I tried to find a derivation of the derivative of $\zeta(s)$ but I couldn't so I went ahead myself and computed it. The answer I obtain seems to be correct but I want to make sure that the computation is flawless. Can one of you confirm it for me? (or find the flaw(s))
\begin{align} \zeta(s)&=\sum_{n=1}^{\infty} \frac{1}{n^s}\nonumber\\ &=\sum_{n=1}^{\infty} \frac{1}{n^{x+iy}}\nonumber\\ &=\sum_{n=1}^{\infty} \frac{1}{n^xn^{iy}}\nonumber\\ \end{align} We can rewrite $n^{iy}$ as $e^{iy\text{log}(n)}$ and apply Euler's formula to obtain: \begin{align}\label{z4} \zeta(s)&=\sum_{n=1}^{\infty} \frac{1}{n^x\text{cos}(y\text{log}(n))+in^x\text{sin}(y\text{log}(n))} \end{align} \textbf{Corollary} Using definition of division in complex numbers, $$\frac{1}{x+iy}=\frac{x-iy}{x^2+y^2}$$ Therefore, equation \ref{z4} becomes, \begin{align} \zeta(s)=\sum_{n=1}^{\infty} \frac{\text{cos}(y\text{log}(n))}{n^x}-i\sum_{n=1}^{\infty} \frac{\text{sin}(y\text{log}(n))}{n^x}. \end{align} Now, since we have the zeta function in the form $f=u+iv$, we can check if $f$ is holomorphic on our domain: \begin{align}\label{a9} \frac{\partial u}{\partial x}&=-\frac{\text{log}(n)\text{cos}(y\text{log}(n))}{n^x} \end{align} \begin{align}\label{a10} \frac{\partial u}{\partial y}&=-\frac{\text{log}(n)\text{sin}(y\text{log}(n))}{n^x} \end{align} \begin{align}\label{a11} \frac{\partial v}{\partial x}&=\frac{\text{log}(n)\text{sin}(y\text{log}(n))}{n^x} \end{align} \begin{align}\label{a12} \frac{\partial v}{\partial y}&=-\frac{\text{log}(n)\text{cos}(y\text{log}(n))}{n^x}, \end{align}
Clearly Cauchy-Riemann equations are satisfied, hence the function is differentiable on the entire domain, $Re(s)>1$. The actual process of differentiation of the series is much easier. Since zeta is just $$\zeta(s)=1+\frac{1}{2^s}+\frac{1}{3^s}+...$$ and $$\frac{d}{dx}\left(\frac{1}{n^x}\right)=-\frac{\text{log}(n)}{k^x},$$ differentiating term by term we obtain \begin{equation} \frac{d\zeta(s)}{ds}=-\sum_{n=2}^{\infty} \frac{\text{log}(n)}{n^s}\text{, }Re(s)>1. \end{equation} Additionally, we can provide a result for a general nth derivative, \begin{equation} \frac{d^k\zeta(s)}{ds}=\left(-1\right)^k\sum_{n=2}^{\infty} \frac{\text{log}^k(n)}{n^s}\text{, }Re(s)>1. \end{equation}
Your work seems fine but $\zeta(s)$ has other representations as well that you could differentiate.
$$\zeta(s)=\prod_{p=primes}\frac{1}{1-p^{-s}}, Re(s)>1$$ $$log(\zeta(s))=-\sum_{p}log(1-p^{-s})$$ Differentiating: $$\frac{\zeta'(s)}{\zeta(s)}=-\sum_{p}\frac{log(p)p^{-s}}{1-p^{-s}}$$
Another representation of the Zeta Function that was derived by Riemann:$$\zeta(s)=2^s\pi^{s-1}sin(\frac{\pi s}{2})\Gamma(1-s)\zeta(1-s), Re(s)<1$$ $$log(\zeta(s))=slog(2)+(s-1)log(\pi)+log(sin(\frac{\pi s}{2}))+log(\Gamma(1-s))+log(\zeta(1-s))$$ Differentiating: $$\frac{\zeta'(s)}{\zeta(s)}=log(2)+log(\pi)+\frac{\pi}{2}cot(\frac{\pi s}{2})-\frac{\Gamma'(1-s)}{\Gamma(1-s)}-\frac{\zeta'(1-s)}{\zeta(1-s)}$$ Replacing $s$ with $1-s$: $$\frac{\zeta'(1-s)}{\zeta(1-s)}=log(2)+log(\pi)+\frac{\pi}{2}tan(\frac{\pi s}{2})-\frac{\Gamma'(s)}{\Gamma(s)}-\frac{\zeta'(s)}{\zeta(s)}$$
Replacing $\frac{\zeta'(s)}{\zeta(s)}$ with the Euler Product derived earlier:$$\frac{\zeta'(1-s)}{\zeta(1-s)}=log(2)+log(\pi)+\frac{\pi}{2}tan(\frac{\pi s}{2})-\frac{\Gamma'(s)}{\Gamma(s)}+\sum_{p}\frac{log(p)p^{-s}}{1-p^{-s}}$$ This is valid for $Re(s)>0$ and we finally have a derivative for zeta valid for negative values of zeta.
EDIT: Another representation of Zeta in terms of Stieltjes Constants that might be easier to differentiate and is valid for $z\neq1$: $$\zeta(s)=\frac{1}{s-1}+ \sum_{n=0}^\infty \frac{(-1)^n}{n!}\gamma_{n}(z-1)^n$$