Backpropagation derivation using Leibniz notation

The cost function $C$ depends on $w_{jk}^l$ only through the activation of the $j$th neuron in the $l$th layer, i.e. on the value of $a_j^l$. Thus we can use the chain rule to expand:

$\frac{\partial C}{\partial w_{jk}^l}=\frac{\partial C}{\partial a_j^l}\frac{\partial a_j^l}{w_{jk}^l}$

We know that $\frac{\partial a_j^l}{w_{jk}^l}={\sigma }^{\prime }(z_j^l)a_k^{l-1}$ because $a_j^l=\sigma ({\sum }_k{w}_{jk}^l{a}^{l-1}+b_j^l)$.

In turn, $C$ depends on $a_j^l$ only through the activations of the $(l+1)$th layer.