high order approximation of the first derivative These are the numerical approximations of the first derivative
$$\begin{gathered}
  f'\left( x \right) \simeq \frac{{f\left( {x + h} \right) - f\left( x \right)}}{h}, \hfill \\
  f'\left( x \right) \simeq \frac{{f\left( {x + h} \right) - f\left( {x - h} \right)}}{{2h}}, \hfill \\
  \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \vdots  \hfill \\ 
\end{gathered} $$
Is there a numerical approximations of the first derivative that has $h^2$ in denominator?
 A: Let $F$ be an antiderivative of $f$ such that $\frac{d}{dx} F = f$ and assume that $f\in C^2$. Then
We approximate $F$ around $x$ using its third order Taylor approximation with Peano remainder,
$$
\begin{align}
F(x + h) &= F(x) + h f(x) + \frac{h^2}{2}f'(x) + \frac{h^3}{6}f''(\xi_+) \tag{I}\label{i} \\
F(x - h) &= F(x) - h f(x) + \frac{h^2}{2}f'(x) - \frac{h^3}{6}f''(\xi_-) \tag{II}\label{ii}
\end{align}
$$
where $\xi_+ \in [x, x+h]$ and $\xi_- \in [x-h, x]$. The first order terms cancel if we subtract \eqref{ii} from \eqref{i},
$$
F(x+h) + F(x-h) = 2F(x) + h^2 f'(x) + \mathcal{O}(h^3)
$$
which we can flip around to arrive at 
$$
f'(x) = \frac{F(x+h) + F(x-h) - 2F(x)}{h^2} + \mathcal{O}(h)
$$
Notice that this is equivalent to computing the second order central difference approxmation of $F$.
A: If you mean that you have several evaluation points (f(x+2 h) or f(x+h), ...) in the numerator and then have $h^2$ in the denominator, the answer is no.
You see this by considering physical units. If f is a length measured in meters and x is a time measured in seconds then $f^\prime$ is a velocity. The units of the finite difference should agree, so it is "evaluation points of f" divided by h.
Maybe you are looking for a finite difference approximation that has higher orders in the error term. One side finite difference is $O(h)$. Central finite difference is $O(h^2)$. It is possible to derive a finite difference of $O(h^3)$. All have linear combinations of $f$ evaluated at different points in the numerator and an $h$ in the denominator.
