Estimating a Derivative From a Table

A table hands you finitely many points, so no limit is available and the best you can do is a secant slope over the tightest interval that brackets the point. Suppose the table gives $f(2) = 3$, $f(4) = 8$, $f(5) = 11$, $f(6) = 15$ and $f(8) = 20$. The rows either side of $x = 5$ give $f'(5) \approx \frac{15 - 8}{6 - 4} = 3.5$.

Straddling the point usually beats leaning to one side of it. The tightest one-sided estimates here are $\frac{11 - 8}{5 - 4} = 3$ and $\frac{15 - 11}{6 - 5} = 4$, and each describes an average rate over an interval lying entirely on one side of $x = 5$. The centred quotient is the length weighted average of those two, so it always lands between them. When the rows sit at equal distance $h$ either side of the point, as they do here with $h = 1$, that quotient takes the symmetric form.

Usually is the honest word. For a smooth $f$ with equally spaced rows the centred estimate carries error of order $h^2$ while a one-sided estimate carries error of order $h$, which is why the centred one is the standard choice; neither is exact, and a one-sided value can land closer. Take $f(x) = x^4$ near $a = 1$ with $h = 1$: the centred quotient gives $\frac{16 - 0}{2} = 8$ and the backward one gives $\frac{1 - 0}{1} = 1$, against the true $f'(1) = 4$.