Theoretical justification of exponential growth differential equation Consider a number of $x_0$ reproducing individuals and ignore death and limiting environmental factors. I've heard that the growth of such a population (of bacteria, insects, humans, etc...) can be modeled by the differential equation
$$\dot{x} = \lambda x , \quad \quad \quad x(0)=x_0\in \mathbb{R}^{+}.$$
This seems to be taken for absolutely granted in nearly every book and every lecture I've seen. But I still don't understand what assumptions I have to make to the reproduction rate of a single individual within the population to RIGOROUSLY derive this equation.
In a book ("Evolutionary Dynamics" by Martin Nowak) I read that the basic assumption behind this differential equation is a stochastic process, namely that the time until one individual $A$ produces another individual $A'$ is exponentially distributed around an average time $t_0>0$. This would mean, that the probability of $A$ producing the offspring $A'$ within the time interval $[0,t]$ is given by $1-\exp(-t/t_0)$.
Then why is the differential equation fully deterministic? Is $x(t)$ the probabilistic expectation value of individuals after time $t$?
Is there somebody who can strictly derive the differential equation by just using the abolve stochastic process? Or is there another way to theoretically justify this equation? And why does it also work for human populations, where there is always TWO individuals producing offspring?
Thanks!
 A: The stochastic process model that you seem to have in mind can be rigorously formulated as follows. When you have a population of size $N$, each member of the population is independently waiting to make another member of the population. If you assume that the process is Markov, then these waiting times must be exponentially distributed. (Note that this Markov hypothesis is an oversimplification, for various reasons.) We assume they all have the same mean $\frac{1}{\lambda}$. (This is also an oversimplification.) Then the time to wait for the next birth is the minimum of $N$ independent exponential variables with mean $\frac{1}{\lambda}$. This is itself exponential with mean $\frac{1}{N \lambda}$. 
Now if the population is currently $N$, the average time to the next birth is $\frac{1}{N \lambda}$, which means that the average growth rate of the population is currently $\lambda N$, as in the exponential growth ODE.
Note that the model goes through the same with slightly different hypotheses. For instance, maybe only some fraction $0<p<1$ of the population is actively trying to reproduce. In this case the model goes through exactly the same except that $\lambda$ is replaced by $\lambda'=p \lambda$.
A: To make it simpler, think of it first in case of discrete time. You have a population of $x_k$, and on the next step each from these population produce on average $r$ kids, so if you don't allow deaths
$$
  x_{k+1} = x_k + rx_k = (1+r)x_k = (1+r)^kx_0
$$
which is obviously a growth function, exponential w.r.t time variable $k$. If you say that a production rate is $\mu\cdot \Delta t$ and a death rate of $\nu\cdot\Delta t$, saying that a $\mu$ fraction of population is added in a negligibly small time period $\Delta t$, and a $\nu$ fraction dies, you get
$$
  x_{t + \Delta t} = x_t + (\mu - \nu))x_t\Delta t
$$
so 
$$
 \dot x_t \approx \frac{x_{t + \Delta t} - x_t}{\Delta t} = (\mu - \nu)x_t.
$$
