Derivative of linear transformation with confusing moment 
After reading this part of Rudin's book i have one question: $A'(\mathbf{x})=A$ seems to me little bit weird because:
1) $A'(x)$ - it's derivative of operator $A$ at point $\mathbf{x}\in \mathbb{R^n}$ and it's number but RHS is function. 
2) Let $A(x)=2x$ then it's linear operator on $\mathbb{R}^1$. Let $x_0=1$ then  $A'(x_0)=2$ but $2\neq 2x$ and $A'(x)=A$ is meaningless.
I searched the same questions on MSE but I can't understand their meaning. I would be very grateful if anyone explain it to me in detail.
 A: Note that the definition of a derivative in this context is slightly different that what you are used to.
Note that $f(x) = x$ is a map from $\Bbb R \to \Bbb R$.  So, its derivative at $x$ should be a linear map from $\Bbb R \to \Bbb R$.  Indeed, we end up with 
$$
[f'(x)](y) = 1\cdot y
$$
which can be neatly represented by the number $1$.  Similarly, the derivative of $f(x) = Ax$ is given by
$$
[f'(x)](y) = A y
$$
which can be neatly represented by the matrix $A$. Let's try a another function, and maybe you'll start getting the idea.  For $f(x) = x^3$ (from $\Bbb R$ to $\Bbb R$), we have
$$
[f'(x)](y) = 3x^2 y
$$
which gives us a linear function of $y$ at any value of $x$.
Hopefully this clarifies things a bit.

Note that the derivative of $f(x)$ is defined to be the (unique) linear function $[f'(x)](y)$ such that
$$
\lim_{h \to 0}\frac{f(x+h) - f(x) - [f'(x)](h)}{\|h\|} = 0
$$
If we start with $f(x) = Ax$, then we can show that $[f'(x)](y) = Ay$ is the derivative of $f$ by this definition.  When we plug in, we have
$$
\lim_{h \to 0}\frac{f(x+h) - f(x) - [f'(x)](h)}{\|h\|} = \\
\lim_{h \to 0}\frac{A(x+h) - A(x) - Ah}{\|h\|} = \\
\lim_{h \to 0}\frac{0}{\|h\|}
$$
so the necessary condition holds.
A: It would be nice if mathematicians were better about saying what the types of things were.
Write $Df$ for $f'$.  Then $D : (\mathbb{R}^n \to \mathbb{R}^m) \to (\mathbb{R}^n \to L(\mathbb{R^n}, \mathbb{R^m}))$.  You don't notice this at first because $L(\mathbb{R},\mathbb{R}) \cong \mathbb{R}$, i.e. in the 1D case from high school, the linear transformation that results is represented by a single number.
So what's happening is $A : L(\mathbb{R}^n, \mathbb{R}^m)$ which is just a special type of function $\mathbb{R}^n \to \mathbb{R}^m$ so $DA : \mathbb{R}^n \to L(\mathbb{R}^n, \mathbb{R}^m)$ so it's still expecting the point about which to make the linear approximation.  So $DA(x) : L(\mathbb{R}^n, \mathbb{R}^m)$ just like $A$.  It will serve you well to think about what the types of things are.
A: The differential of a function defined on $\mathbb{R}^n\rightarrow \mathbb{R}^m$ at $x$ is a LINEAR function $L_x:\mathbb{R}^n\rightarrow \mathbb{R}^m$ such that $\lim\limits_{h\rightarrow 0} (f(x+h)-f(x)-L_x(h))=0$. If $n=1$, the differential of $f:\mathbb{R}\rightarrow \mathbb{R}$ at $x$ is identified with a number since a linear map of $\mathbb{R}$ is of the form $y\rightarrow ay$ and $a$ determines the map. So if $A$ is a linear map and $x\in \mathbb{R}$ $\lim\limits_{h\rightarrow 0}(A(x+h)-A(x)-A(h))=0$ so the differential of $A$ is $A$.
