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
$$
\left(\frac{∂}{∂t}\right)_u(\_) = \left(\frac{∂}{∂s}\right)_t(\_) + \left(\frac{∂}{∂t}\right)_s(\_),\quad
\left(\frac{∂}{∂u}\right)_t(\_) = \left(\frac{∂}{∂s}\right)_t(\_), \\
\left(\frac{∂}{∂t}\right)_s(\_) = \left(\frac{∂}{∂t}\right)_u(\_) - \left(\frac{∂}{∂u}\right)_t(\_),\quad
\left(\frac{∂}{∂s}\right)_t(\_) = \left(\frac{∂}{∂u}\right)_t(\_),
$$
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](https://en.wikipedia.org/wiki/Differentiable_manifold#Differentiation_of_functions) in the sense of differential geometry on [differentiable manifolds](https://en.wikipedia.org/wiki/Differentiable_manifold), since this seems to best fit with and fully encompass the intended usage.

<b>Non-Analytic Usages</b><br/>
There are other usages, devised by analogy, that lie outside differential geometry; e.g. [Differential Algebra](https://en.wikipedia.org/wiki/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 <i>arbitrary</i> 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 <i>Parikh's Theorem</i> $ℜ(M) = ℭ(M)$, 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.