Proof that derivative of Hurwitz Zeta by the first argument is not expressable in terms of Hurwitz Zeta

The set of elementary functions is defined so that it to be closed against operation of differentiation. It is also evidently close against discrete differentiation.

In the discrete calculus there is a similar set of functions, but with one sufficient difference. It appears not to be closed against normal (non-discrete) differentiation.

But I need a proof.

So I am asking for a proof for the following statement regarding Hurwitz Zeta: $$\frac{d}{dq}\zeta(q,p)$$ cannot be expressed in terms of elementary functions and Hurwitz Zeta.

UPDATE

I found the following formula which connects the two functions, but still a question remains whether one of them can be expressed explicitly.

$\zeta '\left(z,\frac{q}{2}\right)-2^z \zeta '(z,q)+\zeta '\left(z,\frac{q+1}{2}\right)=\zeta(z,q)2^{z}\ln 2$

• What is the "similar set of functions" in discrete calculus that plays the role of elementary functions but is not closed under differentiation? – Henry Cohn Apr 7 '12 at 12:55
• @Henry Cohn, Hurwitz Zeta, generalized Bernoulli polynomials, polylogarithm, polygamma. All expressable through each other. – Anixx Jul 7 '15 at 13:27
• Definition... $$\zeta(s,q) := \sum_{n=0}^\infty\frac{1}{(q+n)^s}$$ with analytic continuation. en.wikipedia.org/wiki/Hurwitz_zeta_function – Gerald Edgar Jul 7 '15 at 15:03