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:

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

We know that ${\frac {\partial a_{j}^{l}}{\partial w_{jk}^{l}}}=\sigma '(z_{j}^{l})a_{k}^{l-1}$ because $a_{j}^{l}=\sigma (z_{j}^{l})=\sigma \left(\sum _{k}w_{jk}^{l}a_{k}^{l-1}+b_{j}^{l}\right)$. We have used the chain rule again here.

In turn, $C$ depends on $a_{j}^{l}$ only through the activations of the $(l+1)$th layer. Thus we can write (using the chain rule once again):

$${\frac {\partial C}{\partial a_{j}^{l}}}=\sum _{i=1}^{n}(l+1){\frac {\partial C}{\partial a_{i}^{l+1}}}{\frac {\partial a_{i}^{l+1}}{\partial a_{j}^{l}}}$$

where $n(l+1)$ is the number of neurons in the $(l+1)$th layer.

Backpropagation works recursively starting at the later layers. Since we are trying to compute ${\frac {\partial C}{\partial a_{j}^{l}}}$ for the $l$th layer, we can assume inductively that we have already computed ${\frac {\partial C}{\partial a_{i}^{l+1}}}$.

It remains to find ${\frac {\partial a_{i}^{l+1}}{\partial a_{j}^{l}}}$. But $a_{i}^{l+1}=\sigma (z_{i}^{l+1})=\sigma \left(\sum _{j}w_{ij}^{l+1}a_{j}^{l}+b_{i}^{l+1}\right)$ so we have

$${\frac {\partial a_{i}^{l+1}}{\partial a_{j}^{l}}}=\sigma '(z_{i}^{l+1})w_{ij}^{l+1}$$

Putting all this together, we obtain

$${\begin{aligned}{\frac {\partial C}{\partial w_{jk}^{l}}}&={\frac {\partial C}{\partial a_{j}^{l}}}{\frac {\partial a_{j}^{l}}{\partial w_{jk}^{l}}}\\&=\left(\sum _{i\in \{1,\ldots ,n(l+1)\}}{\frac {\partial C}{\partial a_{i}^{l+1}}}{\frac {\partial a_{i}^{l+1}}{\partial a_{j}^{l}}}\right)\sigma '(z_{j}^{l})a_{k}^{l-1}\\&=\left(\sum _{i\in \{1,\ldots ,n(l+1)\}}{\frac {\partial C}{\partial a_{i}^{l+1}}}\sigma '(z_{i}^{l+1})w_{ij}^{l+1}\right)\sigma '(z_{j}^{l})a_{k}^{l-1}\end{aligned}}$$