If d/dx is an operator, on what does it operate?

Question

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?

Answer 1

(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.

Answer 2

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.

Answer 3

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.)

Answer 4

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 $t \mapsto (x(t), y(t))$.

Answer 5

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 $\frac{d}{dt} f(t,x(t))$...

Answer 6

I believe that the origin or the problem here is the common schizophrenic mixture of Leibniz-time notation with Bourbaki-time foundations of calculus.

When trying to make sense of the traditional Leibniz notation in calculus, it is important to understand that even the notion of function in the 18th century was not what it is now. Now a function is something along the lines of "a set of pairs," "a functional relation," "a triple of a functional relation, a domain, and a co-domain," "a morphism in the category of sets." In the 18th century, to be a function was a property/predicate on variables: one variable could be a function of another, and also two variables could be functions of each other.

So, if you wish to use the notation $dy/dx$ in a non-schizophrenic way, you probably need to return to the "original" approach: you have different quantities, which are denoted by variables, and some of these variables are "functions" of others, because they are related by some identities, and then you can find expressions for differentials of variables, like $dx$ and $dy$, and also to calculate $dy/dx$. Of course, you will not be talking about any "operators" acting on anything this way.

As a work-around solution for teaching, I think of $d/dx$ as a form of a macro, or syntactic sugar, acting on expressions: if $y = x^2$, then $$ \frac{d}{dx}y =\frac{d}{dx}x^2 = 2x. $$

Answer 7

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 ordered 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]

Answer 8

Little late to the game, but when dealing with operators, as I constantly do, I always keep in mind Bernard Friedman's comment in one of his books on differential equations and Green functions to the effect: An operator is ill-defined if you don't specify on what it operates. He goes on to talk about the boundary conditions of differential equations that make them well-defined. (This notion is also manifest in the necessity of inventing and distinguishing among partial derivatives and total / material derivatives, constants w.r.t. to whatever, dependent and independent variables, Lagrangian versus Eulerian frames of reference, active and passive transformations, etc.)

Feynman stated similar sentiments when talking about the grad, or ∇ operator, in Section 2.4 of his lecture "Differential calculus of vector fields". After first talking about the beauty of mathematical abstraction in separating the del op from what it is operating on--"We leave the operators, as Jeans said, 'hungry for something to differentiate'.”--he then says, "You must always remember, of course, that ∇ is an operator. Alone, it means nothing."

In quantum mechanics, there is a similar but more heated situation. There have been skirmishes between different schools of researchers with different interpretations of the math of quantum mechanics, but in the words of Rota, "Le teorie vanno e vengono ma le formule restano." (The theories may come and go but the formulas remain.) ... because the math is fairly clear on what mathematical ops act on what mathematical functions to produce a physical observable. It's the Copenhagen probability wave function versus the de Broglie-Bohm guiding field that's the rub here for some--the interpretation of the function rather than the op.

As far as derivatives go, I can apply one term-by-term to a formal power series in one independent variable, that is divergent except at the origin. Such a series is not even a function yet the term-by-term differentiation is consistent with formal composition and compositional and multiplicative inversion of the series even formal Laplace and Legendre transforms. However, the interpretation of the derivative at a global level as the slope of some curve at some point is nonsensical. Distribution theory sheds no light on the situation. Category theory? Enlighten me. No math god or demon, no ghost of Plato gives any meaning to the differentiation aside from my specification of what it operates on and how.

Even though divergent there are often operations on the series that can give it even a physical significance. As Heaviside said, "It's divergent! Good, then maybe I can do something with it." And, any divergent series is a derivative, in my prescription, of another divergent series. I can truncate the series and assign the usual meanings, but where to truncate it? The theory of divergent series can often give some algorithm for where to truncate it so that the truncation has some significance, but then I'm throwing in another operator and specifying how it, the derivative operator, and the divergent series interact, and that's a cottage industry, assigning meaning to the "work of the Devil".

