Carrying forward the summation symbol when differentiating $E = \dfrac{1}{2} \sum (z - t)^2 $?

Question

My lecture notes state the following:

$z = \dfrac{1}{1 + e^{-s}}$

$E = \dfrac{1}{2} \sum (z - t)^2$

$\therefore \dfrac{ \partial{E} }{ \partial{s} } = (z - t)z(1 - z)$

In this case, $E$ is an error function.

I have calculated that $\dfrac{ \partial{E} }{ \partial{s} } = \dfrac{ \partial{E} }{ \partial{z} } \dfrac{ dz }{ ds } = \left[ \sum (z - t) \right]z(1 - z)$

And so I have two questions that I would appreciate clarification on:

1. Am I correct in saying that $\dfrac{ \partial{E} }{ \partial{s} } = \dfrac{ \partial{E} }{ \partial{z} } \dfrac{ dz }{ ds }$?

2. Why doesn't the summation symbol carry forward?