Calculate and estimate derivatives from tables

Try

A table can contain function values, derivative values, or measurements used to estimate a derivative. Those are different kinds of evidence. In the first table below, the entries are exact supplied mathematical data for differentiable functions f and g. Their inputs and outputs are dimensionless. The table is not a sampled picture from which you must guess slopes.

Input x	f(x)	f′(x)	g(x)	g′(x)
1	2	3	3	−1
2	5	−2	1	2
3	1	4	5	1/2
5	3	1	2	−4

Assume both functions are defined and differentiable on open intervals containing all the inputs used. To find f(g(2)), trace the input through two lookups: g(2) = 1, then f(1) = 2. Mark those two table cells. The second lookup uses input 1, not input 2. Predict whether finding the derivative of this composition will also require changing rows.

Predict

The chain rule says that the rate of the outside function is evaluated at the inside output. Write the route 2 → g(2) → f(g(2)) before substituting any derivative entries. Predict the sign of (f∘g)′(2): a small positive input change makes g increase, and f increases near the resulting input 1. Their composition should increase locally.

Reverse the order and trace 2 → f(2) → g(f(2)). Now f decreases near input 2, and g decreases near the resulting input 5. Predict the sign of the product of these two local rates. A correct sign prediction does not establish that the magnitudes match, so retain both numerical factors.

Experiment

Complete this routing record manually using the supplied table. It connects the intermediate input to the rate lookup rather than treating the chain rule as a multiplication slogan.

Requested derivative	Intermediate value	Outside derivative at that value	Inside derivative	Result
(f∘g)′(2)	g(2) = 1	f′(1) = 3	g′(2) = 2	6
(g∘f)′(2)	f(2) = 5	g′(5) = −4	f′(2) = −2	8

Cover the results and redo the lookups from the first table. Then inspect the incorrect calculation f′(2)g′(2) = −4. Explain which row choice made it wrong for the first composition. Finally, determine whether the table supplies enough information for (f∘g)′(1). It does: g(1) = 3, so the required product is f′(3)g′(1) = 4(−1) = −4. A numerical answer can therefore accidentally agree with an incorrect calculation for a different question; the routing evidence matters.

Understand

For a differentiable composition, (f∘g)′(a) = f′(g(a))g′(a). The hypotheses concern differentiability of g near a and of f near g(a). A finite table of f and g values alone would not supply those hypotheses or the derivative numbers. Here they are explicitly given.

Inverse derivatives reverse a route as well. Use a separate function h, assumed continuously differentiable and one-to-one on an interval containing the listed inputs, with the indicated nonzero derivative at the point in question. The original f table is not being claimed invertible.

Input u	h(u)	h′(u)
1	3	2
2	5	3
4	9	4

To find (h⁻¹)′(5), first locate the input that produces output 5. It is u = 2. Differentiate h(h⁻¹(y)) = y locally: h′(h⁻¹(y))(h⁻¹)′(y) = 1. Consequently (h⁻¹)′(5) = 1/h′(2) = 1/3. Using 1/h′(5) would ask for the wrong input, and that entry is not supplied anyway. The nonzero-derivative condition permits division and the local inverse derivative; one-to-one behavior specifies the branch. If h converts seconds to meters, its derivative has units meters per second and the inverse derivative has seconds per meter.

Now change the evidence type. Suppose only these exact position values are available, with no derivative column:

Time t in seconds	1	2	4
Position s(t) in meters	3	7	19

The left secant rate at the middle time is (7 − 3)/(2 − 1) = 4 m/s. The right secant is (19 − 7)/(4 − 2) = 6 m/s. The secant across the full interval is 16/3 m/s. Because the time gaps differ, neither taking an unweighted average of the neighboring rates nor calling the full secant a centered difference is justified. These are exact average rates from the supplied values, not exact instantaneous derivatives.

Apply

An optional quadratic interpolation model makes the distinction concrete. Using numerical time in seconds and position in meters, q(t) = (2/3)t² + 2t + 1/3 passes through all three measurements. Its derivative at 2 is 14/3 m/s. That value is the derivative of the chosen interpolating quadratic, not a derivative determined uniquely by three observations.

Indeed q(t) + c(t − 1)(t − 2)(t − 4) passes through the same three points for every constant c, but its derivative at 2 is 14/3 − 2c in the corresponding units. Even smooth interpolations can therefore give different derivatives while agreeing with the entire table. With noisy measurements, add uncertainty in the values to this modeling uncertainty. Near a boundary, a one-sided estimate may be all that the data support; do not invent an unobserved row on the other side.

For a fresh practice table, record the intermediate input, derivative row, units and hypothesis for each composition or inverse. For measured values, report the chosen intervals and estimation assumption. If a required function value or derivative row is absent, say what information is missing rather than interpolate silently.

Continue

The chain-rule lesson explains the underlying composition mechanism. Retain its distinction between the input and the intermediate quantity. The next investigation uses measured rate values to estimate accumulated change; it needs equally explicit decisions about intervals, units and what happens between recorded times.