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.

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.
