Regarding the derivative of Euclidean L2 norm, Definition of differentiation in Rudin. I am trying to understand the answer posted by hemanth in this post.
I understand how he derived the derivative $f:\Bbb R^n \rightarrow \Bbb R$ defined as $f(x)=\rVert x \rVert$,
$$
Df(x) = \nabla_x\rVert x\rVert_2= \frac{x}{\lVert x \rVert}.
$$
So we have $Df:\Bbb R^n \rightarrow L(\Bbb R^n,\Bbb R)$, but according to Rudin, $Df(x)$ must be a linear transformation from $\Bbb R^n$ to $\Bbb R$, while I do not see $Df(x)h \in \Bbb R$ for any h $ \in \Bbb R^n$.
I am trying to derive $Df(x)$ such that: $$
\lim_{h\rightarrow0}=\frac{\rVert f(x+h)-f(x)-Df(x)h \rVert}{\rVert h \rVert},$$ so $Df(x)h$ must be a real number.
Could it be that $Df(x)h$ is a product of a column vector and a row vector?
If so, how does
$$
\begin{split}
\frac{\rVert f(x+h)-f(x)-Df(x)h \rVert}{\rVert h \rVert} &= \frac{\Vert\rVert x+h \rVert - \rVert x \rVert -Df(x)h \Vert}{\rVert h\rVert}\\ 
&\le\frac{\rVert\rVert x \rVert+\rVert h \rVert - \rVert x \rVert -Df(x)h \Vert}{\rVert h\rVert}=\frac{\rVert\rVert h \rVert -\frac{x}{\rVert x \rVert}h \rVert}{\rVert h \rVert}\rightarrow 0\;,
\end{split}
$$ as  $h \rightarrow 0$ ?
Anyone can help me with this confusion?
It is my first time asking the question here, so my formatting might be bad.
Thank you for understanding.

I would simply write
$$
Df(\mathbf{x})[\mathbf{h}] = 
\frac{\partial f}{\partial \mathbf{x}}:\mathbf{h}
$$
where the colon operator denotes the inner product in $\mathbb{R}^N$
This says how much (up to first order) $f$ will change when you move from
$\mathbf{x}$ to $\mathbf{x+h}$.
