Fréchet derivative of norm function How to calculate the Fréchet derivative of $f: \mathbb{R}^n \to \mathbb{R}:
(x_1, \dots, x_n) \mapsto \sqrt{x_{1}^{2}+ \cdots + x_{n}^{2}}$?
I dont't know hot to fin a lineat operator such that
$$\lim_{(h_1 , \ldots h_n) \to (0. \ldots 0)} \frac{f((x_1+h_1, \dots, x_n+h_n))-f((x_1, \dots, x_n))-A((h_1, \dots, x_n))}{\sqrt{h_{1}^{2}+ \cdots + h_{n}^{2}}}=0$$
 A: You can also use the following:
Your function F is fréchet differentiable, if there is a linear operator F'(just notation) satisfying
$F(x+h) = F(x) + hF'(x) + o(||h||)$.
Just plug in and try to get to the form of the right hand side, that's the easiest way 
EDIT: 
An example: $F(x,y) = 3x^2 -7y^2.$
Then 
$F(x+h_1,y+h_2)=3(x+h_1)^2 - 7 (y+h_2)^2 = ... 
= 3h_1^2 + 6h_1x - 7h_2^2 - 14h_2y + 3x^2 - 7y^2$.
Reordering gives you
$... = \underbrace{3x^2-7y^2}_{=F(x,y)} + \underbrace{6h_1x-14h_2y}_{ } +\underbrace{3h_1^2 - 7h_2^2}_{o(||h||)}$
and the 2 terms in the middle you define as F' (note that they are what you were looking for)
A: Write $f(x)=:\|x\|$. You easily compute $${\partial f\over\partial x_i}={x_i\over \|x\|}\ (1\leq i\le n)\ ,$$
which is tantamount to $$\nabla f(x)={x\over\|x\|}\qquad(x\ne 0)\ .$$
Now for the Fréchet derivative of $f$ at $x$ we have
$$df(x).X=\nabla f(x)\cdot X={x\cdot X\over\|x\|}\ .$$
In terms of matrices with respect to the standard bases in ${\mathbb R}^n$ and ${\mathbb R}$ this means that
$$\bigl[df(x)\bigr]=\left[{x_1\over\|x\|},\ldots,{x_n\over\|x\|}\right]\ ,$$
an $(1\times n)$-matrix.
A: Decompose $f$ into a composite and use the chain rule: 
define 
$r:\mathbb R^{n}\rightarrow \mathbb R$ by $(x_{1},\cdots ,x_{n})\mapsto x_1^{2}+\cdots +x^{2}_n$. 
Now define 
$s:\mathbb R\rightarrow \mathbb R$ by $x\mapsto \sqrt x$ 
so that $f=s\circ r$ and 
$f'(\vec x):\mathbb R^{n}\rightarrow \mathbb R$ is given by 
$f'(\vec x)(h)=s'(r(\vec x))\circ r'(\vec x)(h)$. 
With $\vec x=(x_1,\cdots ,x_n)$, and $\vec h=(h_1,\cdots ,h_n),\ $substitute to find that 
$s'(r(\vec x))\circ r'(\vec x)(h)=\frac{1}{2\sqrt{r(\vec x)}}\cdot <\nabla r(\vec x), h>=\frac{2x_1h_1+\cdots +2x_nh_n}{2\sqrt{x_1^{2}+\cdots +x_n^{2}}}=\frac{x_1h_1+\cdots +x_nh_n}{\sqrt{x_1^{2}+\cdots +x_n^{2}}}$.
A: I'll provide another answer, more generalizable. I'll assume you know chain rule. Then, consider the function $p_i:(x_1, \ldots, x_d) \mapsto x_i$ (the projection onto factor $i$). This is linear, then its derivative is given by $p_i'(x) \cdot h = h_i.$ As the derivatives of $t \mapsto \sqrt{t}$ and $t \mapsto t^2$ are $\dfrac{1}{2\sqrt{t}}$ for $t \neq 0$ and $t \mapsto 2t$ one gets at once that the derivative of the norm $\displaystyle \|x\| = \sqrt{\sum_{i = 1}^d p_i(x)^2}$ is $\displaystyle\| x \|'\cdot h = \dfrac{1}{2 \|x\|} \sum_{i = 1}^d 2 p_i(x) p_i'(x) \cdot h =  \dfrac{1}{\|x\|} \sum_{i = 1}^d p_i(x) h_i = \dfrac{(x|h)}{\|x\|},$ where $(x|h)$ is the standard inner product in $\Bbb R^d.$
