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.