Let

* $d\in\mathbb N$
* $\Omega\subseteq\mathbb R^d$
* $\mathcal D(\Omega):=C\_c^\infty(\Omega)$

Let $$\frac{\partial p}{\partial x\_i}(\phi):=-p\left(\frac{\partial \phi}{\partial
 x\_i}\right)\;\;\;\text{for }\phi\in\mathcal D(\Omega)$$ for $p\in\mathcal D'(\Omega)$. Is the mapping $$\frac\partial{\partial x\_i}:\mathcal D'(\Omega)\to\mathcal D'(\Omega)\tag 1$$ injective?

At a first glance, I wouldn't think so, cause $\frac\partial{\partial x\_i}:C\_c^1(\Omega)\to C\_c^0(\Omega)$ is not injective, but on the other hand each $\phi\in\mathcal D$ can be written as the $i$-th partial derivative of some $\psi\in\mathcal D(\Omega)$ ...