Here's another perspective--two diff ops that are equivalent when acting on certain beasts and quite different when acting on others. The hungry ops are

$$O_1 = \sum_{n \geq 0} a_n z^n \frac{\partial_{z=0}^n}{n!} $$

and, in umbral notation with

$$(1-a.)^\beta= \sum_{k \geq 0} (-1)^k \binom{\beta}{k}a_k = \triangle_{k \geq0}^\beta a_k,$$

$$O_2 = \sum_{n \geq 0}(-1)^n (1-a.)^n z^n \frac{\partial_z^n}{n!}.$$

When acting on $z^m$ for $m= 0,1,2,...$, they give

$O_1 z^m = a_m z^m$

and

$O_2 z^m = (z - (1-a.)z)^m = (a.z)^m = a_mz^m.$

They are the same, yet acting on $z^{\alpha}$ for $\alpha > 0$ and not an integer,

$O_1 z^\alpha$ diverges

and

$O_2 z^\alpha = (z-(1-a.)z)^{\alpha} = (\triangle_{n \geq 0}^\alpha\triangle_{j=0}^n a_j)z^{\alpha} = a_{\alpha}z^{\alpha},$

a Newton-Gregory series that might or might not diverge. One can assign meaning to $O_1$ via $O_2$ where it gives a convergent result, but I'm assigning that meaning. $O_1$ isn't capable of making that choice by itself.

(A theoretical physicist or a Ramanujan might even see an expression in terms of the discrete integers $n$ and then operate on it with $\frac{\partial}{\partial n}$ and find a meaningful useful interpretation. Whatever works.)

Answer 9

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: $$ \left\langle\frac{du}{dx},\phi\right\rangle_{\mathcal D'(\mathbb R),\mathcal D(\mathbb R)}= -\left\langle u,\frac{d\phi}{dx}\right\rangle_{\mathcal D'(\mathbb R),\mathcal D(\mathbb R)}. $$ 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), $$ \frac{d}{dx}(\ln \vert x\vert)=\operatorname{pv}\frac{1}{x} $$ and many other classical formulas. In particular, you can define the derivative of any $L^1_\text{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 $\frac{d}{i\,dx}$ is selfadjoint.

Answer 10

The question is: how to properly render $$\frac{d}{dx}: (x ↦ f(x)) ↦ (x ↦ f'(x)),$$ so as to make the tie-in with $x$ explicit.

Since $x ↦ f(x)$ is synonymous with $(λx)f(x)$, which is $f$, itself, by the $η$-rule - similarly for $x ↦ f'(x) = (λx)f'(x) = f'$, then we can equate $$\frac{d}{dx}(\_) = Dλx(\_),$$ where $D = (λf)f'$.

So, it's what we might call a "binding" operator, since it implicitly contains a nested $λ$ in it. It's understood that it's only a partial operator, since not all $f$ are differentiable at all points of interest.

For partial derivatives, we might write: $$\frac{∂}{∂x} = λ(x,y)(Dλxf(x,y))(x), \quad \frac{∂}{∂y} = λ(x,y)(Dλyf(x,y))(y).$$ To render this accurately requires extending the syntax for $λ$-expressions to include tuple constructor and deconstructors satisfying the identities: $$H(x,y) = x, \quad I(x,y) = y, \quad (H(z),I(z)) = z.$$ Then, we can write this as: $$\frac{∂}{∂x} = λz(Dλxf(x,I(z)))(H(z)), \quad \frac{∂}{∂y} = λz(Dλyf(H(z),y))(I(z)),$$ or by reusing the earlier definition of total derivatives: $$\frac{∂}{∂x} = λz\left(\frac{d}{dx}f(x,I(z))\right)(H(z)), \quad \frac{∂}{∂y} = λz\left(\frac{d}{dy}f(H(z),y)\right)(I(z)).$$

