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==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]].

==Proof==

==References==

[[Category:Theorem]]
[[Category:Unproven]]

Return to C n^(lambda)'(x)=2lambda C (n+1)^(lambda+1)(x).

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