I need to prove the following:
$$\Omega = \mathbb{Z}\omega +\mathbb{Z}\omega': $$
$$ \wp'(z)= \frac{\sigma(z-\frac{\omega}{2})\sigma(z-\frac{\omega'}{2}) \sigma(z+\frac{\eta}{2})}{\sigma(z)^{3}\sigma(\frac{\omega}{2})\sigma(\frac{\omega'}{2})\sigma(-\frac{\eta}{2})} $$
where $\eta = \omega + \omega'$ and $\wp$ is the Weierstrass p function
I know the following facts:
$$ \zeta(z)= \frac{\sigma'(z)}{\sigma(z)} $$
$$ = \frac{1}{z} + \Sigma_{\omega \in \Omega} (\frac{1}{z-\omega}+\frac{1}{\omega}+\frac{z}{\omega^{2}}) $$
and
$$ \wp'(z) = -\frac{2}{z^{3}}- \Sigma'_{\omega \in \Omega} \frac{2}{(z-\omega)^{3}} $$
But, how do I get from here to what I want to prove?