Chain rule for change of variables to get $\frac{d}{dx} = \frac{d\overline{x}}{dx}\frac{d}{d\overline{x}}$ 
A change of variable from $x$ to $\overline{x} = \frac{x}{\epsilon^\alpha}$ is done, but I do not understand how they obtain $$\frac{d}{dx} = \frac{d\overline{x}}{dx}\frac{d}{d\overline{x}} = \frac{1}{\epsilon^{\alpha}}\frac{d}{d\overline{x}}$$
How is chain-rule applied to the differential operator $\frac{d}{dx}$? I am familiar with chain rule saying basically: $f'(g(x)) = f'(g(x))g'(x)$
 A: To determine what a function does, supply it an input.  Two functions having the same domain are equal if they give the same output for the same input.
To determine what an operator does, supply it a function.  Two operators (having the same domain) are equal if they give the same output for the same input.
The operator $\mathrm{d}/\mathrm{d}x$ is equal to the operator $1/\epsilon^\alpha \,\mathrm{d}/\mathrm{d}\overline{x}$ if they agree for any function supplied. So, if, for all $f$ in the intersection of the domains of the operators $\mathrm{d}/\mathrm{d}x$ and $1/\epsilon^\alpha \,\mathrm{d}/\mathrm{d}\overline{x}$, we have
$$  \frac{\mathrm{d}f}{\mathrm{d}x} = \frac{1}{\epsilon^\alpha} \frac{\mathrm{d}f}{\mathrm{d}\overline{x}}  \text{,}  $$
the operators are equal.
Additionally, you should become familiar with the chain rule in Leibniz's notation.  This form is presented in the introductory paragraph here: https://en.wikipedia.org/wiki/Chain_rule in the discussion of the equation
$$  \frac{\mathrm{d}z}{\mathrm{d}x} = \frac{\mathrm{d}z}{\mathrm{d}y} \cdot \frac{\mathrm{d}y}{\mathrm{d}x}  $$
and exactly where these derivatives are evaluated.
