We know that if $f(x)$ is analytic, i.e $f(x) \in C^\infty$ in an open set $D$ and $f(x)$ has a convergent Taylor series at any point $x_{0}\in D$ for $x$ in some neighborhood of $x_{0}$, we can write
$\left|{\frac {d^{k}f}{dx^{k}}}(x)\right|\leq C^{k+1}k!$
Is there a counterpart for the derivative boundedness for infinite differentiable functions $g(x) \in C^\infty$?
$\left|{\frac {d^{k}g}{dx^{k}}}(x)\right|\leq h(k)?$
Given any sequence $h_n$, you can find a $C^\infty$ function $g$ such that $g^{(n)}(0) = h_n$. This implies the answer is "no."
Consider $$g(x) = \sum_{k=0}^\infty g_k(x), \qquad g_k(x) = \frac{h_k}{k!} x^k \alpha(c_k x),$$ where $\alpha:\mathbb R \to [0,1]$ is a $C^\infty$ function which is $1$ on $[-1,1]$ and $0$ on $\mathbb R \setminus (-2,2)$, and $c_k$ will be chosen later.
So let's try to show that $\sum_k g_k^{(n)}$ converges uniformly for each $n$. Using Leibnitz' formula, we see that for $k \ge n$ $$ {\|g_k^{(n)}\|}_\infty \le \frac{|h_k|}{k!} \sum_{m=0}^n \binom nm \frac{k!}{(k-m)!}\big[\sup_{c_k x \in [-2,2]}|x|^{k-m}\big] c_k^{n-m} {\|\alpha^{(n-m)}\|}_\infty \le K_n |h_k| 2^kc_k^{n-k} $$ where $K_n$ depends only on $n$ and $\alpha$. So if $c_k \ge 3 |h_k|^{1/k}$, then $\sum_k {\|g_k^{(n)}\|}_\infty$ converges.
Hence $g = \sum_k g_k$ converges in $C^\infty$. Then it can be easily shown that $g^{(n)}(0) = h_n$.