 Squared-error backpropagation

For squared loss $E={\sum }_k(t_k-z_k{)}^2/2$, let $e_k=(t_k-z_k)g_k^{\prime }(x_k)$ and $e_j=g_j^{\prime }(x_j){\sum }_k{w}_{jk}{e}_k$. The chain rule gives $\partial E/\partial w_{jk}=-e_k{z}_j$ and $\partial E/\partial w_{ij}=-e_j{z}_i$. A gradient descent step of size $\eta$ uses $\Delta w_{jk}=\eta e_k{z}_j$ and $\Delta w_{ij}=\eta e_j{z}_i$, with every derivative evaluated at the same pre-update weights.