Are all zeros of ζ^{k}(s)±ζ^{k}(1−s) on the critical line (k=k-th derivative)? The non-trivial zeros of $\zeta^{k}(s)$, with $k=k^{th}$ derivative, do not lie on a line and seem to be distributed randomly in the region $\sigma > \frac12$. However the non-real zeros in the critical strip of:
$$\zeta^{k}(s) \pm \zeta^{k}(1-s)$$
all appear to reside on the critical line (with maybe a finite number of exceptions lying outside the critical strip). Could this be proven with similar techniques as outlined here $\zeta(s)-\zeta(1-s)$ ?
The reason I ask is that Speiser(1934), Levinson & Montgomery (1974) and recently Yildirim have proven that assuming RH, $\zeta^{1}(s)$, $\zeta^{2}(s)$ and $\zeta^{3}(s)$ have no zeros in $0 < \Re(s) < \frac12$, but also that the number of zeros of $\zeta^{k}(s)$ residing in the region $\Re(s) < \frac12$, must be finite (there is actually only one pair found for $\zeta^{2}(s)$ and $\zeta^{3}(s)$ in $\Re(s) < 0$).   
Now suppose $k=1..3$ and it can indeed be proven that all zeros of $\zeta^{k}(s) \pm \zeta^{k}(1-s)$ must lie on the critical line, then the only possibility for a zero of $\zeta^{k}(s)$ to hide in $0 < \Re(s) < \frac12$ (and thereby falsifying the RH), is when $\zeta^{k}(s)=\zeta^{k}(1-s)=0$. This then immediately raises the second question on whether contrary to $\zeta(s)$ and the absence of a reflexive functional equation for its derivatives, it could be shown that when $\zeta^{k}(s)$ is a zero, $\zeta^{k}(1-s)$ cannot be one?   
 A: Just to share what I have found about the second part of my question.
From Tom Apostol's paper I derived the following formula for $\zeta^{(1)}(1-s)$:
$$A(s)= \Gamma  \left( s \right) \cos \left(\frac12\pi s \right) 2 \left( 2\pi \right) ^{-s}$$
$$\zeta^{(1)}(1-s) = A(s)\left( \zeta  \left( s \right) \left( \ln 
 \left( 2\pi \right) +\frac12\pi \tan \left(\frac12\pi s \right) 
  -\Psi \left( s \right)\right)  -\zeta^{(1)}(s)  \right)$$
It is easy to isolate $\zeta(s)$ and to obtain a closed form related to its derivatives:
$$\zeta(s) = \frac{\frac {\zeta^{(1)}(1-s)}{A(s)} +\zeta^{(1)}(s)} {\ln \left( 2\pi \right) +\frac12\pi \tan \left(\frac12\pi s \right)  -\Psi(s)}$$
When $s=\rho$, and knowing that $\zeta(\rho) = \zeta(1-\rho)$ then from:
 $$\zeta(\rho)=\frac{\zeta^{(1)}(1-\rho)}{A(\rho)} +\zeta^{(1)}(\rho)=0$$ 
it follows that when $\zeta^{(1)}(1-\rho)=\zeta^{(1)}(\rho)=0$, then $\zeta(\rho) = \zeta(1-\rho) =0$. This obviously doesn't answer my second question, but maybe does provide a small clue.
Since it is also true that $\zeta(s) A(s) = \zeta(1-s)$, the following equation can be derived:
$$\frac {\zeta^{(1)}(1-\rho)}{A(\rho)} +\zeta^{(1)}(\rho)=\zeta^{(1)}(1-\rho)+A(\rho)\zeta^{(1)}(\rho)$$
that can be simplified into:
$$\frac{\zeta^{(1)}(1-\rho)}{\zeta^{(1)}(\rho)} + A(\rho)=0$$
This equation correctly reproduces all the known $\rho$, but also adds a single new (actually quite natural) one at $\rho'=\frac12 \pm 6.2898359888369027796...$. This is a value I did encounter before see other question and I believe it is the point where:
$$\displaystyle \lim_{\sigma \to \frac12} |\dfrac{\zeta(\sigma+ti)}{\zeta(1-(\sigma+ti))}|=1$$
