Define a distribution $u$ on $(0,\infty)$ by $u(\phi) = \partial^n\phi(1)$, what's the order of this distribtion?
It's supposed to be $n$ since $|u(\phi)|=|\partial^n\phi(1)|\leq \sup |\partial^n\phi|$ obviously and others doesn't seem fit, but how to prove it?
(Definition of the order of $u$: If given any compact set $K\subset (0,\infty)$, there exists a real number $C_K\geq 0$, such that for all $\phi\in C^{\infty}_c(0,\infty)$ with support in $K$, we have $|u(\phi)|\leq C_K\sum_{k=1}^{N} \sup(|\partial^k\phi|)$. The smallest $N$ possible is defined to be the order of $u$.)
For simplicity I will instead consider $u$ with $u(\phi) = \partial^n\phi(0).$
Take some $\phi \in C_c^\infty$ such that $\partial^n\phi(0) \neq 0$. Then consider $\phi_\epsilon(x) = \epsilon^n \phi(x/\epsilon)$. This satisfies $\partial^k \phi_\epsilon(x) = \epsilon^{n-k} \partial^k\phi(x/\epsilon)$.
Assume that $u$ is of lower order than $n$, i.e. there is $C>0$ such that $|u(\phi)| \leq \sum_{k=0}^{n-1} \sup |\partial^k \phi|$. We then have $$ 0 < |\partial^n\phi(0)| = |u(\phi)| \leq C \sum_{k=0}^{n-1} \sup |\partial^k \phi_\epsilon| = \epsilon^{n-k} \, C \sum_{k=0}^{n-1} \sup |\partial^k \phi| . $$ But by choosing $\epsilon$ small enough, the last expression can be made smaller than $|\partial^n\phi(0)|$ which results in a contradiction. Thus, $u$ has order $n$.