# Derivative of Riemann Zeta Function

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## Theorem

The derivative of the Riemann zeta function is:

- $\displaystyle \map {\zeta'} z = \frac {\d \zeta} {\d z} = -\sum_{n \mathop = 2}^\infty \frac {\map \ln n} {n^z}$

## Proof

\(\displaystyle \frac {\d \zeta} {\d z}\) | \(=\) | \(\displaystyle \map {\frac \d {\d z} } {\sum_{n \mathop = 1}^\infty n^{-z} }\) | |||||||||||

\((1):\quad\) | \(\displaystyle \) | \(=\) | \(\displaystyle \sum_{n \mathop = 1}^\infty \map {\frac \d {\d z} } {n^{-z} }\) | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \sum_{n \mathop = 1}^\infty \paren {-\map \ln n n^{-z} }\) | Derivative of Exponential Function | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle -\sum_{n \mathop = 1}^\infty \frac {\map \ln n} {n^z}\) | |||||||||||

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

$\blacksquare$