Approximation Reasoning I can't understand one step in the following problem. We start with a function $f(x)=x^\alpha$ on the interval $(0,1)$ where $\alpha>0$ is a constant. We pick two points $x_1<x_2$ from this interval so that we have $x_1^\alpha=x_2^\alpha-L$, where $L>0$ depends on $x_1$ and $x_2$. Upon rearranging, this may be expressed as $$x_1=x_2(1-\frac{L}{x_2^\alpha})^\frac{1}{\alpha}.$$
I have difficulties understanding the next claim that through a first order approximation, the above equation may be written as $$x_1\approx x_2-\frac{L\cdot x_2}{x_2^\alpha}\cdot\frac{1}{\alpha}.$$ 
My first thought was to use a Taylor series expansion but taking the first derivative doesn't get rid of $\frac{1}{\alpha}$ in the exponent. Or is there some entirely different trick involved?
In the end, we're interested in the order of the difference $x_2-x_1$, but if the approximation is valid, this difference is of the order $\frac{L}{x_2^{\alpha-1}}.$ 

let $$y = x^\alpha, x = y^{1/\alpha}.$$  you are given $$y_1 = x_1^\alpha, y_2 = x_2^\alpha, y_2 - y_1 = L $$  
we need to approximate $x_1 - x_2$ for small $L.$ we have 
$$\begin{align} x_1 - x_2 &= y_1^{1/\alpha} - y_2^{1/\alpha}\\ 
&=  (y_2-L)^{1/\alpha} - y_1^{1/\alpha} \\
&=-\frac 1\alpha y_2^{(1-\alpha)/\alpha}L + \cdots\\
&=  -\frac 1\alpha y_2^{1/\alpha}\frac L{y_2} + \cdots\\
&=  -\frac 1\alpha \frac {Lx_2}{x_2^\alpha} + \cdots\\\end{align} $$
