Let $\delta_{j} \in D'(\mathbb{R})$ denote the dirac delta distribution at $j \in \mathbb{N}$ (which is defined as $\delta_j(\phi)=\phi(j) $ for any $\phi \in C^{\infty}_c$). Then, the order of the distribution $T_j=\frac{d^j \delta_j}{dx^j}$ is $j$.
I showed that the order is $\leq j$. I don't know how to prove the $\geq$ part. Can you provide a proof for that $\geq$ inequality?
Definition: Let $\Lambda \in D'(\Omega): \exists m \in \mathbb{N} \ \ \forall K \subset \Omega$ compact $\exists C>0: \forall \phi \in D_K=\{\phi \in C^{\infty}(\Omega):supp(\phi) \subset K \}$ it holds $|\Lambda(\phi)| \leq C p_m^K(\phi)$ where $p_m^K(\phi)=sup \{|D^{\alpha}\phi(x)|:x \in K, |\alpha| \leq m \}$ are the seminorms in $D_K$. The order of the distribution $\Lambda$ is the infimum over all such m.