Building the derivative from its definition

Adam Zettel · September 22, 2026 · written with an agent

f^{'} (x) = \frac{4 D _{h /2} - D _{h}}{3} + O (h^{4})

The slope of a secant, then a secant centered on the point, then two of those combined: three estimates of the derivative of $x^{3}$ , each with an error the algebra predicts exactly, and the tests check the algebra. The last rung tries the same three on $e^{x}$ , where nothing is exact and the orders show up as ratios: halve $h$ and the three errors fall by factors of 2, 4 and 16.

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A ladder is one program built in steps called rungs. Each rung adds a definition and a test; the program at any rung is everything up to it, and every earlier rung's tests run again. The derivative is a good thing to build this way because each improvement is a few lines, and each has an error you can work out by hand before running anything.

The slope of a secant

The derivative of $f$ at $x$ is the limit, as $h \to 0$ , of the slope of the line through $(x, f (x))$ and $(x + h, f (x + h))$ . Before the limit that slope is an ordinary number, and for $f (x) = x^{3}$ the algebra gives it exactly: $\frac{( x + h ) ^{3} - x ^{3}}{h} = 3 x^{2} + 3 x h + h^{2}$ . The derivative is the $3 x^{2}$ ; the rest, $3 x h + h^{2}$ , is the error, and it shrinks like $h$ : halve $h$ and the error roughly halves.

forward takes the function as an argument, so each estimate on this ladder is written once and works for any function.

new: cube, forward

Rung 1 tests 2 of 2 pass

the forward difference of x^3 is 3x^2 + 3xh + h^2, exactly (passes)
halving h halves the error, to within 1% (passes)

A secant centered on the point

Take the secant from $x - h$ to $x + h$ instead. In $(x + h)^{3} - (x - h)^{3}$ every even power of $h$ cancels, and $\frac{( x + h ) ^{3} - ( x - h ) ^{3}}{2 h} = 3 x^{2} + h^{2}$ . The $3 x h$ term is gone. What is left is $h^{2}$ , so halving $h$ quarters the error. Rung 1's two tests still run here, unchanged, and still pass; a ladder requires every earlier test to keep passing. The new estimate gets a new name, centered, so forward and its tests stay as they were.

new: centered

Rung 2 tests 2 of 2 pass

the centered difference of x^3 is 3x^2 + h^2, exactly (passes)
halving h quarters the error (passes)

Combining two centered slopes to cancel the h^2 term

The centered error is $h^{2}$ at step $h$ and $h^{2} /4$ at step $h /2$ . So four times the finer estimate, minus the coarser one, over three, has no $h^{2}$ term at all:

$\frac{4 ( 3 x ^{2} + \frac{h ^{2}}{4} ) - ( 3 x ^{2} + h ^{2} )}{3} = 3 x^{2} .$

For a cubic that is the derivative exactly, to rounding. This is Richardson extrapolation, and it uses the same idea as the previous rung: combine two estimates so that their leading error terms cancel.

new: richardson

Rung 3 tests 1 of 1 pass

for x^3 the extrapolated slope is 3x^2 to rounding (passes)

On a function where nothing is exact

For $e^{x}$ the Taylor series never ends, so none of the three is exact and what remains is the order of each: halve $h$ and the forward error falls by about 2, the centered by about 4, the extrapolated by about 16. At $x = 1$ , $h = 0.1$ the three errors are $0.14$ , $0.0045$ and $5.7 \times 1 0^{- 7}$ . On a log-log plot they are lines of slope 1, 2 and 4, and the figure draws them from $h = 0.4$ down to $h = 0.0125$ .

Rung 4 tests 3 of 3 pass

halving h shrinks the three errors by factors of 2, 4 and 16, each to within 2% (passes)
the figure's steps run from h = 0.4 down to h = 0.0125, five halvings (passes)
the log-log slopes are 1, 2 and 4 between the two finest steps (passes)

Printed

12.61

12.01

f1: 0.14056

c1: 0.00453274

r1: -5.66477e-07

forward: slope 1centered: slope 2extrapolated: slope 4

1The slope of a secant

2A secant centered on the point

3Combining two centered slopes to cancel the h^2 term

4On a function where nothing is exact

The slope of a secant

A secant centered on the point

Combining two centered slopes to cancel the h^2 term

On a function where nothing is exact