Definition. Let $R$ be a commutative algebra over $\mathbb{C}$. A linear operator $D\colon R\to R$ is an operator of order zero if, for every $r\in R$, the operator $$[D,r]=Dr-rD\colon R\to R$$ is equal to the zero operator.
Definition. A linear operator $D\colon R\to R$ is an operator of order $k$ or less, where $k\ge 1$ is an integer, if for every $r\in R$ the operator $[D,r]$ is an operator of order $k-1$ or less. The set of all operators on $R$ of order $k$ or less is denoted by $\mathcal{D}_{k}(R)$.
The alebra of differential operators on $R$ is defined as $$\mathcal{D}(R):=\bigcup_{k\ge 0} \mathcal{D}_k(R).$$
Theorem. Prove that $\mathcal{D}_0(\mathbb{C}[x])=\mathbb{C}[x].$
What I have done so far:
Let $f\in R=\mathbb{C}[x]$. We can think of $f$ as an operator on $R$ as follows: for any $g\in R$, we set $$f(g):=fg.$$ Then $f$ is an operator of order zero because $[f, g]=fg-gf=0$ for all $g\in R$. This means that $f\in \mathcal{D}_0(R)$, so that $R\subseteq \mathcal{D}_0(R)$.
I am having trouble proving the other inclusion and don't have an idea for how to go about doing so.