weak differentiability of log log function I want to understand why the following function has a weak derivative in two or three dimensions:
$w(x) = \ln |\ln|x|| , x \in B_{1/2}(0)$. 
Can I say that if I have a strong derivative (except for the point 0), then it is shown? If yes, why is the dimension important?
I computed the following
derivative $\nabla w(x) = \frac{x}{(\ln|x|) |x|^2}$. Is it right?
Thanks a lot!
The definition of a weak derivative is: 
Let $w \in L^1_{\text{loc}}$. Then $u \in L^1_{\text{loc}}(\Omega)$ is called partial weak derivative if $\forall \phi \in D(\Omega)$:
$\int_\Omega w \partial_i \phi  = -\int_\Omega \partial_i u \phi $.
If this holds for all partial derivatives, $u$ is called weak derivative.
 A: The double log has a weak derivative in all dimensions $\ge 2$. The restriction to $n=2,3$ is not necessary.

Can I say that if I have a strong derivative (except for the point 0), then it is shown?  

Depends on what you mean by strong derivative (there is such a term for differentiation of $L^p$ functions); I think you meant the pointwise derivatives, in which case the answer is no. Having a pointwise derivative, even at every point, is not enough to conclude that the weak derivative exists. 
I will outline a different approach: write $w$ as an $L^1$ limit of  functions $w_n$ for which  the $x_i$  partial derivative, denoted $u_{i,n}$, exists, and converges in $L^1$ as $n\to\infty$. Then you can pass to the limit on both sides in $$\int_\Omega w_n \partial_i \phi  = -\int_\Omega   u_{i,n} \phi$$
A natural choice for $w_n$ is truncation $w_n=\min(w,n)$. To use it, you  need to know that Lipschitz functions have weak derivatives. But the convergence part is straightforward.
Otherwise, you can use $w_n=\eta_n w$ where $\eta_n(x)=\eta(2^n x)$ and $\eta$ is a smooth   function on $\mathbb R^d$ that vanishes when $|x|<1/2$ and is $1$ for $|x|>1$. Then there are longer estimates for establishing convergence of derivatives in $L^1$.  
