Derivative of a linear transformation $f: \mathbb{R^n} \to \mathbb{R^m} $ In multivariable calculus we have that , $f: \mathbb{R^n} \to \mathbb{R^m}$ is differentiable at $a \in \mathbb{R^n}$, if there exists a linear transformation $\mu : \mathbb{R^n} \to \mathbb{R^m} $ such that $$\lim_{h \to 0} \frac{|f(a+h)-f(a)-\mu(h)|}{|h|} = 0$$

then $\mu$ is the derivative of $f$ at $a$ and we denote it by $Df(a)$. According to this definition it is easy to see that, if $f$ is a linear transformation, then $Df(a)=f$ for all $a \in \mathbb{R^n}$.

Having this definition in mind, I am confused by this simple example: Let $f: \mathbb{R} \to \mathbb{R}$ be defined by $f(x)=2x$, is a linear transformation, 

I can't believe that  $Df(a)=f$, which sends any real $x$ number to $2x$

but clearly $Df(a)=2$, is the Jacobian $1 \times 1$ matrix, I kind of see that, it is just the same thing, this also sends any real $x$ number to $2x$.

What is happenning here ?!?!
Thanks for your help in advance
Think of derivative of $f$ at $x$ as a map that applies to vectors based at $x$. In one dimension we can say that $$Df(x)=(2)$$ a $1\times 1$ matrix which stretches these vectors  by the factor of $2$. The analog in two  dimensions is 
$$Df(x) = \begin{pmatrix} 2 &amp; 0 \\ 0 &amp; 2 \end{pmatrix} $$ which does the same. 

It is misleading to say that $Df(x)=f$. The original map $f$ operates on the points in $\mathbb R^n$, the derivative $Df(x)$ operators on the vectors based at $x$. We can write
$$Df(x)\,v = \begin{pmatrix} 2 &amp; 0 \\ 0 &amp; 2 \end{pmatrix}v $$
to indicate this. But 
$$Df(x)\, x = \begin{pmatrix} 2 &amp; 0 \\ 0 &amp; 2 \end{pmatrix}x $$
is a notational confusion. 

Same in one dimension: what we usually write as $f'(x)=2$ can be written as $Df(x)\,v = 2v $  for compatibility with multivariable notation.
Suppose $f:R^n\rightarrow R^m$ linear map

and $Df(x)=f$

Now $D^2f(x)=0$

Can you tell me where 0 should be?