Consider the function $f :z\mapsto (z^2 -1)^{ \frac12}$. Now Here proved that it has branch such that $f$ is analytic in $|z|>1$.Where branch cut is $[-1,1]$.
Now f is composition of two function namely
$$ G : z \mapsto z^2-1$$ and $$ K: z\mapsto z^{ \frac12}$$ Now $G\left([-1,1]\right)=[-1,0]$ .If we choose any branch of $z^{ \frac12}$ then we have discontinuity on some ray ( say $\alpha$ ) joining 0 and $\infty$. which is certainly larger (i.e never exact equal to [-1,0]) than [-1,0]. Now $f$ is differentiable at $z_0$ if $G$ is differentiable at $z_0$ and $K$ is differentiable at $G(z_0)$.
So my question is How $f$ can be differentiable on point $z$ such that $|z|>1$ and $G(z)$ lies on ray $\alpha$ .
One way to approach this is to notice that $z^2-1 = (z-1)(z+1)$ so that you can write the multivalued function $f=(z^2-1)^{1/2}$ as the product $(z-1)^{1/2}(z+1)^{1/2}$, and hence obtain a branch of $f$ by choosing a branch of $f_1 = (z-1)^{1/2}$ and $f_2 = (z+1)^{1/2}$. As you cross the branch cut for your choice of $f_1$ it will change sign (and hence be discontinuous) but if your cuts are arranged to overlap, then crossing a cut for a branch of $f_1$ and $f_2$ will change the sign by $(-1)^2$, that is, the product of the branches will be continuous even though each branch by itself has a discontinuity.