Hello, I asked this question before, but didn't get any response, so I took the liberty of asking once again , with slightly modified version of the question:

Consider an orientation-preserving quasiconformal homeomorphism $f$ of the open unit disk $\mathbb{D}\subset \mathbb{C}$ with the complex dilatation/Beltrami cofficient $\mu, ||\mu||_{L^\infty(\mathbb{D})}<1, \mu \in C^0(\mathbb{\bar{D}})$ , i.e. $\mu$ is continuous on the closed unit disk $\mathbb{\bar{D}}$. Is it true that the restriction of $f$ on the boundary $S^1$ has continuous (ordinary) derivative on $S^1$, i.e., is $f\in C^1(S^1)$ ?

I guess one might start with continuously extending $\mu$ to all of $\mathbb{C}$, then solve the Beltrami equation on $\mathbb{C} $ with that extended $\mu$, but then I am getting stuck:because I guess the solution to this new equation might not be $C^1(\mathbb{C})$ ?? ( Look at Examples 15.1 in the book "Elliptic PDE and Quasiconformal Mappings" by K. Asltala, T. Iwaniec and G. Martin.

http://books.google.com/books?id=5aOgM9XRiXUC&printsec=frontcover&hl=sl#v=onepage&q&f=false

May be, to start with, we can ALSO assume that $\mu=0$on $S^1$. Then is the answer to my question yes ?

Could you please cite a reference of your proof to my question or give a counterexample ? Thanks a lot !!