Do I need commutativity of $A$ to say that $Der_\mathbb{K}(A)$ is a left $A$-module? Let $A$ be an associative $\mathbb{K}$-algebra, with $\mathbb{K}$ a field.
A derivation of $A$ is a $\mathbb{K}$- linear map $D:A\to A$ such that for each $a,b\in A$ we have $D(ab)=D(a)b+a D(b)$ $\quad[1]$
For each $a\in A$ and for each derivation $D$ define $aD:A\to A$ with $(aD)(b):=a \cdot D(b)$.
I want to show that $aD$ is still a derivation of $A$.
In particular let's prove the property $[1]$.
Let $b,c\in A$, we have $(aD)(bc)=a\cdot D(bc)=a (D(b)c+bD(c))=a \cdot (D(b)\cdot c)+abD(c)=$ (associativity) $=(aD)(b) \cdot c + abD(c)$.
Do I nedd commutativity of the multiplication in $A$ to continue? I would like to say that $abD(c)=b (aD)(c)$ but without commutativity I can't "exchange" the position of $a$ and $b$ rigth?
 A: You are correct that $Der(A)$ will not be an $A$ module, for example take $A = k\langle a,b,c\rangle$ the free algebra and the derivation determined by $$D(a) = a, D(b) = b, D(c) = 1.$$ Then $abD(c) \neq baD(c)$ because $ab \neq ba$ in the free algebra.
In general the natural structure on $Der(A)$ is a $Z(A)$-Lie algebra and in particular a bimodule over $Z(A)$.

More generally you can consider derivations in a bimodule $M$, and ask when $Der(A,M)$ is an $A$ module. The same problem arises that $abD(c) \neq baD(c)$. One condition that suffices to make $Der(A,M)$ an $A$-module is that $M$ is a central bimodule. This means that the left and right actions are the same, i.e. for all $a \in A, m \in M$ then $a.m = m.a$. This is equivalent to saying that $[a,b].m = 0$ for all $a,b\in A$. The data of a central bimodule $M$ is the same as that of a module over the abelianization of $A$, the commutative ring $A_{ab} = A/F^1A$ where $F^1A = A[A,A]A$ is the two-sided ideal generated by commutators of elements of $A$.
(There is another notion of central bimodule which says instead only that $z.m = m.z$ for $z \in Z(A)$. If $M$ is this kind of central bimodule, then $Der(A,M)$ is a central $Z(A)$-module.)
