Derivatives of a function with compact suport have compact support

Question

Let $u \in C_0^k(\mathbb{R}^n)$ be a function with compact support than all of its derivatives $D^{\alpha}u$ have compact support as well. Where $$D^{\alpha}= \frac{d^k}{dx_1^{\alpha_1}\cdots dx_k^{\alpha_k}} \qquad \alpha \in \mathbb{N}.$$ I think I am confusing myself a little here. We have defined the support of a function as $$\text{supp}(u)=\overline {\{ x \in \mathbb{R^n} : u(x) \neq 0 \}} \; .$$ What I have got : If we consider $\frac{d}{dx_i}u$. Since $u=0$ on $A=\mathbb{R^n} \setminus \text{supp}(u)$, we have $\frac{d}{dx_i}u =0$ on $A$ as well because $A$ is an open set and hence $\text{supp}(\frac{d}{dx_i}u) \subset \text{supp}(u)$. By induction the result should hold for $D^{\alpha}u$ as well.

Would appreciate it if anyone could tell me if I got an error in my attempt