Since the beginning of the year, our maths teacher showed us the Binomial Theorem in $\mathbb{R}$\, then in $\mathbb{C}$\, in $M_n(\mathbb{K)}$ with two matrices which commute, and now the Leibniz Theorem for derivatives.
Each time the proof of the theorem was very similar, by induction, but not exactly the same.
Isn't there a "big" theorem which would gather all these previous theorems as particular cases?
Or isn't there a proof which would be the same for all of these?
I think you probably want something like
Let $A,B:M \to M$ be linear operators acting on a module $M$ over an abelian ring $R$ (with identity $1_R$), such that $AB=BA$. Then we have the identity $$ (A+B)^n = \sum_{k=0}^n \binom{n}{r}_R A^r B^{n-r}, $$ where the coefficients are defined recursively by $$ \binom{n}{0}_R = \binom{n}{n}_R = 1_R, \\ \binom{n}{r}_R = \binom{n-1}{r-1}_R+\binom{n-1}{r}_R. $$
(I'm not sure this result has a name in this general form, but it'll probably involve at least one of the words "binomial" and "Leibniz".)
This obviously has the usual inductive proof. It contains specialisations to multiplying by $x$ and $y$ over the real and complex numbers, commuting matrices on a field, and derivatives on rings of smooth functions. This last may not be obvious, so I'll just point out an algebraic way of doing it.
As you know, the derivative acts by the product/Leibniz rule as $$ D(f \cdot g)=(Df) \cdot g+f \cdot (Dg), $$ omitting variables. But that means that we can define the action on $f$ and $g$ as $$ D = \partial_f+\partial_g, $$ i.e. the sum of a partial derivative which only acts on $f$ and one which only acts on $g$. You can see that the operators $\partial_f,\partial_g$ are commutative, and then the general Leibniz rule follows from the above result.
In the binomial form, this is probably the best you can do (you need the additive and multiplicative structure of the module, for example).
