Proof of Chain Rule for One-Dimensional Derivatives

Throughout, $|\cdot|$ is the absolute value on $\mathbb{R}$. Write $s_0=\gamma(t_0)$, $A=g'(s_0)$ and $B=\gamma'(t_0)$, the derivatives supplied by the hypotheses.

Step 0 (two elementary facts).

(a) If $0\le c$ and $p\le q$ then $cp\le cq$. If $c=0$ both products are $0$; if $0<c$ and $p=q$ they are equal; and if $0<c$ and $p<q$ then $cp<cq$ by claim 10 of Elementary Order Arithmetic in an Ordered Field.

(b) For every $x\in\mathbb{R}$ one has $0<|x|+1$, the inverse $(|x|+1)^{-1}$ exists and is positive, and $|x|\,(|x|+1)^{-1}\le 1$. By claim 1 of Properties of the Absolute Value in an Ordered Field, $0\le|x|$, and $0<1$ by claim 6 of Elementary Order Arithmetic in an Ordered Field; claim 3 of that lemma, applied to the strict inequality $0<1$ and the inequality $0\le|x|$, gives $0<1+|x|=|x|+1$, so claim 7 supplies the positive inverse. Claim 1 of that lemma turns $0<1$ into $|x|<|x|+1$, and fact (a) with $c=(|x|+1)^{-1}$ gives $|x|\,(|x|+1)^{-1}\le(|x|+1)(|x|+1)^{-1}=1$.

Step 1 (the increment function of $g$). Define $\Phi:J\to\mathbb{R}$ by

the inverse existing because $s\ne s_0$ gives $s-s_0\ne0$.

(1a) For every $s\in J$, $g(s)-g(s_0)=\Phi(s)\,(s-s_0)$. For $s\ne s_0$ this is field arithmetic, using $(s-s_0)^{-1}(s-s_0)=1$; for $s=s_0$ both sides are $0$.

(1b) For every $\varepsilon\in\mathbb{R}$ with $0<\varepsilon$ there is $\eta\in\mathbb{R}$ with $0<\eta$ such that every $s\in J$ with $|s-s_0|<\eta$ satisfies $|\Phi(s)-A|<\varepsilon$. By hypothesis $s_0$ is an interior point of $J$ at which $g$ is differentiable with derivative $A$, so the definition of the derivative supplies $\eta$ with $0<\eta$ such that every $k\in\mathbb{R}$ with $0<|k|<\eta$ and $s_0+k\in J$ satisfies

Let $s\in J$ with $|s-s_0|<\eta$. If $s=s_0$ then $\Phi(s)-A=A-A=0$, and $|0|=0<\varepsilon$ by claim 1 of Properties of the Absolute Value in an Ordered Field. Otherwise $s-s_0\ne0$, hence $0<|s-s_0|$ by claim 1 of that lemma; taking $k=s-s_0$, for which $s_0+k=s\in J$, the displayed estimate is exactly $|\Phi(s)-A|<\varepsilon$.

Step 2 (a bound on the difference quotients of $\gamma$). For $h\in\mathbb{R}$ with $h\ne0$ and $t_0+h\in I$ set

Since $0<1$ by claim 6 of Elementary Order Arithmetic in an Ordered Field, differentiability of $\gamma$ at $t_0$ with derivative $B$ supplies $\delta_1$ with $0<\delta_1$ such that every $h$ with $0<|h|<\delta_1$ and $t_0+h\in I$ satisfies $|Q(h)-B|<1$. For such $h$, claim 3 of Properties of the Absolute Value in an Ordered Field gives $|Q(h)|-|B|\le\bigl||Q(h)|-|B|\bigr|$, which is at most $|Q(h)-B|$ by claim 7 of that lemma; so $|Q(h)|-|B|<1$ by claim 2 of Elementary Order Arithmetic in an Ordered Field, and adding $|B|$ to both sides, by claim 1 of that lemma,

Step 3 (the estimate). Let $\varepsilon\in\mathbb{R}$ with $0<\varepsilon$. By claim 8 of Elementary Order Arithmetic in an Ordered Field, $0<\varepsilon\cdot2^{-1}$ and $\varepsilon\cdot2^{-1}+\varepsilon\cdot2^{-1}=\varepsilon$. By Step 0 (b) the inverses $(|B|+1)^{-1}$ and $(|A|+1)^{-1}$ exist and are positive, so by claim 5 of that lemma the numbers

are positive.

