The tutorial you posted here is actually doing it wrong. I double checked it against Bishop’s two standard books and two of my working implementations. I will point out below where exactly.
An important thing to keep in mind is that you are always searching for derivatives of the error function with respect to a unit or weight. The former are the deltas, the latter is what you use to update your weights.
If you want to understand backpropagation, you have to understand the chain rule. It’s all about the chain rule here. If you don’t know how it works exactly, check up at wikipedia – it’s not that hard. But as soon as you understand the derivations, everything falls into place. Promise! 🙂
∂E/∂W can be composed into ∂E/∂o ∂o/∂W via the chain rule. ∂o/∂W is easily calculated, since it’s just the derivative of the activation/output of a unit with respect to the weights. ∂E/∂o is actually what we call the deltas. (I am assuming that E, o and W are vectors/matrices here)
We do have them for the output units, since that is where we can calculate the error. (Mostly we have an error function that comes down to delta of (t_k – o_k), eg for quadratic error function in the case of linear outputs and cross entropy in case for logistic outputs.)
The question now is, how do we get the derivatives for the internal units? Well, we know that the output of a unit is the sum of all incoming units weighted by their weights and the application of a transfer function afterwards. So o_k = f(sum(w_kj * o_j, for all j)).
So what we do is, derive o_k with respect to o_j. Since delta_j = ∂E/∂o_j = ∂E/∂o_k ∂o_k/∂o_j = delta_k ∂o_k/o_j. So given delta_k, we can calculate delta_j!
Let’s do this. o_k = f(sum(w_kj * o_j, for all j)) => ∂o_k/∂o_j = f'(sum(w_kj * o_j, for all j)) * w_kj = f'(z_k) * w_kj.
For the case of the sigmoidal transfer function, this becomes z_k(1 – z_k) * w_kj. (Here is the error in the tutorial, the author says o_k(1 – o_k) * w_kj!)