How to differentiate the functional equation of $\zeta(s)$

Question

Given the functional equation $$ \zeta\left(s\right) = 2^{s}\pi^{s - 1}\sin\left(\pi s \over 2\right) \Gamma\left(1 - s\right)\zeta\left(1 - s\right) $$ we can differentiate it at $s = -2k$, $k$ integer. Then we show that the derivative is non zero at those points to show that the trivial zeros are simple $\left(~\mbox{order}\ 1\right)$. How do I differentiate it ?. I guess, product rule will work, but what is the derivative of $\zeta\left(1 - s\right)$. I feel likt it is a circular argument.

Answer 1

$$\zeta'(s)=2^s\pi^{s-1}[\log 2\pi \sin(\pi s/2) \Gamma(1-s) \zeta(1-s)+\color{red}{\frac\pi2\cos\frac{\pi s}2\Gamma(1-s)\zeta(1-s)}- $$

$$-\sin(\pi s/2) \Gamma'(1-s) \zeta(1-s)- \sin(\pi s/2) \Gamma(1-s) \zeta'(1-s)]$$

so if $\;s=-2k\,,\,\,k\in\Bbb Z\;$ ,we get:

$$\zeta(-2k)=2^{-2k}\pi^{-2k-1}\frac\pi2\cos\pi k\,\Gamma(1+2k)\zeta(1+2k)=(-1)^k\,2^{1-2k}\,\pi^{-2k}(2k)!\,\zeta(1+2k)$$

and we're done as $\;\zeta(s)\neq0\;$ for $\;s>1\;$ (all the time real $\;s$)

Answer 2

The trivial zeroes of the $\zeta$ function are simple since all the zeroes of the sine function are simple and $2^{1-s},\pi^{-s},\Gamma(s)$ are positive and log-convex functions on $\mathbb{R}^+$: there is no need for explicit differentiation.