Well, this is simply by taking logarithms on either side of the Euler product representation to get,
\log{\zeta(s)} = -\sum_{p} \log{(1 - p^{-s})}
where p[/tex] is the set of primes.<br />
Differentiating then gives,<br />
<br />
\frac{\zeta&#039;(s)}{\zeta(s)} = -\sum_{p} (p^{s} - 1)^{-1} \log{p}<br />
<br />
This gives the zeta-regularized sum (and hence product) on primes (which looks curious as it is, unless special values of s[/tex] are used), but generally its more convenient to consider,<br />
<br />
\prod_{n} \lambda_{n} = \exp{-\zeta_{\lambda}&amp;#039;(0)}<br />
<br />
for a zeta function defined on a sequence (\lambda_{n})_{n \geq 1}[/tex].&lt;br /&gt;
&lt;br /&gt;
If its a prime regularization you&amp;#039;re after, look for this paper by Munoz Garcia and Perez Marco called &amp;#039;Super Regularization of Infinite Products&amp;#039;.&lt;br /&gt;
&lt;br /&gt;
Never mind, here&amp;#039;s the link to the preprint pdf-&lt;br /&gt;
http://inc.web.ihes.fr/prepub/PREPRINTS/M03/M03-52.pdf