In contrast, this can't really be considered a decisive answer, since you actually want to render the kind of distinction seen in the following example: $$f(t, u) = F(t + u, t)\quad⇔\quad F(s,t) = f(t, s - t),$$ where $$ \left(\frac{∂}{∂t}\right)_u(\_) = \left(\frac{∂}{∂s}\right)_t(\_) + \left(\frac{∂}{∂t}\right)_s(\_),\quad \left(\frac{∂}{∂u}\right)_t(\_) = \left(\frac{∂}{∂s}\right)_t(\_),$$ and $$ \left(\frac{∂}{∂t}\right)_s(\_) = \left(\frac{∂}{∂t}\right)_u(\_) - \left(\frac{∂}{∂u}\right)_t(\_),\quad \left(\frac{∂}{∂s}\right)_t(\_) = \left(\frac{∂}{∂u}\right)_t(\_), $$ in as direct of a way as possible.

I think that in this case, it would be more fruitful to treat the operators as differential operators in the sense of differential geometry on differentiable manifolds, since this seems to best fit with and fully encompass the intended usage.

Non-Analytic Usages
There are other usages, devised by analogy, that lie outside differential geometry; e.g. Differential Algebra. This can be generalized to semi-rings as in the following example.

Consider the following context-free grammar $$S → u L v | x, \quad L → λ | S L,$$ over a monoid $M$ for which $u,v,x ∈ M$, where $λ ∈ M$ denotes the identity. (Footnote: the notion of context-free languages and context-free grammar can be defined for arbitrary monoids, not just for free monoids). As an algebraic system, it can be expressed as: $$S ≥ u L v + x, \quad L ≥ 1 + S L,$$ over suitably-defined algebra $ℜ(M) ⊆ ℭ(M)$, that contains an idempotent sum $+$ (i.e. $x + x = x$) and multiplicative identity $1$ in place of $λ$. More precisely, $ℜ(M)$ and $ℭ(M)$ are, respectively, the rational and context-free subsets of $M$, with singletons $\{m\}$, for $m ∈ M$ denoted as just $m$, the identity $1 = \{λ\}$, the sum $A+B = A∪B$, for $A,B⊆M$ and $≥$ denoting inclusion of subsets.

This is a system of fixed-point inequations, in which the desired solution is the least fixed point. (Another way of saying the same is: it's a non-numeric "optimization problem").

If $M$ is a free commutative monoid (that is: a free object in the category of commutative monoids), then by Parikh's Theorem $ℜ(M) = ℭ(M)$, the simplest proof arising directly by way of an analogue differential calculus. The least fixed point of $x ≥ f(x)$, written as $(μx)f(x)$ is $x = f'(f(0))^* f(0)$, where $A ↦ A^*$ is the Kleene star operator - the same as what's used with regular expressions. This applies even in the multivariate case - but with partial differential operators.

Thus, $$L ≥ 1 + S L\quad⇒\quad(μL)(1 + SL) = \left(\frac{∂}{∂L}(1 + LS)\right)^*(1 + 0S) = S^*.$$ Substituting $f(S) = (μL)(1 + SL)$, we obtain: $$S ≥ u L v + x\quad⇒\quad(μS)(1 + xf(S)y) = \left(\frac{∂}{∂S}(1 + xf(S)y)\right)^*(1 + xf(0)y).$$ For commutative Kleene algebras, $x↦x^*$ behaves as the exponential function. Therefore, $0^* = 1$, $d(A^*)/dA = A^*$. Thus, $$(μS)(1 + xf(S)y) = \left(xf'(0)y)\right)^*(1 + x(f(0))y) = (xy)^*(1 + xy) = (xy)^*.$$ (In Kleene algebra $A^*(1 + f(A)) = A^*$ for any polynomial $f(A)$.) Together, this yields the least fixed point solution $L = {(xy)^*}^* = (xy)^*$ and $S = (xy)^*$, since ${A^*}^* = A^*$ in Kleene algebra.