Proof of Chain Rule Along an Affine Path

Write $a=x+\tau_{0}h$ and $L=\sum_{i=1}^{n}\partial_{i}f(a)h_{i}$, where $\partial_{i}f$ abbreviates $\partial f/\partial x_{i}$, and let $|\cdot|$ be the absolute value. Claim numbers refer to Elementary Order Arithmetic in an Ordered Field and to Properties of the Absolute Value in an Ordered Field as indicated. By An Open Interval is an Interval All of Whose Points Are Interior the set $(p,q)$ is an interval all of whose points are interior, so differentiability of $F$ at $\tau_{0}$ is meaningful.

We use repeatedly that squares are strictly monotone on nonnegative reals: if $0\le\alpha$, $0\le\beta$ and $\alpha<\beta$, then $0<\beta$ by claim 2, so $\beta\alpha<\beta\beta$ by claim 10, while $\alpha\alpha\le\beta\alpha$; claim 2 gives $\alpha^{2}<\beta^{2}$. Consequently $\alpha\le\beta$ holds whenever $\alpha^{2}\le\beta^{2}$, since $\beta<\alpha$ would give $\beta^{2}<\alpha^{2}$. Write $\lVert z\rVert$ for the Euclidean distance from $z\in\mathbb{R}^{n}$ to the origin, the nonnegative real with $\lVert z\rVert^{2}=\sum_{i=1}^{n}z_{i}^{2}$.

Case 1: every $h_{i}$ is $0$. Then $\tau h=0$ for every $\tau$, so $F$ is the constant function with value $f(x)$, and $L=0$. For $\kappa\ne0$ with $\tau_{0}+\kappa\in(p,q)$ the difference quotient of $F$ at $\tau_{0}$ is $0$, so the quantity $|0-L|=|0|=0$ is less than any $\varepsilon$ with $0<\varepsilon$; taking $\delta=1$, positive by claim 6, shows $F$ is differentiable at $\tau_{0}$ with $F'(\tau_{0})=0=L$.

Case 2: some $h_{i_{0}}\ne0$. Then $h_{i_{0}}^{2}$ is positive by claim 5 applied to $|h_{i_{0}}|$, using claim 4 of Properties of the Absolute Value in an Ordered Field, and every other $h_{i}^{2}$ is nonnegative, so $\lVert h\rVert^{2}$ is positive by claim 3 and hence $0<\lVert h\rVert$. So $\lVert h\rVert^{-1}$ exists and is positive by claim 7.

Let $\varepsilon\in\mathbb{R}$ with $0<\varepsilon$. Put $\varepsilon_{1}=\varepsilon\cdot2^{-1}\cdot\lVert h\rVert^{-1}$, positive by claims 8, 7 and 5. Since $f$ is differentiable at $a$, applied with $m=1$ and with $\varepsilon_{1}$ in place of $\varepsilon$, there is $\delta_{1}$ with $0<\delta_{1}$ such that every $k=(k_{1},\dots,k_{n})\in\mathbb{R}^{n}$ with $0<\sum_{i}k_{i}^{2}<\delta_{1}^{2}$ and $a+k\in U$ satisfies

Put $\delta=\delta_{1}\lVert h\rVert^{-1}$, positive by claim 5. Let $\kappa\in\mathbb{R}$ satisfy $0<|\kappa|<\delta$ and $\tau_{0}+\kappa\in(p,q)$, and set $k=\kappa h$, so $k_{i}=\kappa h_{i}$ for each $i$ and $a+k=x+(\tau_{0}+\kappa)h$, which lies in $U$ by hypothesis. Then

using claim 4 of Properties of the Absolute Value in an Ordered Field. This is positive, since $|\kappa|$ and $\lVert h\rVert$ are positive and claim 5 applies; and $|\kappa|\lVert h\rVert<\delta\lVert h\rVert=\delta_{1}$ by claim 10, so $\sum_{i}k_{i}^{2}<\delta_{1}^{2}$ by the monotonicity of squares. Hence the displayed estimate applies to this $k$.

Since $f(a+k)=F(\tau_{0}+\kappa)$, $f(a)=F(\tau_{0})$ and $\sum_{i}\partial_{i}f(a)k_{i}=\kappa L$, the estimate reads

and both $\bigl|F(\tau_{0}+\kappa)-F(\tau_{0})-\kappa L\bigr|$ and $\varepsilon_{1}|\kappa|\lVert h\rVert$ are nonnegative with $\bigl|F(\tau_{0}+\kappa)-F(\tau_{0})-\kappa L\bigr|^{2}$ equal to the left side, by claim 4 of Properties of the Absolute Value in an Ordered Field. By the monotonicity of squares,

Dividing by the positive number $|\kappa|$, that is multiplying by $|\kappa|^{-1}$ and using claim 4 of Properties of the Absolute Value in an Ordered Field together with claim 10,

which is less than $\varepsilon$ by claim 8; so the quantity is less than $\varepsilon$ by claim 2.

As $\varepsilon$ was arbitrary, $L$ satisfies the defining condition of differentiability of $F$ at $\tau_{0}$. Hence $F$ is differentiable at $\tau_{0}$ with $F'(\tau_{0})=L$.