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
Best Answer
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
$$ g(x^{\alpha}) = \log {\dot { (x^{\alpha})}} = \frac{\alpha x^{\alpha - 1}\dot x}{x^{\alpha}} = \frac{\alpha \dot x}{x} = \alpha g(x)$$