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
$$\dot{v}(t) \le -c v(t)$$ ?

I don't think you can. Look at something like $v(t) = e^{-t}\sin(g(t) t)$. Then $v'(t) = e^{-t}(\sin(g(t)t) + g(t)\sin(g(t)t)$. If you choose $g(t)$ so that it grows fast enough, your original function will be bounded from above by exponential decay, but you can make the derivative grow as fast as you like.
Edit: This reply doesn't look at the positivity condition, but this should be fixable by replacing $\sin(g(t) t)$ with something like $2 + sin(g(t) t)$
