Let $\mathcal{O}_{\mathbb{R},0}$ be the ring of germs of $C^{\infty}$ funcitons on the real line. A derivation of $\mathcal{O}_{\mathbb{R},0}$ is a $\mathbb{R}$-linear map $\partial:\mathcal{O}_{\mathbb{R},0}\to\mathcal{O}_{\mathbb{R},0}$ that satisfies the Leibniz rule $\partial fg=f\partial g+g\partial f$ for all $f,g\in\mathcal{O}_{\mathbb{R},0}.$ The set $Der_{\mathbb{R}}(\mathcal{O}_{\mathbb{R},0},\mathcal{O}_{\mathbb{R},0})$ of all derivations is a (left) $\mathcal{O}_{\mathbb{R},0}$-module in the obvious way.
Question: Let $x$ be a coordinate on the real line and let $\frac{d}{dx}$ be a derivation $f\mapsto\frac{df}{dx}.$ Is it true that $Der_{\mathbb{R}}(\mathcal{O}_{\mathbb{R},0},\mathcal{O}_{\mathbb{R},0})=\mathcal{O}_{\mathbb{R},0}\frac{d}{dx}$ as an $\mathcal{O}_{\mathbb{R},0}$-module?
Motivation: If $\mathcal{O_{\mathbb{C},0}}$ is the ring of germs of complex analytic functions and $z$ is a complex coordinate then $Der_{\mathbb{C}}(\mathcal{O}_{\mathbb{C},0},\mathcal{O}_{\mathbb{C},0})=\mathcal{O}_{\mathbb{C},0}\frac{d}{dz}.$ Unfortunately the proof of this fact is based on the observation that if $\mathfrak{m}\subset\mathcal{O}_{\mathbb{C},0}$ is the unique maximal ideal then $\bigcap_{n\geq0}\mathfrak{m}^{n}=(0).$ This is not true in case of $\mathcal{O}_{\mathbb{R},0}$.
EDIT: Following the comments I changed the notation from $\mathcal{O}$ to $\mathcal{O}_{\mathbb{R},0}.$ By "the ring of germs of $C^{\infty}$ funcitons on the real line" I meant the ring of germs at $x=0$ of $C^{\infty}$ functions defined on open neighbourhoods of the origin of the real line.
