TITLE: Closed form for derivatives $\zeta^{(n)}(1/2)$
QUESTION [7 upvotes]: According to mathworld
 41,42. "Derivatives $\zeta^{(n)}(1/2)$ can also be given in closed form"
 with example for the first derivative.
What is the closed form? References?
The motivation is that this question
expresses $\zeta(3)$ in terms of $\zeta(1/2)$ and the first 3 derivatives,
so closed form possibly might result in closed form for zeta(3)
(unless the closed form is derived by the linked question).
Particaluraly intersted in the second and third derivatives.
On what the derivatives would depend helps too.

REPLY [3 votes]: Edit: My original answer was incorrect.
You can evaluate $\zeta'(\frac{1}{2})$ recursively in terms of $\zeta(\frac{1}{2})$ using the symmetric form of the functional equation:
$$ \zeta(s)\Gamma(\tfrac{s}{2}) \pi^{-s/2} = \zeta(1{-}s)\Gamma(\tfrac{1-s}{2}) \pi^{(s-1)/2}. $$
Differentiating both sides sides of the equation, plugging in $s=\frac{1}{2}$, and then solving for $\zeta'(\frac{1}{2})$, I get the value listed on the MathWorld website.
As Noam Elkies points out, taking higher derivatives, this process allows you to write $\zeta^{(2n+1)}(\frac{1}{2})$ in terms of the smaller even derivatives $\zeta(\frac{1}{2}),\zeta''(\frac{1}{2}), \zeta^{(4)}(\frac{1}{2}), \ldots, \zeta^{(2n)}(\frac{1}{2})$.