Proof of Chain Rule Along an Affine Path

Write $a=x+\tau_0h$, which lies in $U$ because $\tau_0\in J$, and set

Preliminaries. By hypothesis and the definition of differentiability at a point there is a real matrix $A$ with one row and $n$ columns such that $f$ is differentiable at $a$ with derivative matrix $A$. By claim 1 of A Derivative Matrix is the Jacobian Matrix, and is Unique, $A_{1i}=\partial_i f(a)$ for every $i\in\{1,\dots,n\}$, so by the definition of the matrix-vector product the single coordinate of $A\,k$ is $\sum_{i=1}^{n}\partial_i f(a)\,k_i$ for every $k\in\mathbb{R}^n$. Moreover, for a point $v$ of $\mathbb{R}^1$ the Euclidean norm $\lVert v\rVert$ equals $|v_1|$: both are nonnegative, and they have the same square by claim 1 of Elementary Properties of the Euclidean Norm on $\mathbb{R}^n$ and claim 1 of Properties of the Absolute Value in an Ordered Field. Hence the defining estimate of differentiability at $a$ reads: for every real $\varepsilon'>0$ there is a real $\delta'>0$ such that every $k\in\mathbb{R}^n$ with $0<\lVert k\rVert<\delta'$ satisfies $a+k\in U$ and

Finally, for $\eta\in\mathbb{R}$ the point $a+\eta h$ has $k$th coordinate $x_k+\tau_0h_k+\eta h_k=x_k+(\tau_0+\eta)h_k$, so $a+\eta h=x+(\tau_0+\eta)h$; consequently $F(\tau_0+\eta)=f(a+\eta h)$ whenever $\tau_0+\eta\in J$, and $F(\tau_0)=f(a)$.

Case $h=0$. Every coordinate of $h$ is $0$, so $L=0$ and $x+\tau h=x=a$ for every $\tau\in\mathbb{R}$; hence $F$ takes the constant value $f(a)$ on $J$. Given any real $\varepsilon>0$, take $\delta=1$: for every $\eta\in\mathbb{R}$ with $0<|\eta|<\delta$ and $\tau_0+\eta\in J$ the quotient $\bigl(F(\tau_0+\eta)-F(\tau_0)\bigr)/\eta$ is $0$, so its distance to $L=0$ is $0<\varepsilon$. By Derivative at an Interior Point, $F$ is differentiable at $\tau_0$ with $F'(\tau_0)=0=L$.

Case $h\ne0$. The Euclidean norm is nonnegative and, by claim 3 of Elementary Properties of the Euclidean Norm on $\mathbb{R}^n$, vanishes only at the origin, so $\lVert h\rVert>0$. Let $\varepsilon\in\mathbb{R}$ with $\varepsilon>0$, and apply $(\ast)$ with the positive number $\varepsilon'=\varepsilon/(2\lVert h\rVert)$, obtaining a real $\delta'>0$ as above. Put $\delta=\delta'/\lVert h\rVert$, a positive real number.

Let $\eta\in\mathbb{R}$ satisfy $0<|\eta|<\delta$ and $\tau_0+\eta\in J$, and set $k=\eta h$, the point of $\mathbb{R}^n$ with coordinates $\eta h_i$. By claim 5 of Elementary Properties of the Euclidean Norm on $\mathbb{R}^n$, $\lVert k\rVert=|\eta|\,\lVert h\rVert$, and since $0<|\eta|<\delta$ and $\lVert h\rVert>0$ this gives $0<\lVert k\rVert<\delta\,\lVert h\rVert=\delta'$. So $(\ast)$ applies to this $k$; using $f(a+k)=F(\tau_0+\eta)$, $f(a)=F(\tau_0)$ and $\sum_{i=1}^{n}\partial_i f(a)\,k_i=\eta L$, it reads

By claim 4 of Properties of the Absolute Value in an Ordered Field the left-hand side equals $|\eta|\cdot\bigl|\bigl(F(\tau_0+\eta)-F(\tau_0)\bigr)/\eta-L\bigr|$, so dividing by the positive number $|\eta|$ gives

Since $\varepsilon>0$ was arbitrary and $\delta>0$ was produced from it, Derivative at an Interior Point gives that $F$ is differentiable at $\tau_0$ with $F'(\tau_0)=L$, which is the assertion. $\blacksquare$