Everyone learns the chain rule the same way: to differentiate a composition, multiply the derivative of the outside by the derivative of the inside. It's fast, it's mechanical, and after a few dozen practice problems it becomes automatic. What almost nobody sees is the proof — and that's a shame, because the proof contains a genuinely sneaky trap, the kind of thing that separates "using calculus" from "understanding why calculus is true."
The Proof That Looks Finished
Start from the definition of the derivative of a composition f ∘ g at a point a:
(f ∘ g)′(a) = limh→0 [f(g(a+h)) − f(g(a))] / h
The natural move is to multiply and divide by g(a+h) − g(a):
limh→0 [f(g(a+h)) − f(g(a))] / [g(a+h) − g(a)] · [g(a+h) − g(a)] / h
As h approaches 0, the second factor approaches g′(a) by definition. And the first factor looks exactly like the difference quotient for f, evaluated at g(a), which should approach f′(g(a)). Multiply the two limits together and you're done: (f ∘ g)′(a) = f′(g(a)) · g′(a). It looks complete. It is not a proof.
The Hidden Trap
The problem is a single silent assumption: that g(a+h) − g(a) is never zero for h near 0. If it is zero, you've just divided by zero — and calculus does not forgive that, no matter how small the mistake looks.
Here's a function where this actually happens, even though g is not constant anywhere near a. Let a = 0 and
g(x) = x² sin(1/x) for x ≠ 0, and g(0) = 0.
This function oscillates faster and faster as x approaches 0, crossing zero infinitely many times in every interval around 0, no matter how small. That means g(0 + h) − g(0) = g(h) equals exactly 0 for infinitely many values of h arbitrarily close to 0 — so the "obvious" proof above tries to divide by zero infinitely often, right in the limit we most care about. And this isn't some exotic pathology cooked up to break things on purpose: g is differentiable everywhere, including at 0. The naive proof fails on a perfectly well-behaved function.
The Fix: Build a Function That Doesn't Care
The way out is a small piece of engineering. Define a new function φ that equals the troublesome difference quotient when it's safe to compute, and equals f′(g(a)) — the value we want — when it isn't:
φ(h) = [f(g(a)+k) − f(g(a))] / k if k = g(a+h) − g(a) ≠ 0, and φ(h) = f′(g(a)) if k = 0.
The whole proof now rests on one claim: φ is continuous at 0. That takes some careful epsilon-delta work (differentiability of f at g(a) guarantees φ behaves the first way; continuity of g at a — which follows from differentiability — keeps k small whenever h is small), but once it's established, the payoff is immediate. For every h ≠ 0, whether or not g(a+h) − g(a) happens to be zero, it's simply true that:
[f(g(a+h)) − f(g(a))] / h = φ(h) · [g(a+h) − g(a)] / h
Check the case k = 0 separately and you'll see both sides are just 0 — so the equation holds unconditionally, no illegal division anywhere. Take the limit as h → 0 of the right-hand side, using the continuity of φ at 0, and you get exactly f′(g(a)) · g′(a). The Chain Rule falls out clean.
Why This Trick Is Worth Remembering
This is a pattern that shows up again and again once you start reading real proofs in analysis: a natural argument breaks on a division-by-zero edge case that looks too rare to matter, and the fix is to build an auxiliary function that absorbs the bad case by definition, then show that function is continuous so the limit goes through anyway. It's a small piece of craftsmanship, and it's the kind of move you only ever see if a book is willing to prove the Chain Rule instead of just handing it to you.
This is exactly the level of care Michael Spivak's Calculus takes with every rule you learned to use quickly in a computational course — the product rule, the quotient rule, and especially the Chain Rule, which the book calls "the most important" differentiation formula and treats accordingly.
If you want the full epsilon-delta argument for why φ is continuous at 0 (it's short, and satisfying once you see it), Spivak's Calculus, 4th Edition works through it in Chapter 10.
Had you seen a real proof of the Chain Rule before, or just the rule itself? Let us know in the comments.
