# Derivative of Riemann Zeta Function

$\displaystyle \zeta' \left({z}\right) = \frac{\mathrm d \zeta}{\mathrm dz} = -\sum_{n \mathop = 2}^\infty \frac{\ln \left({n}\right)}{n^z}$
 $\displaystyle \frac{\mathrm d \zeta}{\mathrm dz}$ $=$ $\displaystyle \frac{\mathrm d}{\mathrm dz} \left({\sum_{n \mathop = 1}^\infty n^{-z} }\right)$ $(1):\quad$ $\displaystyle$ $=$ $\displaystyle \sum_{n \mathop = 1}^\infty \frac{\mathrm d}{\mathrm dz} \left({n^{-z} }\right)$ $\displaystyle$ $=$ $\displaystyle \sum_{n \mathop = 1}^\infty \left({- \ln \left({n}\right) n^{-z} }\right)$ from Derivative of Exponential Function $\displaystyle$ $=$ $\displaystyle -\sum_{n \mathop = 1}^\infty \frac{\ln \left({n}\right)}{n^z}$ $\displaystyle$ $=$ $\displaystyle -\sum_{n \mathop = 2}^\infty \frac{\ln \left({n}\right)}{n^z}$ as $\ln 1 = 0$
$\blacksquare$