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