One-sided Derivative Question Let's say we define $$D_{+}f(x):=\lim_{h\to 0^+}\frac{f(x+2h)-f(x+h)}{h}$$ to be the "right-handed" derivative. This way the function does not have to exist (or equal what it 'should') at the point $x$ itself.
Is this definition ever used? Is there a name for this quantity? Thanks.
 A: Often in numerical computations, it is useful to write down approximations to the derivative as a linear combination of function values along a set of points. For example, we can construct one such approximation as follows
\begin{align}
  f(x_{0} + h) &= f(x_{0}) + h f^{\prime}(x_{0}) + \frac{h^{2}}{2} f^{\prime\prime}(x_{0}) + \frac{h^{3}}{6} f^{(3)}(x_{0}) + o(h^{4})\\
  f(x_{0} - h) &= f(x_{0}) - h f^{\prime}(x_{0}) + \frac{h^{2}}{2} f^{\prime\prime}(x_{0}) - \frac{h^{3}}{6} f^{(3)}(x_{0}) + o(h^{4})
\end{align}
notice that by subtracting the two expressions and some rearrangement we get
\begin{align}
f^{\prime}(x_{0}) = \frac{f(x_{0}+h)-f(x_{0}-h)}{2h} +o(h^{3}).
\end{align}
In that case, if we used this formula instead of the derivative at a point, we could get a good approximation to the derivative. 
Suppose we let $f(x) = e^{x}$ and $x_{0} = 0$. Then for $h\in \{0.2,0.1,0.05, 0.025\}$ Then our errors using this method respectively is $E(0.2) = 0.00668$, $E(0.1)  = 0.001667$, $E(0.05) = 0.000417$ and $E(0.025) = 0.0001042$.
In your case, you could construct the formula you have in the same manner:
\begin{align}
  f(x_{0} + 2h) &= f(x_{0}) + 2h f^{\prime}(x_{0}) + \frac{4h^{2}}{2} f^{\prime\prime}(x_{0}) + o(h^{3})\\
  f(x_{0} + h) &= f(x_{0}) + h f^{\prime}(x_{0}) + \frac{h^{2}}{2} f^{\prime\prime}(x_{0}) +o(h^{3})
\end{align}
Then subtracting these would give you
\begin{align}
\frac{f(x_{0} + 2h)-f(x_{0}+h)}{h} + o(h^{2}) = f^{\prime}(x_{0}).
\end{align}
Another method using a similar approach would be to consider the one-sided formula
    $$f^{\prime}(x_{0}) = C_{0}f(x_{0}) + C_{1}f(x_{0} + h) + C_{2}f(x_{0} + 2h)  + o(h^{2})$$
where in this case we are constructing an approximation that is $o(h^{2})$. Then, taking the Taylor series expansions:
    \begin{align*}
  f(x_{0} + h) &= f(x_{0}) + hf^{\prime}(x_{0}) + \frac{h^{2}}{2} f^{\prime\prime} (x_{0})  + o(h^{2})\\
  f(x_{0} + 2h) &= f(x_{0}) + 2hf^{\prime}(x_{0}) + 2h^{2}f^{\prime\prime}(x_{0}) + o(h^{2})
    \end{align*}
We can then substitute these expressions in and rearrange the equations to get
        $$f^{\prime}(x_{0}) = (C_{0} + C_{1} + C_{2}) f(x_{0}) + (C_{1} + 2C_{2} ) hf^{\prime}(x_{0}) + \left( \frac{1}{2} C_{1} + 2C_{2}) h^{2} f^{\prime\prime} (x_{0}\right) + o(h^{2})$$
Equating these coefficients we have that for $f^{\prime}(x_{0})$ we obtain the following system
    \begin{align*}
  \begin{cases}
  C_{0} + C_{1} + C_{2} &= 0\\
  h(C_{1} + 2C_{2})  &= 1\\
  h^{2}(\frac{1}{2} C_{1} + 2C_{2}) &= 0
  \end{cases} \implies f^{\prime}(x_{0}) = f^{\prime}(x_{0}) + o(h^{2})
 \end{align*}
Solving this system of equations, we get $C_{0} = -\frac{3}{2h}$, $C_{1} = \frac{2}{h}$, and $C_{2} = -\frac{1}{2h}$. This yields an approximation to the derivative as 
    $$f(x_{0}) \approx -\frac{3}{2h} f(x_{0}) + \frac{2}{h}f(x_{0}+h) -\frac{1}{2h}f(x_{0} + 2h).$$
Of course, there are other ways you can use this idea such as Richardson's extrapolation
