Definition clarifications on functions of vectors and their derivatives Definition clarifications would be appreciated:
How do I interpret the following ? For $f: R^n\to R^m$, $Df(\vec\xi)(\vec{x})$ in differentiation of a vector function. I know it as a function that maps $\vec x$ to some other vector and this function's definition depends on the point of evaluation $\vec\xi$. But I am still slightly confused. Is $Df(\vec\xi)$ some sort of gradient or something? Because I was told that $Df(\vec\xi)(\vec x) \circ \vec x=0$, where $\circ$ is the dot product.
Also, how does it act on $\vec x$? Just multiplying by $\vec x$? Since surely $Df(\vec\xi)$, having a given point of evaluation takes a "value". And what happens if $Df(\vec\xi)$ contains a vector? Then do we automatically use the dot product to define its "acting on" $\vec x$?
 A: Here I assume that $f$ is a map from some normed vector space $X$ to another normed vector space $Y$.
$Df(y)$ is a linear map from $X$ to $Y$, depending on the point $y \in X$. In the case where $X=R$, it is the map $x \rightarrow f'(y) x$. In the case where $Y=R$, it is a linear form on $X$ and therefore may be represented as $x \rightarrow (u,x)$ for some vector $u$, if $X$ has a scalar product. Then $u$ is the gradient of $f$ at $y$. But in general it is just a linear map, taking the vector $x \in X$ as an argument.
Actually, it is much better to think of $y$ as a point (a "reference position") and of $x$ as a vector (a "perturbation", or a "movement"): you can have for example $f$ defined only on an open subset $U$ (and then $y$ must belong to $U$) but $Df(y)$ must always be defined on the whole $X$.
Also this way of thinking will make it easier if you ever learn about manifolds ;)
A: Think about differentiation in one variable for a second. The derivative of the one-dimensional vector function (from $\mathbb R$ to $\mathbb R$) gives you the slope at each point. Another way to look at it is that at each point $x \in \mathbb R$ you associate to it a linear transform, $f'(x) : \mathbb R \to \mathbb R$ which for $y$ gives you the number $f'(x)y$ (here this is just standard number multiplication). The transform way is the right way to see things in $\mathbb R^n \to \mathbb R^m$ and furthermore from $V \to W$ where $V$ and $W$ are arbitrary normed vector spaces.
Now when you think of the gradient, say $f : \mathbb R^n \to \mathbb R$, you have a differentiation operator $\nabla : \mathbb R^n \to \mathrm{Hom}(\mathbb R^n, \mathbb R)$ (the latter means all linear applications from $\mathbb R^n$ to $\mathbb R$) which maps every $\mathbb x \in \mathbb R^n$ to $\nabla f(x)$, but instead of seeing $\nabla f(x)$ as a vector, think of it as a linear transform : $\nabla f(x) : \mathbb R^n \to \mathbb R$ and $\nabla f(x)(y) = \nabla f(x) \cdot y$, where on the left you apply the function $\nabla f(x)$ on $y$ and on the right this is the scalar product of both vectors.
When $f : \mathbb R^n \to \mathbb R^m$, exactly the same thing happens : you have a differentation operator $D : \mathbb R^n \to \mathrm{Hom}(\mathbb R^n, \mathbb R^m)$ and $x \in \mathbb R^n$ gets mapped to $Df(x)$ (here the old $\nabla$ became the more general $D$), the matrix with the partial derivatives in it. When you compute $Df(x)(y)$, this is just matrix multiplication : $Df(x)$ is the $m \times n$ matrix and $y$ is a $n \times 1$ column vector. 
Note that it isn't true that $Df(\vec{\xi})(\vec x) \cdot \vec x = 0$ in general : think of the case where $f$ is the identity function, thus $Df(\vec{\xi})$ is the identity matrix for every $\vec{x}$, hence $Df(\vec{\xi})(\vec x) \cdot \vec x = \vec x \neq 0$ as long as $\vec x \neq 0$. Also note that the $\circ$ you had put up there is not the scalar product but rather matrix multiplication when $f : \mathbb R^n \to \mathbb R^m$.
Hope that helps,
