= Solution
First suppose each $g_j$ is <differentiable> at $x_j$. Write $g_j(x_j+t)=g_j(x_j)+v_jt+r_j(t)$, where $v_j\in\mathbb R^m$ and $r_j(t)=o(|t|)$. Define the <linear map> $Lh=\sum_jv_jh_j$. Given $\varepsilon>0$, choose a common sufficiently small neighbourhood so that $\|r_j(h_j)\|\le\varepsilon|h_j|$ for all $j$, with $r_j(0)=0$. Then
$$
\left\|f(x+h)-f(x)-Lh\right\|\le\varepsilon\sum_j|h_j|\le\varepsilon\sqrt n\,\|h\|.
$$
This proves <differentiability>, with
$$
\boxed{Df(x)h=\sum_{j=1}^n g_j'(x_j)h_j.}
$$
Conversely, suppose $f$ is <differentiable> with <Fréchet derivative> $L$. Along the $j$th coordinate direction $e_j$, the other summands cancel exactly:
$$
g_j(x_j+t)-g_j(x_j)=f(x+te_j)-f(x)=tLe_j+o(|t|).
$$
Thus $g_j$ is <differentiable> at $x_j$, with $g_j'(x_j)=Le_j$. \b[The equivalence works because each summand varies in only one coordinate; it is stronger than merely having partial derivatives.]
Back to article page
