\[\operatorname{Ai}'\left(z\right)=-\pi^{-1}(z/\sqrt{3})K_{\pm 2/3}\left(\zeta%
\right)=(z/3)\left(I_{2/3}\left(\zeta\right)-I_{-2/3}\left(\zeta\right)\right)%
=\tfrac{1}{2}(z/\sqrt{3})e^{-\pi i/6}{H^{(1)}_{2/3}}\left(\zeta e^{\pi i/2}%
\right)=\tfrac{1}{2}(z/\sqrt{3})e^{-5\pi i/6}{H^{(1)}_{-2/3}}\left(\zeta e^{%
\pi i/2}\right)=\tfrac{1}{2}(z/\sqrt{3})e^{\pi i/6}{H^{(2)}_{2/3}}\left(\zeta e%
^{-\pi i/2}\right)=\tfrac{1}{2}(z/\sqrt{3})e^{5\pi i/6}{H^{(2)}_{-2/3}}\left(%
\zeta e^{-\pi i/2}\right),\]