Answers · The mathematics underneath
Why do you multiply derivatives when functions are composed?
In shortBecause a small change at the input is scaled once by the inner function and again by the outer function, so the two scale factors multiply.
Because a small change at the input is scaled twice before it reaches the output: once by the inner function, then by the outer function. The overall rate is the product of those two scale factors. Take a nudge h at x. The inner function turns it into a middle change Δy, and the outer function turns that middle change into an output change Δz. For small h, Δy is about g'(x)h, and Δz is about f'(y)Δy, where y=g(x). Substituting the first into the second gives Δz≈f'(y)g'(x)h. Divide by h and let h shrink: the ratio tends to f'(y)g'(x). That is the multiplication.
One nudge, two scalings
Use g(x)=3x+1 and f(y)=y^2, so z=(3x+1)^2. At x=2 the middle is y=7 and the output is z=49. Push x by h=0.1. The middle becomes 7.3, so Δy=0.3. The output becomes 53.29, so Δz=4.29. The middle-to-output ratio is 4.29/0.3=14.3. The input-to-middle ratio is 0.3/0.1=3. Their product is 42.9, which is also Δz/h. Halve the nudge and the ratios close in.
| h | Δy | Δz | Δz/Δy | Δz/Δx |
|---|---|---|---|---|
| 0.1 | 0.3 | 4.29 | 14.3 | 42.9 |
| 0.01 | 0.03 | 0.4209 | 14.03 | 42.09 |
| 0.001 | 0.003 | 0.042009 | 14.003 | 42.009 |
The table is not a proof, but it shows the two ratios are stable and their product is stable. The outer ratio is converging to the outer derivative at the middle value. The inner ratio is converging to the inner derivative at the input. The product converges to the product of those limits.
Read the last row: the outer ratio is heading for 14, which is f'(7)=2×7, and the inner ratio is exactly 3, which is g'(2). Their product is 42. In symbols, the rate is (2y)(3)=6(3x+1), and at x=2 that is 6×7=42. The factor 14 is not the rate of squaring at 2. The rate of squaring at 2 is 4. The outer rule is being applied to 7.
A rate is a local scale factor. If a length is doubled and then tripled, the final length is six times the original, not five times. Nesting does the same thing to changes: the inner rate scales the nudge, and the outer rate scales the already scaled nudge. Addition would describe two separate causes acting on the same change. Here there is one cause passing through two stages, so the stages compose by multiplication. This is also why the chain rule is written as if the middle change cancelled: Δz/Δx=(Δz/Δy)(Δy/Δx), with the same Δy appearing in both places. The resemblance to cancelling fractions is a memory aid, not the proof.
The outer rate is read at the middle
The natural guess is to differentiate both functions at x and then multiply or add. That fails because the outer function is not evaluated at x. In the example, squaring at 2 has rate 4, but the squaring is applied to 7, where its rate is 14. Using 4×3=12 gives the wrong answer. Using 4+3=7 is even further away. The outer factor belongs to the middle value, and the inner factor belongs to the input. Once both are read in the right place, their product is the only combination that matches the measured ratios.
Multiplying two functions is a different operation. If u and v both depend on x, the product uv changes because u changes while v is held at its current size, and because v changes while u is held. Those two contributions add: (uv)'=u'v+uv'. Both rates are read at the same x because neither function is feeding the other. Composition has one path through both functions, so the scale factors multiply. A product has two separate paths, so the contributions add. Asking whether one function feeds the other or sits beside it settles which rule applies.
Where the cancellation stops
The symbols dy and dx are not ordinary numbers, and the chain rule is not a licence to cancel them as fractions. They are limits of fractions. The step that inserts Δy requires Δy to be nonzero for the small nudges under consideration. If the inner function is constant near the point, so that the middle change is zero, that step has to be replaced by a separate argument. If the outer function is not differentiable at the middle value, the product may not describe the rate at all. Under the standard hypotheses, g differentiable at x and f differentiable at g(x), the product formula holds; the delicate part is exactly the vanishing middle change.
Worth keeping
- A composed function scales a small input change by the inner rate and then by the outer rate, so the overall rate is the product.
- The outer rate must be evaluated at the middle value, not at the original input.
- When two functions are multiplied rather than composed, their rates add as u'v + uv' because each factor changes independently.
Where this goes next
This question is one step of Change and Accumulation, a course that works through its subject from the beginning. The page that teaches this part of it is One Rule Inside Another, and it reads the whole way through whether or not you have an account.
What an account changes is the practice: problems standing under every idea, marked with a reason rather than a tick, in a copy of the course that is yours.
Read that pageThe whole courseOther questions people ask about this
- How do you find the parabola that best fits a function near a point?Match the function's height, slope and bending at the chosen point; those three conditions single out exactly one parabola.
- How can there be a slope at a single point if slope needs two points?A slope at one point is the single value that slopes between nearby pairs of points settle on as the pair closes up; it is not a slope measured between two identical points.
- How can you add up infinitely many infinitely thin slices and get a finite number?You add small positive pieces, then take the finite total they approach as the pieces shrink; the pieces never all become zero before the adding.
- Why is a function thought of as a rule rather than just a formula?A function is the rule that gives exactly one output for each input; a formula is only one way of writing that rule down.
- What is the difference between variance and standard deviation?They measure the same spread: variance is the average squared distance from the mean, standard deviation is its square root, so you report the standard deviation and calculate with the variance.
- How do you make a vague claim precise enough to tell whether it's true?Fix what is being talked about, what is asserted, which cases are covered and the cut-off, then name an observation that would prove it false.