So I know that for a general function $g$ that depends on the variables $a_1,a_2...$, that each have relative uncertainties $\epsilon_1,\epsilon_2...$, the propagation error is given by $$\xi_p=\sum_{i=j}^n\left|\frac{a_j}{g}\frac{\partial g}{\partial a_j}\right|\epsilon_j$$
My prolem is to show that for a central difference approximation of $f'(x)$, i.e $$f'_n(x)=\frac{f(x+\Delta x)-f(x-\Delta x)}{2\Delta x}$$ the propagation error becomes $$\xi_p=\frac{|f(x)|\epsilon}{|f'(x)|\Delta x}$$ where $\epsilon$ is the machine accuracy.
I'm really stuck here. What I've done myself is basically substituting $g$ with the central difference expression for $f'(x)$ in the formula for the propagation error (i.e letting $g=f'_n$), letting $a_1=f(x+\Delta x)$ and $a_2=-f(x-\Delta x)$ and $\frac{\partial g}{\partial a_1}=\frac{1}{2\Delta x}$, $\frac{\partial g}{\partial a_2}=-\frac{1}{2\Delta x}$. This yields the following expression $$\xi_p=\left|\frac{f(x+\Delta x)}{f(x+\Delta x)-f(x-\Delta x)}\right|\epsilon_1+\left|\frac{f(x-\Delta x)}{f(x+\Delta x)-f(x-\Delta x)}\right|\epsilon_2$$ It looks promising if I do a first order Taylor expansion from here, but the absolute values become problematic, as well as how $\epsilon_1$ and $\epsilon_2$ become a single $\epsilon$. I would be extremely thankful for any guidance.