Second derivative of a vector field I wonder how to treat the "second derivative" of a vector field. For example, imagine we have a vector field $f:\mathbb{R}^n \rightarrow \mathbb{R}^n$. Then we evaluate the derivative at two points $Df(a)$ and $Df(b)$ which are matrices! Now,
$$D[Df(a)Df(b)] = D^2f(a)Df(b)+Df(a)D^2f(b).$$
My question is, what is $D^2f(a)$? How can I treat this? I imagine is something identifiable with $\mathbb{R}^{n\times n \times n}$. In such a case, if I wish to compute the "matrix" norm of $D[Df(a)Df(b)]$ (as the sum of all entries) is this then the sum of all possible combinations of
$$\frac{\partial}{\partial x_i}\frac{\partial}{\partial x_j} \frac{\partial}{\partial x_k} f(a)  \ ?$$
Thank you very much for your help!
 A: As already answered by SAUVIK, if $f:E\longrightarrow F$,
$$Df(x)\in L(E,F)\text{  and  } Df:E\longrightarrow L(E,F).$$
So,
$$D^2 f:E\longrightarrow L(E,L(E,F)).$$
Now, the space $L(E,L(E,F))$ can be identified of the space of bilinear functions form $E$ to $F$ via the isomorphism
$$g\to\hat g,\qquad\hat g(x,y) = (g(x))(y).$$
The trick can be obviously extended to higher orders.
A: Suppose $f'(x)$ denotes the derivative of $f$ at $x$, which is practically a linear transformation. 
Now as you vary $x$ you can think of $f'$ is a mapping from $\mathbb{R^n}$ to $L(\mathbb{R^n},\mathbb{R^n})$, the set of all linear transformation from $\mathbb{R^n}$ to $\mathbb{R^n}$. 
Now how do you give metric structure on L($\mathbb{R^n}$,$\mathbb{R^n}$)?
Define for $A \in L(\mathbb{R^n},\mathbb{R^n})$: $|A|=\sup A(x)$ where $|x|\le 1$ (it's a compact set and linear maps are continuous so the supremum exists).
You can easily check that it's a metric on $L(\mathbb{R^n},\mathbb{R^n})$.
So now of course you can talk about continuity and differentiability of $f':\mathbb{R^n} \to L(\mathbb{R^n},\mathbb{R^n})$.
