How to derive the growth rate rules from the log growth equation? These are approximations, which, I guess, only apply for small growth rates $g$ ($g$ is a function that denotes the change in its argument).
$$\begin{align}
g(xy) &= g(x) + g(y) \\
g(x/y) &= g(x) − g(y) \\
g(x^\alpha) &= \alpha g(x)
\end{align}$$
And there are also these rules
$$\dfrac{\Delta(x*y)}{xy} = \dfrac{\Delta x}x + \dfrac{\Delta y}y$$
and 
$$\dfrac{\Delta(x^\alpha)}{x^\alpha} = z*\dfrac{\Delta x}x$$
I guess $g(x)$ equals $\dfrac{\Delta x}x$? So these rules are actually the same as the above ones.
However, I don't see how this can be derived from a log growth equation (which one?). I know a natural log function of a percentage change (e.g. $1.02$) approximates a percentage change that represents growth rates very close to zero (e.g. $0.02$), but ...
Thank you
babi
 A: Let $ \dot x $ be the time derivative of $x(t)$.  That is $$ \dot x  = \frac{d}{dt} x(t) $$
We define the growth rate of $x$ as $$ g = \frac{\dot x}{  x(t)} $$
To see this "roughly" in a discrete example, let $x(t_0) = 100 $ and suppose at the next instant $x(t_1) = 110$.  Then g = .1 or 10%.
Lets try to compute your formula using our definition:
\begin{align} g(xy) &= \frac{ \dot {(xy)} }{xy} \\ &= \frac{ \dot x y + x \dot y } {xy}\\ & = \frac{\dot x y}{xy} + \frac{x \dot y}{xy} \\ &= \frac{\dot x}{x}  + \frac{\dot y}{y} \\ &= g(x) + g(y) \end{align} Where the second inequality came from applying the product rule to $\dot {(xy)}  $
Also note that by definiton $$ g = \dot {log (x)} = \frac{\dot x}{x} $$  That is to say our growth rate is just the time derivative of the logged variable.
We can use this definition to do another of your questions
$$ g(x^{\alpha}) =  \log {\dot { (x^{\alpha})}} = \frac{\alpha x^{\alpha - 1}\dot x}{x^{\alpha}} = \frac{\alpha \dot x}{x} = \alpha g(x)$$
A: Not explicitly an answer to your question, but perhaps helpful.
I took a quick look at the link. I think it's trying to explain exponential growth without using calculus. The other questions you've asked suggest that you are learning calculus now. I think you'd be better off trying to understand this economics from scratch using calculus. If $f$ is growing at a constant rate (thinking of the independent variable as time $t$) the
$$
f(t) = f(0)e^{rt};
$$
the growth rate is $r$. Its units are $1/\text{time}$. 
Suppose population is growing at $2\%$ per year you can use $r=0.02$ in that formula if that rate is interpreted as "continuous compounding".  If the $2\%$ is the increase in one year then in the formula you'd need
$$
r = \ln(1.02) = 0.0198
$$
which is (as you understand) not much less than $0.02$.
Then you could write the function as
$$
f(t) = f(0)e^{0.0198t}  
$$
or
$$
f(t) =   f(0)(1.02)^t .
$$
The first is classier mathematics, the second captures the idea of $2\%$ annual growth" somewhat better.
