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