actually the restriction of a functor on the category of pointed smooth manifolds and pointed maps, to the subcategory consisting of 1-dimensional vector spaces over $\mathbb{R}$. The idea of coordinate-independence is captured in the principle of equivalence (the violation of which used to jokingly be called "evil" by some people), in that mathematics can't tell the difference between diffeomorphic manifolds, and so whatever we call this operator, $\frac{\mathrm{d}}{\mathrm{d}x}$ or $\frac{\mathrm{d}}{\mathrm{d}t}$ or what-have-you, they are all naturally isomorphic, and so indistinguishable in the 1-variable case. I do take the point, hashed out in the comments below, that in the multi-variable case things are more subtle, and I cede to Golderstern's answer.

But the punchline is that the category of manifolds can be defined in many different ways: from material sets, from structural sets, via synthetic differential geometry or via Fermat theories, so I contend there is not a single answer to (my reading of the) question.

I find the statement

"(differentiable) functions" (i.e., variable-agnostic sets of ordered pairs)

exceeding peculiar. A differentiable function is a certain arrow in the category of smooth manifolds, and even better, it's a arrow in the category where objects are finite-dimensional $\mathbb{R}$-vector spaces $E^n$ (for all $n$) with the usual topology. The tangent bundle functor takes a smooth function $f\colon E^n \to E^m$ and returns a smooth function $df\colon TE^n \to TE^m$ (the tangent bundle of $E^n$ is diffeomorphic to $E^{2n}$, hence again a vector space). Let us say we are in the case $n=1$. We can restrict this function to the tangent space of $E^1$ at $0$ and get a smooth function $E \to E^m$. No coordinates were chosen here.

But how did you get this category of manifolds? I hear you ask. Well, I started with the category of sets and did the usual thing. But how did this category of sets turn up? Well, to give the short answer, ETCS. The longer answer is that the category of sets (or rather, a category of sets strong enough to formalise all of undergraduate calculus and in fact most of mathematics) can be defined in terms of a first order theory. (Aside, if it irks you to miss out of the more hard-core parts of ZFC, use the foundational theory SEAR-C instead - it likewise doesn't define functions as sets of ordred pairs.)

At no point did I define a function to be a set of ordered pairs, and everything is independent of choices of coordinates.

Alternatively, we just say that $d/dx$ is an operation in the Fermat theory of $C^\infty$-rings. In this sense, smooth functions can be seen as models for a theory which is far more focussed than set theory, and there is no flab, in that this theory only talks about smooth functions.

[If you are asking questions that assume $df(t)/dt$ and $df(x)/dx$ are somehow distinguishable, and bringing foundational definitions into basic calculus, then expect answers that answer with a similar level of chutzpah]

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    $\begingroup$ I think it is quite natural that (possibly naive) foundational questions are asked in a basic calculus course. Of course it is difficult to answer them at the right level. I think that the idea "a relation is a set of ordered pairs, a function is a special kind of relation" can be understood and applied by first year students, even if they may reject it when (or if) they later take a category-theoretic viewpoint. $\endgroup$
    – Goldstern
    Dec 5, 2012 at 13:40
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    $\begingroup$ I agree with Joel that the question is more about careful use of syntax (e.g., the proper manipulation of free vs. bound variables), and maybe not so much about things like set-theoretic foundations. It's connected with a familiar kind of abuse of notation seen in calculus courses where one writes $f(x)$ for a function when one really means $\lambda x. f(x)$. Here's an illustration of the trouble one can get into: we sometimes express FTC in the form $\frac{d}{d x} \int_a^x g(t) dt = g(x)$. Now suppose you ask a student to differentiate the function $F(x) = \int_a^x g(t) - g(x) dt$. (cont.) $\endgroup$
    – Todd Trimble
    Dec 5, 2012 at 14:35
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    $\begingroup$ Do you just substitute $x$ for $t$ in the integrand and get $g(x) - g(x) = 0$? No? Then how do you explain the rule properly? This would get into an interesting discussion of how variables are treated. $\endgroup$
    – Todd Trimble
    Dec 5, 2012 at 14:36
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    $\begingroup$ @quid: I'm glad you understand me better. But as I see it, the main issue the OP has is: what is the precise role of the notation "$x$" when we write $\frac{d}{dx}$, and this is the syntactic issue that I (and I think Joel) want to address. He or she only brings in the bit about "differentiable functions as sets of ordered pairs" to say that that doesn't address the problem at hand (so let's not get deflected by that). A careful treatment of what I think is worrying OP would center on how the correct handling of variables, which is an important issue made more piquant by invoking (cont.) $\endgroup$
    – Todd Trimble
    Dec 5, 2012 at 19:37
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    $\begingroup$ @quid: Concerning "more 'classroom' than 'computer algebra development'": I've found that calculus classrooms often include students who want to compute by explicit, precise rules, just like a computer algebra system (only less sophisticated), and who get very confused when they are told (or they merely get the impression) that the correct manipulation of symbols in mathematics depends on thinking about what the symbols mean. Such thinking is, for them, a mystery; they want (at least) that this "meaning" be unambiguously inferrable from what is written. I think that underlies this question. $\endgroup$ Dec 6, 2012 at 0:59

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