C n^(lambda)'(x)=2lambda C (n+1)^(lambda+1)(x)
From specialfunctionswiki
Theorem
The following formula holds: $$\dfrac{\mathrm{d}}{\mathrm{d}x} C_n^{\lambda}(x)=2\lambda C_{n+1}^{\lambda+1}(x),$$ where $C_n^{\lambda}$ denotes Gegenbauer C.
