 Boundary-distance derivative bound for a conformal bijection

For a conformal bijection $f:D\to \tilde{D}$ between proper planar domains, put $d=\mathrm{dist}(z,\partial D)$ and $\tilde{d}=\mathrm{dist}(f(z),\partial \tilde{D})$. Apply the Koebe quarter theorem to $(f(z+dw)-f(z))/(d{f}^{\prime }(z))$ on the unit disc to obtain $d|f^{\prime }(z)|/4\le \tilde{d}$. Apply it again to $f^{-1}$ to get the opposite bound. Thus $\boxed{\tilde{d}/(4d)\le |f^{\prime }(z)|\le 4\tilde{d}/d}$. The domains need not be simply connected domains. For the whole plane, the boundary distances are infinite and these ratios are undefined.