Neglecting higher order terms in expansion Suppose we have a function $v$ of $x$ with a minimum at $x=0$. We have, for $x$ close to zero, $$v'(x) = v'(0) +xv''(0) +\frac{x^2}{2}v'''(0)+\cdots$$ Then as $v'(0)=0$ $$v'(x)\approx xv''(0)$$ if $$|xv'''(0)|\ll v''(0)$$
Which is fine. I am unable to understand this statement: 

Typically each extra derivative will bring with it a factor of $1/L $
  where $L$ is the distance over which the function changes by a large
  fraction. So $$x\ll L$$

This is extracted from a physics derivation, and I cannot get how they tacked on a factor of $1/L$

If each derivative contributes $\frac{1}{L}$,  then $|xv'''| << v'' \implies  x(\frac{1}{L})^3 << (\frac{1}{L})^2$.  Divide both sides by $(\frac{1}{L})^3$ and this becomes  $x << L$.  
That $\frac{1}{L}$ term is refering to the change in the function according to the difference method of derivatives (Definition via difference quotients) given in Wikipedia. If you calculate out the quotient between the second and third derivatives (or first and second), it should approximate to the result above given the context.
