I want to find an upper bound for $\zeta'(s)$ along a vertical line $\Re(s)=b$, where $1<b<0$.
One way to do this is using $$\frac{\zeta'(b+iT)}{\zeta(b+iT)}=O_b(\log T)$$ and $$\zeta(b+iT)=O_{b,\varepsilon}(T^{1/2b+\varepsilon})$$ for each $\varepsilon>0$.
Multiplying them gives us $$\zeta'(b+iT)=O_{b,\varepsilon}(T^{1/2b+\varepsilon}\log T)$$ as $T\rightarrow\infty$. I want to know if there is a bound, for fixed $b$, that is sharper.

2$\begingroup$ The functional equation expresses $\zeta'(b+iT)$ in terms of $\zeta(1biT)$, $\zeta'(1biT)$, and the complex Gamma function and its derivative on the lines of real part $b$ and $1b$. If $1<b<0$ then $1 < 1b < 2$ so $\zeta(1biT)$ and $\zeta'(1biT)$ are bounded and it's just a matter of the estimating the relevant Gamma and $\Gamma'$ factors. $\endgroup$– Noam D. ElkiesMar 10 '14 at 3:34

7$\begingroup$ Use the functional equation connecting $\zeta(b+iT)$ to $\zeta(1biT)$ and differentiate. Then use Stirling's formula. This gives the bound $\zeta^{\prime}(b+it)= O_b(T^{1/2b}\log T)$ (your expression seems to have a typo). This is also best possible since $\zeta(1biT)$ is bounded away from zero. $\endgroup$– LuciaMar 10 '14 at 3:34

$\begingroup$ Yes, you are correct about the typo, so I fixed that. Thanks for letting me know that this is the best possible estimate. $\endgroup$– B.W.Mar 10 '14 at 3:40

$\begingroup$ Here you can find some bounds math.univlille1.fr/~ramare/TMEEMT/Articles/Art06.html $\endgroup$– user21574Mar 10 '14 at 13:27
I'm not sure if you need me to go further, but if I were you I'd start out with the functional equation. Take the derivative of both sides of the functional equation and you can derive:
$\frac{\zeta'(1s)}{\zeta(1s)} = log(2\pi)+\frac{\pi}{2}\tan(\frac{\pi s}{2})  \psi(s)\frac{\zeta'(s)}{\zeta(s)}$.
Letting $s = 1biT$, the left hand side will be exactly what you're looking at.
I am some what confused. You say the vertical line where $Re(s)=b$ where $0<b<1$. If it's a vertical line, then wouldn't $T$ vary and $b$ be held fixed?
If $T$ is allowed to vary (i.e. a vertical line), then wouldn't the answer depend on whether the line crosses the real axis? I believe $\tan(\frac{\pi}{2}s)$ blows up in magnitude when $b$ is really small in absolute value.
If $T$ is held fixed, and you allowed $b$ to vary, then believe the $\tan$ term is bounded since you'd be a fixed distance away from the problem point $b=0$. If this is the case, Lucia is correct I believe.
Perhaps I'm wrong...it is awful late and my mind is worn out from all day of flying across the country. I apologize if I made a mistake somewhere in my logic.
If I am wrong in my logic, could someone explain? I'm very interested as well.