Given $N\in \mathbb Z_{\geq0}$, $0<C$, $x\in \mathbb R$, does there exist a compact support $f\in C^{\infty}_c(\mathbb R)$ such that $\partial^{N+1}f(x) > C\sum_{k=1}^{N}\sup|\partial^{k} f|$ ?
In general I think it might exist since the criterion doesn't seem that strict to me, can assume $x=0$ to simplify, I tried something like $f=\int\cdots\int g$ then $g = \partial^{N+1}f$ but I don't know how to construct $g$.
Let $N \in \mathbb N_0$, $C>0$, $x\in \mathbb R$ be given. W.l.o.g. we can assume that $x=0$.
Take $f \in C_c^\infty(\mathbb{R})$ such that $\partial^{N+1}f(0) \neq 0$ and set $f_\epsilon(x) = \epsilon^N f(x/\epsilon).$ Then $\partial^k f_\epsilon(x) = \epsilon^{N-k} \partial^k f(x/\epsilon).$ Now, $$ \sup |\partial^k f_\epsilon(x)| \to \begin{cases} 0, & (k<N) \\ \sup |\partial^N f|, & (k=N) \\ \infty, & (k>N) \\ \end{cases} $$
Therefore, $C \sum_{k=0}^{N} \sup |\partial^k f_\epsilon| \to C \sup |\partial^N f|$ as $\epsilon \to 0$, while $|\partial^{N+1} f_\epsilon(0)| \to \infty$. Thus, for $\epsilon$ small enough, $$ |\partial^{N+1} f_\epsilon(0)| > C \sum_{k=0}^{N} \sup |\partial^k f_\epsilon| . $$