Soft Question - Generalizations of the Derivative

Question

This is a soft question. I'm asking for any interesting and rather unknown generlizations of the derivative. I know it is generalized through derivations which are functions $\delta$ satisfying $$\delta(uv) = v\delta(u) + u\delta(v)$$ Some of them include for example, q-derivatives used in quantum calculus $$\left(\frac{d}{dx}\right)_qf(x) = \frac{f(qx)- f(x)}{qx-x}$$ The arithmetic derivative used in number theory $$D(n) = n\sum_{p\mid n}\frac{\nu_p(n)}{p}$$ The Hasse derivative $$D^{(r)}X^n = {n\choose r}X^{n-r}$$ and many more... I'm asking if anyone knows any interesting and perhaps relatively unknown generalizations of the derivative/examples of interesting derivations. Feel free to share anything you feel is interesting

Answer 1

Two made-up cases.

1) Acting on elements one at a time

$d$ applies a function $f$ on elements of a subset $A$, outputting a set of copies of $A$ where in each copy, one $x$ is replaced by $f(x)$.
Example: $d(\{a,b,c\}) = \{\{f(a), b, c\}, \{a, f(b), c\}, \{a, b, f(c)\}\}$
Quite intuitive, but the precise definition is somewhat tedious:
For any $f: E \to E$, define
$d: \mathcal P(E) \to \mathcal P (\mathcal P(E))$ with $d(A) = \{(A \setminus \{x\}) \cup f(x) , x \in A\}$
Then $d(A \sqcup B) = (d(A) \otimes B) \cup (d(B) \otimes A) $
where $\otimes: (\mathcal P (\mathcal P(E)), \mathcal P(E)) \to \mathcal P (\mathcal P(E))$, $X \otimes Y = \{ x \cup Y, x \in X \}$.

A similar one, cutting one element at a time:
$d: \mathcal P(E) \to \mathcal P (\mathcal P(E))$ with $d(A) = \{(A \setminus \{x\}), x \in A\}$
Example: $d(\{a,b,c\}) = \{\{b,c\}, \{a,b\}, \{a,c\}\}$
This also verifies $d(A \sqcup B) = (d(A) \oplus B) \cup (d(B) \oplus A) $

This "cutting elements one at a time" is close to the derivative of simplices in algebraic topology: quickly said, the boundary of a $p$-simplex $S$ is the union of $p-1$ simplices, each one being $S$ minus one vertex. Except here we use plain sets, while simplices are oriented sets, and the boundary operator (https://en.wikipedia.org/wiki/Simplex#Algebraic_topology), is an anti-derivation (https://en.wikipedia.org/wiki/Derivation_(differential_algebra)#Graded_derivations).

Note: working with sets means working with union, which is painfully not cancellative. Instead, we could work with multisets, or $\mathbb Z$-modules as in algebraic topology.

The principle of "cutting one element at a time" is at work on polynomials of $\mathbb C [X]$: decompose into a product of prime polynomials, and cross them one at a time.
Example: $((X-a)(X-b)(X-c))' = (X-b)(X-c) + (X-a)(X-c) + (X-a)(X-b)$.
The same process can be used for integers: this is the arithmetic derivative quoted in the OP.
$(30)' = (2 \times 3 \times 5)' = 3 \times 5 + 2 \times 5 + 2 \times 3 = 31$

2) Multisets in $\mathbb R ^n$ (more fancy and conjectural)

Let $P_n$ be the set of finite multisets of points in $\mathbb R ^n$. Define $d: P_n \to P_n$,
$d(A) = \{$ equilibrium points of a set of forces in $\frac 1 r$ from points in $A \;\}$
(With the convention that for any point in $A$ with multiplicity $m$, there is an equilibrium point at the same place, with multiplicity $m-1$).

For $n=2$, identifying $\mathbb R^2$ with $\mathbb C$, if $A$ is the multiset of roots of polynomial $P$, then $d(A)$ is the multiset of roots of $P'$ (Bôcher's theorem: https://en.wikipedia.org/wiki/B%C3%B4cher%27s_theorem). So it ensures $d$ behaves as a derivative.
Example: $d(\{1,i,-i\}) = \{\frac {1-i\sqrt2} 3, \frac {1+i\sqrt2} 3 \}$
Because $\{1,i,-i\}$ are the roots of $(X-1)(X^2+1)=X^3-X^2+X-1$, whose derivative is $3X^2-2X+1$, whose roots are $\{\frac {1-i\sqrt2} 3, \frac {1+i\sqrt2} 3 \}$.

We would like some kind of Leibniz rule: $d(A\otimes B) = d(A)\otimes B \oplus A\otimes d(B)$
As $\otimes$ corresponds to multiplication of polynomials, its definition is clear: it is union of multisets.
As $\oplus$ corresponds to addition of polynomials, however, it has no simple definition: roots of the sum of two polynomials have no simple relationship with the polynomials roots, except in specific cases (e.g. with two degree-1 polynomials, root of the sum is mean of the roots weighted by the coefficient on $X$). And we need to add a weight to our multisets, to account for the polynomial's highest degree coefficient.

But when $n>2\,$? Not even sure that $p$ points, each one with a force in $\frac 1 r$, have $p-1$ equilibrium points: I cannot find a proof nor a reference.
Of course for $n>2$ it works with $p$ points when $p\le 3$, or when the $p$ points are coplanar (in a 2-dimension plane); and probably also when the points are nearly coplanar, by a continuity argument.

Actually this could be extended to any distribution of masses in any subset (not necessarily finite) of $\mathbb R^n$; but how does the equilibrium set of an $\frac 1 r$ force from these masses would behave?