After deriving this with forward differencing formula & Taylor series:
$$ D_1 f = f'(x_{i}) = \frac{f(x_{i+1}) - f(x_{i})}{h} - \frac{f''(x_{i})h}{2!} $$
which could perhaps also be written as:
$$\tag 1 f'(x) \approx \dfrac{f(x+h)-f(x)}{h}$$
I need to apply Richardson Extrapolation to it to find a formula with a smaller error value. From the tutorials I have seen, i understand that I need to use central differnecing formula but I am unable to understand how as the examples I have seen online are quite different.
The next step that I could think of was this:
$$\tag 2 f'(x) \approx 2\dfrac{f(x+h/2)-f(x)}{h}$$
but I don't know how to proceed.
Could explain step by step or give reference to a relevant example?
You have a formula for approximating $f'(x)$ with step-size $h$: $$ f'(x)=\frac{f(x+h)-f(x)}{h}+Kh+O(h^2)=[f](h)+Kh+O(h^2)$$ where $K$ is some constant (all we are saying is that the error is proportional to $h$ and has higher-order terms) and I've used $[f](h)$ to mean $(f(x+h)-f(x))/h$. Now we also have,
$$ f'(x)=\frac{f(x+h/2)-f(x)}{h/2}+\frac{Kh}{2}+O(h^2)=[f]\left(\frac{h}{2}\right)+\frac{Kh}{2}+O(h^2) $$ and so we can combine $[f](h)$ and $[f](h/2)$ to eliminate the $Kh$ terms: $$\begin{align*}-f(x)+2f'(x)&=-\left([f](h)+Kh+O(h^2)\right)+2\left([f](h/2)+\frac{Kh}{2}+O(h^2)\right) \\ f'(x)&=2[f]\left(\frac{h}{2}\right)-[f](h)+O(h^2) \end{align*} $$ so $$2[f]\left(\frac{h}{2}\right)-[f](h)=4\frac{f(x+h/2)-f(x)}{h}-\frac{f(x+h)-f(x)}{h}=\frac{-f(x+h)+4f(x+h/2)-3f(x)}{h}$$ is an $O(h^2)$ approximation to $f'(x)$.