I've a doubt about the total derivative or differntial of an application. If $F : U \subset \mathbb{R}^n \to \mathbb{R}^m$ has differential $dF_{p}$ at each point $p \in U$ then the differential acts on a vector giving the instantaneous rate of change of $F$ in that direction or the increment in the linearization in that direction?
This question arose when I thought on the case $n = m = 1$, i.e.: a function $f : U \subset \mathbb{R} \to \mathbb{R}$. In that case, the differential matrix will have only one entry $f'(x)$ and a vector in $U$ will be simply a scalar $v = x_f - x$. In that case, the action of the differential in the vector will be: $df_{x}(v) = f'(x)(x_f - x)$ and in that case the matrix multiplication produces an increment on the linearization of $f$ instead just the derivative of $f$.
However, if the vector is the unit vector in that direction, then it'll be just $v = 1$ and $df_{x}(v) = f'(x)$, and it is indeed the derivative. So, when the vector is a unit vector de differential produces a rate of change and when it's not a unit vector it produces an increment to the linearization?
Sorry if I've said something silly, and if I wasn't clear enough ask me to explain better. Thanks in advance.