Problem in understanding the partial differentiation I was reading the book on Neural Network and got stuck up on this equation given below.
\begin{eqnarray}
  z^{l+1}_k = \sum_j w^{l+1}_{kj} a^l_j +b^{l+1}_k = \sum_j w^{l+1}_{kj} \sigma(z^l_j) +b^{l+1}_k.
\tag{43}\end{eqnarray}
here w and b are constant vectors. The equation was then differentiated and below is the result. Where did the summation go?
\begin{eqnarray}
  \frac{\partial z^{l+1}_k}{\partial z^l_j} = w^{l+1}_{kj} \sigma'(z^l_j).
\tag{44}\end{eqnarray}
If you want to see the complete problem then visit http://neuralnetworksanddeeplearning.com/chap2.html eqn 43 and 44

The $z_k$'s are defined over a sum of $z_j$'s.
$$
z^{l+1}_k = \sum_j w^{l+1}_{kj} a^l_j +b^{l+1}_k = \sum_j w^{l+1}_{kj} \sigma(z^l_j) +b^{l+1}_k.
$$
You are differentiating with respect to only one of the $z_j$'s. So, for each $j$, all the other terms in the sum go to zero: 
$$
\begin{align}
  \frac{\partial z^{l+1}_k}{\partial z^l_j} &=  \frac{\partial }{\partial z^l_j} \left[\sum_j w^{l+1}_{kj} \sigma(z^l_j) +b^{l+1}_k\right]\\
&= \sum_j \frac{\partial }{\partial z^l_j}\left[w^{l+1}_{kj} \sigma(z^l_j) +b^{l+1}_k\right]\\
&= w^{l+1}_{kj} \sigma'(z^l_j).
\tag{44}\end{align}
$$
