Proposition 3

*For each index $i$ ($1 \leq i \leq p$) let $f_i$ be a function defined on an interval $I \subset \mathbf{R}$, taking values in $E_i$, and differentiable at the point $x_0 \in I$. Then the function

defined on $I$ with values in $F$ has a derivative equal to

at $x_0$.

Let us put $h(x) = [f_1(x).f_2(x) \ldots f_p(x)]$; then, by the identity

we can write

On multiplying both sides by $\frac{1}{x - x_0}$ and letting $x$ approach $x_0$ in $I$, we obtain the expression (1), since both the map

and addition in F are continuous.

$^1$ Recall (Alg., II, p. 265) that a map $f$ of $E_1 \times E_2 \times \cdots \times E_p$ into F is said to be multilinear if each partial mapping

from $E_i$ into F ($1 \leq i \leq p$) is a linear map, the $a_j$ for indices $j \neq i$ being arbitrary in $E_j$. We note that if the $E_i$ are finite dimensional over $\mathbf{R}$ then every multilinear map of $E_1 \times E_2 \times \cdots \times E_p$ into F is necessarily continuous; this need not be so if some of these spaces are topological vector spaces of infinite dimension.

When some of the functions $f_i$ are constant, the terms in the expression (1) containing their derivatives $f'_i(x_0)$ are zero.

Let us consider in detail the particular case $p = 2$, the most important in applications: if $(\mathbf{x}, \mathbf{y}) \mapsto [\mathbf{x}.\mathbf{y}]$ is a continuous bilinear map of $E \times F$ into $G$, (E, F, G being topological vector spaces over $\mathbf{R}$), and $\mathbf{f}$ and $\mathbf{g}$ are two vector functions, differentiable at $x_0$, with values in E and F respectively, then the vector function $x \mapsto [\mathbf{f}(x).\mathbf{g}(x)]$ (which we denote by $[\mathbf{f}.\mathbf{g}]$) has a derivative equal to $[\mathbf{f}'(x_0).\mathbf{g}(x_0)] + [\mathbf{f}(x_0).\mathbf{g}'(x_0)]$ at $x_0$. In particular, if $\mathbf{a}$ is a constant vector, then $[\mathbf{a}.\mathbf{f}]$ (resp. $[\mathbf{f}.\mathbf{a}]$) has a derivative equal to $[\mathbf{a}.\mathbf{f}'(x_0)]$ (resp. $[\mathbf{f}'(x_0).\mathbf{a}]$) at $x_0$.

If $\mathbf{f}$ and $\mathbf{g}$ are both differentiable on I then so is $[\mathbf{f}.\mathbf{g}]$, and we have

(2)