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

No. I will show a non-negative function which does the job (it's easy enough to turn it into a positive one). Take $v(x)=e^{-x}\cos^2(e^x)$. Clearly we have $v(x)\leq v(0)e^{-x}$. The derivative, which is $-e^x\sin(2e^x)$, is not bounded.
