for a function $f(x)=e^{2x}-\cos(2x)$,
at grid points $x \in {-0.3,-0.2,-0.1,0}$
I perform a central difference for the derivative at $x=-0.2$
$$\frac{df}{dx}=\frac{f(-0.1)-f(-0.3)}{2*(-0.1--0.3)}=0.28795$$ The derivative of this function (per mathematica) is $0.561803$ So the error is $$Abs[0.561803-0.28795]=0.273853$$
I have a formula for the upper bound of the error, and that formula is $$\frac{M^*\Delta x^2}{3!}=\frac{8*(-0.1--0.3)}{6}=0.0533333\\\text{Where }M^*=\text{Max}(f''(x)|x\in[-0.3,0])=f(0)=8$$
The issue is the error, $0.27$, is greater than the supposid error bound, $0.05$. When I graph this function with its central difference though, it looks correct.
Any idea why my derivatives are higher than the upper bound? The error does seem pretty high.
The problem lies in your definition of $\Delta x$, remember that
$$ f'_{FD}(x) \approx \frac{f(x + \Delta x) - f(x - \Delta x)}{2\Delta x} $$
Taking $x = -0.2$ and $\Delta x = 0.1$ we get
$$ f'_{FD}(-0.2) \approx \frac{f(-0.1) - f(-0.3)}{2 \cdot 0.1} = 0.575941 \tag{1} $$
So the actual error is
$$ \epsilon = |f'_{FD}(-0.2) - f'(-0.2)|= 0.0141374 \tag{2} $$
Now, the prediction for the error is
$$ \epsilon_{\max} = \frac{|f^{(3)}(c)|\Delta^2 x}{6} \stackrel{c=-0.3}{=} 0.0148461 \tag{3} $$
You can indeed see that
$$ \epsilon < \epsilon_\max $$