Past exam of the mathematics course of the University of Cambridge / 2006 / iii / Paper 84 / 2 / a / Solution

Apply the chain rule to the squared loss, using the same pre-update weights throughout a forward and backward pass. At the output, $\partial E/\partial z_k=z_k-t_k$ and $\partial z_k/\partial x_k=g_k^{\prime }(x_k)$. ThereforeFor a gradient descent step of size $\eta >0$, define ${\delta }_k=\eta (t_k-z_k)g_k^{\prime }(x_k)$. This gives $\boxed{\Delta w_{jk}={\delta }_k{z}_j}$.

A hidden activation affects all output units to which it connects. ThusandThe learning rate is already contained in ${\delta }_k$ and hence in ${\delta }_j$; do not multiply by it a second time. For a sigmoid function, $g^{\prime }(x)=g(x)[1-g(x)]$. Thresholds can be treated as negative biases with an extra constant input and updated by the same derivative rule.

This is squared-error backpropagation. The forward pass computes activations, while the backward pass transports output-loss derivatives through the outgoing weights and the derivatives of the activation functions, yielding an error signal for every hidden unit. That reverse flow of derivatives explains the name backpropagation.