Derivatives of functions satisfying Euler-ish inequality $f(x+y)\le f(x)f(y)$. 
Conjecture. Let $f:\mathbb R\to\mathbb R_{>0}$ be a continuously differentiable function such that $$f(x+y)\le f(x)\cdot f(y)$$ for all reals $x,y$. Then $$f'(x)\le c_1\cdot\exp(c_2\cdot x)$$ for some constants $c_1,c_2$; and all $x\in\mathbb R$.

Is this conjecture true? How can we (dis-)prove it?

Remarks


*

*My conjecture originates from the fact that all continuous functions satisfying $g(x+y)=g(x)\cdot g(y)$ are of the type $g(x)=\exp(c\cdot x)$. In particular, $g'(x)=c\exp(cx).$

*My first instinct was to use  $$f(x+y)-f(x)\le f(x)\cdot(f(y)-1)$$ 
so if we could conclude that $f(x)\le \exp(cx)$ for all $x$, then
$$f'(x)=\lim_{y\to0}\frac{f(x+y)-f(x)}{y}\le \exp(cx)\lim_{y\to0}\frac{\exp(cy)-1}y=\exp(cx)\cdot c.$$
However, $f(x)\le \exp(cx)$ is not always true (consider $f(x)=2\cdot\exp x$).
 A: This is unfortunately not true.
The issue is that your condition essentially only restricts the behaviour
of $f$ "in the large", but not locally.
The derivative of $f$, however, is determined by the local behaviour.
What exactly I mean by this will be made clear by the counterexample.
Define
$$
  f : R \to (0,\infty), x \mapsto e^x \cdot (4 + \sin(e^{x^{2}})) .
$$
Since $-1 \leq \sin(y) \leq 1$, it is then easy to see
$3 e^x \leq f(x) \leq 5 \cdot e^x$ for all $x \in \Bbb{R}$,
and hence
$$
  f(x+y)
  \leq 5 \cdot e^{x+y}
  \leq 3 e^x \cdot 3 e^y
  \leq f(x) \cdot f(y).
$$
However, we have
$$
  f'(x) = e^x \cdot (4 + \sin(e^{x^{2}}))
          + e^x \cdot \cos(e^{x^{2}}) \cdot e^{x^{2}} \cdot 2x.
$$
Now, for $n \geq 5$, let $x_n := \sqrt{\ln (2 \pi n)}$.
Observe $\ln(2 \pi n) \geq \ln(e) = 1$, and thus $e^{x_n} \geq e \geq 1$.
Furthermore, $e^{x_n^2} = e^{\ln(2 \pi n)} = 2\pi n$,
and thus $\cos(e^{x_n^2}) = \cos(2\pi n) = 1$.
Overall, this implies
$$
  f'(x_n)
  \geq 3 \cdot e^{x_n}
       + e^{x_n} \cdot \cos(e^{x_n^2}) \cdot e^{x_n^2} \cdot 2 x_n
  \geq 2 \pi n \cdot 2 x_n
  \geq 4 \pi n.
$$
However, if your desired inequality was true, we would have
$$
  2 e^{x_n^2} = 4 \pi n \leq f'(x_n) \leq C \cdot e^{c x_n},
$$
and thus
$$
  e^{x_n^2 - c x_n} \leq C / 2,
$$
meaning that $x_n^2 - c x_n = x_n \cdot (x_n - c)$ is bounded,
which is clearly not the case.
