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 ${\frac {\partial a_{j}^{l}}{w_{jk}^{l}}}=\sigma '(z_{j}^{l})a_{k}^{l-1}$ because $a_{j}^{l}=\sigma \left(\sum _{k}w_{jk}^{l}a^{l-1}+b_{j}^{l}\right)$.

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