Let $B(0,1)$ be the open unit ball in Euclidean space $\mathbb R^2$ and $(a_n)_{n=1}^\infty$ be a dense subset of $B(0,1)$.
I wish to show for fixed $s\in (0,1)$ the function $$ f(x)=\sum_{n=1}^\infty \frac{1}{2^n \|x-a_n \|^s} \textrm{ for } x\in B(0,1) $$ has the weak partial derivatives of the first order $\partial_i f$.
Notice that the function $$ u(x) = \frac{1}{|x|^s} $$ is in $W^{1,p}$ for all $p<2/(s+1)$ (if i'm not wrong!). Every summand in the series has norm bounded by $||u||/2^n$ and hence the entire sum $f$ is in $W^{1,p}$. So $f$ has weak derivatives.