Proving that norm function is continuously differentiable Let $B:=\mathbb{R}^n$. Consider the function $f:B\backslash\{0\} \to \mathbb{R}$ defined as $f(x)=\|x\|$. I want to prove that $f$ is continuously differentiable on $B$. One way is to use single-variable calculus and find the general partial derivative of $f$ on $B$ explicitly and then observe that it is continuous on $B$. But I decided to use the definition of the directional derivative and use $f^2(x)=\|x\|^2$ instead. This gives that $\partial_i f(x)=2x_i$ for $1\le i \le n$.
But does the fact that $f^2$ has a continuous partial derivative also imply that $f$ has it? The issue is that for $x=\vec{0}$, we can't observe the fact that $f$ has no continuous partial derivative at this point, but $f^2$ does have one.
I'd appreciate if someone could please clarify this matter for me.
 A: Compositions of $C^\infty$ functions are $C^\infty.$ So if $U\subset \mathbb R^n$ is open, $g:U \to (0,\infty),$ and $g\in C^\infty(U),$ then $g^{1/2}\in C^\infty(U).$ This is because $t\to t^{1/2}$ is in $C^\infty((0,\infty)),$ which implies $g^{1/2}$ is the composition of two $C^\infty$ functions. For your particular case we would take $g(x) = |x|^2, U = \mathbb R^n \setminus \{0\}.$
A: Let $C^1$ denote "continuously differentiable."
The sum or product of any finite set of $C^1$ functions from $\mathbb R^n$ \ $\{0\}$  to $\mathbb R$  is also  $C^1.$
If $f:\mathbb R^n$ \  $\{0\}\to \mathbb R$ is $C^1$ and the image of $f$ is $\mathbb R^+$, and if $g:\mathbb R^+\to \mathbb R$ is  $C^1$ then the composite  function $gf$  is $C^1.$ 
For  any fixed $\in \{1,...,n\}$ and for $x=(x_1,...x_n)\in \mathbb R^n$ \ $\{0\}$ let $P_j(x)=x_j.$  
All you have to do is prove that each $P_j$ is $C^1$.
Then each product $f_j(x)=P_j(x)\cdot P_j(x) =x_j^2$ is $C^1.$ 
Then the sum  $f(x)=\sum_{j=1}^nf_j(x)=\sum_{j=1}^nx_j^2$ is $C^1.$
Now $g(y)=\sqrt y\;$ is $C^1$ from $\mathbb R^+$ to $\mathbb R.$ Therefore $\|x\|=g(f(x))$ is $C^1$ on $\mathbb R^n$ \ $ \{0\}.$
A: The norm is Lipschitz with rank 1 hence is differentiable almost everywhere (this is not trivial to prove).
However, in general a norm is not differentiable at every $x \neq 0$.
For example, $f(x) = \|x\|_\infty = \max(|x_1|,...,|x_n|)$ is not differentiable whenever there are two or more indices for which the $\max$ is attained.
