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The chain rule

Composition of functions was (1.3). Differentiation of a composite is the same chain, walked backwards, with a factor for each step.

The right-hand side is a product of numbers, not a composition of functions. I keep wanting to write ${𝑔}^{\prime }\circ {𝑓}^{\prime }$, which is a different object.

Why a naive cancellation is not a proof

The difference quotient of the composite is

when $𝑓(𝑎+h)\neq 𝑓(𝑎)$. The second factor tends to ${𝑓}^{\prime }(𝑎)$. The first looks like the difference quotient of $𝑔$ at $𝑓(𝑎)$, with increment $𝑓(𝑎+h)-𝑓(𝑎)$, which tends to $0$ by Proposition 2.1. That almost gives ${𝑔}^{\prime }(𝑓(𝑎)){𝑓}^{\prime }(𝑎)$.

The gap is the vanishing of the denominator: $𝑓$ may be locally constant, or may return to $𝑓(𝑎)$ for a sequence of ${h}_{𝑛}\to 0$. Then (2.5) is not even written. The repair is to define an auxiliary function that agrees with the difference quotient of $𝑔$ when the increment is nonzero and equals ${𝑔}^{\prime }(𝑓(𝑎))$ when it is zero. Continuity of that auxiliary function at $0$ saves the product. I am following the usual argument; a careful write-up is in Spivak (1967) and Rudin (1976).

Walking a long chain

The packed composition (1.5) is three functions. Theorem 2.3 twice, plus associativity (1.4), gives

Each factor is the derivative of one layer, evaluated at the output of the layers to its right. That is the same reversed order as Proposition 1.2: undoing a composite, or differentiating it, both peel from the outside.

If a factor has no derivative, the composite may still fail even when the outer function is smooth: composing $𝑥\mapsto |𝑥|$ with the identity is absolute value, which is not differentiable at $0$ by Exercise 2.1. Conversely, a composite can succeed when I already know the closed form, as in Exercise 2.2, without the hypotheses of Theorem 2.3 applying to a particular factorization.