Let $\eta$ be as in Step 1 (1b) for $\varepsilon_1$. Differentiability of $\gamma$ at $t_0$ supplies $\delta_2$ with $0<\delta_2$ such that every $h$ with $0<|h|<\delta_2$ and $t_0+h\in I$ satisfies $|Q(h)-B|<\varepsilon_2$. The number $\eta\,(|B|+1)^{-1}$ is positive by claim 5. Applying claim 9 twice, let $\delta$ be a least element among $\delta_1$, $\delta_2$ and $\eta\,(|B|+1)^{-1}$; since $\delta$ equals one of them, $0<\delta$.

Let $h\in\mathbb{R}$ with $0<|h|<\delta$ and $t_0+h\in I$, and put $s=\gamma(t_0+h)$, an element of $J$ by hypothesis. By claim 2 of Elementary Order Arithmetic in an Ordered Field we have $|h|<\delta_1$, $|h|<\delta_2$ and $|h|<\eta\,(|B|+1)^{-1}$.

First, $s-s_0=h\,Q(h)$, so $|s-s_0|=|h|\,|Q(h)|$ by claim 4 of Properties of the Absolute Value in an Ordered Field. Since $h\ne0$, claim 1 of that lemma gives $0<|h|$; so Step 2 and claim 10 of Elementary Order Arithmetic in an Ordered Field give $|h|\,|Q(h)|<|h|\,(|B|+1)$, and claim 10 again, with the positive multiplier $|B|+1$, gives $|h|\,(|B|+1)<\eta\,(|B|+1)^{-1}(|B|+1)=\eta$. By claim 2 we conclude $|s-s_0|<\eta$, so Step 1 (1b) yields

Second, by Step 1 (1a), the identity $s-s_0=h\,Q(h)$ and $h^{-1}h=1$,

By field arithmetic,

so by claims 5 and 4 of Properties of the Absolute Value in an Ordered Field,

For the first summand, Step 0 (a) with $c=|\Phi(s)-A|$, which is nonnegative by claim 1 of Properties of the Absolute Value in an Ordered Field, together with $|Q(h)|\le|B|+1$ from Step 2, gives $|\Phi(s)-A|\,|Q(h)|\le|\Phi(s)-A|\,(|B|+1)$; and claim 10 of Elementary Order Arithmetic in an Ordered Field, with the positive multiplier $|B|+1$, turns $|\Phi(s)-A|<\varepsilon_1$ into $|\Phi(s)-A|\,(|B|+1)<\varepsilon_1(|B|+1)=\varepsilon\cdot2^{-1}$. By claim 2 the first summand is smaller than $\varepsilon\cdot2^{-1}$. For the second summand, Step 0 (a) with $c=|A|$ gives $|A|\,|Q(h)-B|\le|A|\,\varepsilon_2=\bigl(|A|\,(|A|+1)^{-1}\bigr)\,\varepsilon\cdot2^{-1}$, and Step 0 (b) with Step 0 (a) applied with $c=\varepsilon\cdot2^{-1}$ bounds this by $\varepsilon\cdot2^{-1}$.

Adding the two bounds by claim 3 of Elementary Order Arithmetic in an Ordered Field, and using claim 2 of that lemma with the displayed triangle-inequality bound,

Since $\varepsilon$ was an arbitrary positive real number and $t_0$ is an interior point of $I$, the real number $AB$ witnesses the definition of the derivative of $g\circ\gamma$ at $t_0$. Hence $g\circ\gamma$ is differentiable at $t_0$ and $(g\circ\gamma)'(t_0)=AB=g'\bigl(\gamma(t_0)\bigr)\,\gamma'(t_0)$. $\blacksquare$