If $f(x,v) = A(x)\cdot v$ then $f'(x,v)\cdot (h,k) = (A'(x)\cdot h)\cdot v + A(x)\cdot k$ 
Let $A:U\to L(\mathbb{R}^m; \mathbb{R}^n)$ differentiable in the open
  $U\subset\mathbb{R}^p$. Define $f:U\to\mathbb{R}^m\to\mathbb{R}^n$ by
  $f(x,v) = A(x)\cdot v$. Prove that $f$ is differentiable, with
  $f'(x,v)\cdot (h,k) = (A'(x)\cdot h)\cdot v + A(x)\cdot k$

First of all, what does it mean for $A$ to be from $U\subset\mathbb{R}^p$ to $L(\mathbb{R}^m; \mathbb{R}^n)$? I've only worked with functions of the form $g:\mathbb{R}^m\to\mathbb{R}^n$, not to this $L(\mathbb{R}^m; \mathbb{R}^n)$ thing. On the other side, $f$ is a function that I know how to work with. Normally, we'd take the derivative $f'$ by doing:
$$f((x_0, v_0) + (x,v)) = f(x_0,v_0) + f'(x_0,v_0)\cdot (x,v) + r((x,v))$$
So by using $f(x,v) = A(x)\cdot v$ the thing above should look something like
$$A(x_0+x)\cdot(v_0+v) = A(x_0)\cdot v_0 + f'(x_0,v_0)\cdot(x,v) + r(v)$$
I think I should call $(x,v)$ as $(h,k)$ or something like that.
 A: The function $A$ is a matrix value function i.e. $A(x)$ is $n \times m$  matrix, for any $x \in U \subseteq \Bbb R^p .$ So its derivative is $3-$dimension matrix $A'(x)$, which is a $n \times m \times p$ matrix, or one can interpret it as linear transformation from $\Bbb R^p$ to $ \Bbb R^{n \times m }$.  The function $f(x,v) = A(x) . v$ is a product of matrix and vector $v$ in $\Bbb R^m$. Thus to differentiate $f$ at $(x,v)$ we need apply product rule.
So   $$f' (x,v) = \bigl(\begin{smallmatrix}
 A'(x) . v &, ~ 0_{m \times m}
\end{smallmatrix}\bigr) + \bigl(\begin{smallmatrix}
 0_{p \times p}&, ~ A(x) \Bbb I_m
\end{smallmatrix} \bigr) = \bigl(\begin{smallmatrix}
 A(x).v &, ~ A(x) \Bbb I_m
\end{smallmatrix}\bigr) $$
Note that $A'(x) . v$ is now $n \times p$ matrix. To find the directional derivative of $f$ at point $(x,v)$ in direction $(h.k)  \in \Bbb R^p \times \Bbb R^m$ just do dot multiplication 
$$  f' (x,v) . (h,k) = (A'(x) . v ) .h + A(x) . k    $$
dot multiplication here means (linear transformation) multiplying matrix and vector in appropriate way (their)  order must be matched ! for example $A(x) . k $ really means $ A(x) k $, and we do not distinguish between $(A'(x) . v ) .h$ and $(A'(x) . h ) .v$ which is a vector in $\Bbb R^n.$
