Derivative of prime zeta

Theorem

The following formula holds: $$P'(z)=-\displaystyle\sum_{p \mathrm{\hspace{2pt} prime}} \dfrac{\log(p)}{p^z},$$ where $P$ denotes the prime zeta function and $\log$ denotes the logarithm.

Proof

Recall the definition of the prime zeta function: $$P(z)=\displaystyle\sum_{p \hspace{2pt} \mathrm{prime}} \dfrac{1}{p^z}.$$ Rewrite $$\dfrac{1}{p^z} = p^{-z} = e^{-z \log(p)}.$$ Differentiate term-by-term using the derivative of the exponential function and the chain rule: $$\begin{array}{ll} \dfrac{\mathrm{d}}{\mathrm{d}z} P(z) &= \displaystyle\sum_{p \hspace{2pt} \mathrm{prime}} \dfrac{\mathrm{d}}{\mathrm{d}z} \dfrac{1}{p^z} \\ &= \displaystyle\sum_{p \hspace{2pt} \mathrm{prime}} \dfrac{\mathrm{d}}{\mathrm{d}z} e^{-z \log(p)} \\ &= \displaystyle\sum_{p \hspace{2pt} \mathrm{prime}} e^{-z\log(p)} (-\log(p)) \\ &= -\displaystyle\sum_{p \hspace{2pt} \mathrm{prime}} \dfrac{\log(p)}{p^z}, \end{array}$$ as was to be shown.