Is a $\mathscr{C}^{\infty}$ function with negative derivatives decreasing?

If I have some $\mathscr{C}^{\infty}$ real function such that all its derivatives are negative at a point $a$, can I conclude that this function is decreasing for all $x\geq a$? I feel like analyticity is required for this to work but I equally cannot find a counterexample.

No. One of the exercises in Rudin Real and Complex Analysis shows that given any sequence $(a_n)$ there exists $f$ infinitely differentiable with compact support such that $$f^{(n)}(0)=a_n$$ for every $n$. In particular, knowing that $f^{(n)}(0)>0$ for all $n$ says nothing about $f(x)$ for $x>0$.

Edit: About that exercise:

Lemma. Suppose $k$ is a positive integer and $a\in\Bbb C$. There exists a sequence $(\phi_n)\subset C^\infty_c$ such that $D^j\phi_n(0)=0$ for $j<k$, $D^k\phi_n(0)+a$ and $\lim_nD^j\phi_n=0$ uniformly on compact sets for $j<k$.

Proof: Choose $\psi\in C^\infty_c$ such that $D^j\psi(0)=0$ for $j<k$ and $a$ for $j=k$. (For example let $\psi=ph$ where $p$ is a polynomial and $h\in C^\infty_c$ satisfies $h=1$ on $(-1,1)$.) Let $$\phi_n(x)=n^k\psi(x/n).$$Taylor's Theorem shows that $|\psi(x)|\le c|x|^k$, giving the lemma.

Now say $(a_n)$ is an arbitrary sequence of complex numbers. Using the lemma, recursively choose $f_n\in C^\infty_c$ such that $$D^jf_{n+1}(0)= \begin{cases}0,&(0\le j<n+1),\\ a_{n+1}-\sum_{l=1}^n D^l f_l(0),&(j=n+1)\end{cases}$$and $$|D_jf_k(x)|\le 2^{-k}\quad(j<k, |x|\le k).$$

Now $\sum_nD^kf_n$ converges uniformly on compact sets, so $f=\sum f_n$ is infinitely differentiable and we can switch differentiation and summation.