The differential produces directional derivatives or increments?

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.

• Those two things read the same for me at first. $dF_p$ acts on a curve through $p$ by giving it "the increment in the linearization in the direction". This is a term I've never heard of but I know what you mean. Commented Aug 31, 2012 at 1:43

1.) the differential $df_a$ is actually a linear transformation which takes vectors at the tangent space at $a$ and, in the case of $f$ being real-valued, gives the rate of change of $f$ in the direction. However, to make a standard rate of change it is customary to consider only unit-vectors. In advanced calculus texts this is often dropped and the differential takes vectors of arbitrary length as inputs.
2.) the derivative at a point is the matrix of the differential. In the case of a real valued function this reduces to $df_a(v) = (\nabla f)(a) \cdot v$. In the case of functions on $\mathbb{R}$ it reduces to $df_a(v) = f'(a)v$. This can be cutely written as $df_a(dx)=f'(a)dx = \frac{dy}{dx}dx$, so the notation is consistent with the idea $dy=df_a(dx)$.
$$L(a+h) = f(a)+df_a(h)$$