Solution
= Solution
The map is <differentiable> at $x$ if there is a <linear map> $L:\mathbb R^n\to\mathbb R^m$ such that
$$
\lim_{h\to0}\frac{\|f(x+h)-f(x)-Lh\|}{\|h\|}=0.
$$
Equivalently, $f(x+h)=f(x)+Lh+o(\|h\|)$. This linear map is unique and is the <Fréchet derivative> $Df_x$. Its matrix in the standard bases is the <Jacobian matrix>, whose $(i,j)$ entry is $\partial f_i/\partial x_j$ at $x$. A function is differentiable on its domain if this condition holds at every point.