Can derivative of Hurwitz Zeta be expressed in Hurwitz Zeta? Can the derivative of Hurwitz Zeta function by the first argument be expressed in terms of Hurwitz Zeta and elementary fuctions?
There is a formula which expresses Hurwitz Zeta through its derivative:
$$\zeta '\left(z,\frac{q}{2}\right)-2^z \zeta '(z,q)+\zeta '\left(z,\frac{q+1}{2}\right)=\zeta(z,q)2^{z}\ln 2$$
So I wonder whether the opposite can be done?

Note that the derivative of the Hurwitz Zeta Function can be calculated as following (similiar to calculating the derivative of the Riemann Zeta):
\begin{align} \frac{d}{ds} \zeta(s,q) &= \frac{d}{ds} \sum_{n=0}^{\infty} (n+q)^{-s} \\ &=  \sum_{n=0}^{\infty}  \frac{d}{ds} (n+q)^{-s} \\ & = \sum_{n=0}^{\infty}  - \ln(n+q) (n+q)^{-s} \\ &=  -\sum_{n=0}^{\infty}  \frac{\ln(n+q)}{(n+q)^s} \end{align}
Because $\ln(n+q)-\ln(n+q-1) = \ln\left(\frac{n+q}{n+q-1}\right)$, we actually have:
$$\zeta'(s,q) = - \ln(q) \zeta(s,q) + \sum_{n=1}^{\infty} \ln\left(\frac{n+q-1}{n+q}\right) \zeta(s, n+q)$$
I doubt a nicer relation exists. 
