The chain rule, in one line from Carathéodory: if $g$ is differentiable at $c$ and $f$ is differentiable at $g(c)$, then $f\circ g$ is differentiable at $c$ with $(f\circ g{)}^{\mathrm{\prime }}(c)=f^{\mathrm{\prime }}(g(c))g^{\mathrm{\prime }}(c)$

Let $A,B\subseteq \mathbb{R}$, let $g:A\to \mathbb{R}$ with $g[A]\subseteq B$ and let $f:B\to \mathbb{R}$, so that the composite $f\circ g:A\to \mathbb{R}$ is defined. Let $c\in A$ be a limit point of $A$ (Limit point, isolated point, adherent point, derived set, and dense subset of $\mathbb{R}$) at which $g$ is differentiable (The derivative $f^{\mathrm{\prime }}(c)={\lim }_{x\to c}\frac{f(x)-f(c)}{x-c}$ of $f:A\to \mathbb{R}$ at a point $c\in A$ that is a limit point of $A$, and differentiability on a set), put $b:=g(c)$, and suppose $b$ is a limit point of $B$ at which $f$ is differentiable. Then $f\circ g$ is differentiable at $c$ and

$(f\circ g{)}^{\mathrm{\prime }}(c)=f^{\mathrm{\prime }}(g(c))g^{\mathrm{\prime }}(c)\mathrm{.}$

Both limit-point hypotheses are needed, and neither is automatic. That $c$ is a limit point of $A$ is what makes $g^{\mathrm{\prime }}(c)$ and $(f\circ g{)}^{\mathrm{\prime }}(c)$ defined symbols; that $b=g(c)$ is a limit point of $B$ is what makes $f^{\mathrm{\prime }}(b)$ one. Nothing forces the second: $g$ may be differentiable at $c$ and send $c$ to an isolated point of $B$, and there $f^{\mathrm{\prime }}(b)$ is not defined and the formula asserts nothing.

No case analysis appears anywhere. The naive difference-quotient proof writes $\frac{f(g(x))-f(g(c))}{g(x)-g(c)}\cdot \frac{g(x)-g(c)}{x-c}$ and then has to say what happens where $g(x)=g(c)$, which may occur at points arbitrarily close to $c$. Carathéodory's factorisation never divides by the inner increment, so the difficulty does not arise.

Limit point, isolated point, adherent point, derived set, and dense subset of $\mathbb{R}$

The derivative $f^{\mathrm{\prime }}(c)={\lim }_{x\to c}\frac{f(x)-f(c)}{x-c}$ of $f:A\to \mathbb{R}$ at a point $c\in A$ that is a limit point of $A$, and differentiability on a set

• Where the classical proof goes wrong, precisely. It divides by $g(x)-g(c)$, which may vanish at points arbitrarily close to $c$ even when $g$ is differentiable at $c$ with $g^{\mathrm{\prime }}(c)=0$; the usual repair defines an auxiliary function equal to the outer quotient off the bad set and to $f^{\mathrm{\prime }}(b)$ on it, and then proves that auxiliary function continuous. That auxiliary function is $\phi \circ g$, and Carathéodory's characterisation: $f$ is differentiable at $c$ if and only if there is $\phi :A\to \mathbb{R}$, continuous at $c$, with $f(x)-f(c)=\phi (x)(x-c)$ for every $x\in A$, and then $\phi$ is unique and $\phi (c)=f^{\mathrm{\prime }}(c)$ is the observation that it exists before any repair is attempted.

• What is composed is continuity, not differentiability. The only theorem about composites used above is A composite of continuous functions is continuous, with no side hypothesis of the kind the composition of limits needs, and it needs no side hypothesis, unlike the corresponding statement for limits. That is the whole reason the proof has no cases.

• The formula is about the point $g(c)$, not about $g$ near $c$. Both derivatives on the right are taken at single points, and the theorem says nothing about $f$ on the image of any neighbourhood of $c$. In particular no hypothesis is placed on $g[A]$ beyond its lying in $B$.