function bounded by an exponential has a bounded derivative? here's the question. I want to be sure of that. Let $v:[0,\infty) \rightarrow \mathbb{R}_+$ a positive function satisfying
$$\forall t \ge 0,\qquad v(t)\le kv(0) e^{-c t}$$
for some positive constants $c$ and $k$. Can I conclude that
$$\dot{v}(t) \le -c v(t)$$ ?

The answer is no.
Take $v(t) = e^{-t} (1+\sin(e^x) )$
$v'(t) = \cos(e^t)-e^{-t} (1+\sin(e^t))$
and v'(t) doesn't converge to 0
