# Derivative of Riemann Zeta Function

## Theorem

The derivative of the Riemann zeta function is:

- $\displaystyle \zeta' \left({z}\right) = \frac{\mathrm d \zeta}{\mathrm dz} = -\sum_{n \mathop = 2}^\infty \frac{\ln \left({n}\right)}{n^z}$

## Proof

\(\displaystyle \frac{\mathrm d \zeta}{\mathrm dz}\) | \(=\) | \(\displaystyle \frac{\mathrm d}{\mathrm dz} \left({\sum_{n \mathop = 1}^\infty n^{-z} }\right)\) | $\quad$ | $\quad$ | |||||||||

\((1):\quad\) | \(\displaystyle \) | \(=\) | \(\displaystyle \sum_{n \mathop = 1}^\infty \frac{\mathrm d}{\mathrm dz} \left({n^{-z} }\right)\) | $\quad$ | $\quad$ | ||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \sum_{n \mathop = 1}^\infty \left({- \ln \left({n}\right) n^{-z} }\right)\) | $\quad$ from Derivative of Exponential Function | $\quad$ | |||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle -\sum_{n \mathop = 1}^\infty \frac{\ln \left({n}\right)}{n^z}\) | $\quad$ | $\quad$ | |||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle -\sum_{n \mathop = 2}^\infty \frac{\ln \left({n}\right)}{n^z}\) | $\quad$ as $\ln 1 = 0$ | $\quad$ |

$\blacksquare$