Percentage change relationship Given $Z=X^{\alpha}$ how can one prove that the percentage change in $Z$ is simply $\alpha$ times the percentage change in $X$?
This was given as a 'simple mathematical rule' in an economics text book. So I tried to test it using simple numbers:
Let $X=4$ and $\alpha=0.5$
$$Z=4^{0.5}=2$$
Now let $X=9$, so the percentage change in $X$ is given as
$$\frac{9-4}{4}100\%=125\%$$
According to the rule, $Z$ should experience a $0.5*125\%=62.5\%$ change.
But
$$Z=9^{0.5}=3$$
so
$$\frac{3-2}{2}100\%=50\%$$
What am I doing wrong here? (this wasn't my attempt to prove the relationship, I was simply trying to test the rule)
Thanks
 A: I think the problem here is the rule concerns "compounded changes".  Say you have the two values $Z_0$ and $Z_1$ that are supposed to be related to $X_0$ and $X_1$.  The zero subscript means the original values, and the one corresponds to the next value.  Then
$$\frac{Z_1}{Z_0} = \frac{X_1^\alpha}{X_0^\alpha} = \left(\frac{X_1}{X_0}\right)^\alpha$$
Now if we take the $\ln$ of both sides we have
$$\ln\left(\frac{Z_1}{Z_0}\right) = \ln\left(\frac{X_1}{X_0}\right)^\alpha = \alpha \ln\left(\frac{X_1}{X_0}\right)$$
So it is the compounded changes that are related.  Try your example again with $\ln(9/4)$ and $\ln(3/2)$.
A: I think the rule is only true for small changes of x. Vor very small changes of x you can say that
$\huge{\varepsilon }=\frac{\frac{\partial Z}{Z}}{\frac{\partial X}{X}}=\frac{\partial Z}{\partial X} \cdot \frac{X}{Z}$
This is the definition of elasticity for differentiable functions. $\Delta X \to 0$
$\frac{\partial Z}{\partial X}=\alpha \cdot X^{\alpha -1}$
Thus ${\huge{\varepsilon}}=\alpha \underbrace{\cdot X^{\alpha -1}\cdot X}_{X^{\alpha}} \cdot \frac{1}{X^{\alpha}} \Rightarrow \boxed{{\huge{\varepsilon}}=\alpha}$
If you want to calculate it with numbers for the changes of X ($\Delta X $), the changes of X have to be  small, i.e. $\Delta X =0.0001$. 
$\huge{\varepsilon_v}=\frac{\frac{\Delta Z}{Z}}{\frac{\Delta X}{X}}$
The smaller $\Delta X $ is, the closer is $\huge{\varepsilon_v}$ to $\alpha$.

Calculation for $\alpha=0.5 , \ \Delta X =0.0001 $ and $ X=9$:
${\huge{\varepsilon_v}}=\frac{9.0001^{0.5}-9^{0.5}}{9^{0.5}} \cdot \frac{9}{0.0001}=0.49999...\approx 0.5$
