Awesome Testing

Learning lesson 05 · Send credit backward

How does a final loss reach every weight that helped produce it?

examplepredictionlossgradientchanged parameters

01 · Smallest useful mechanism

Backpropagation reuses local derivatives to assign credit through a computation graph.

A network is a composition of weighted sums and nonlinear functions. The chain rule says that sensitivity through a composition is the product of local sensitivities. Reverse-mode differentiation starts at one scalar loss, walks the recorded graph backward, and accumulates a gradient for every parameter that contributed to the result.

Values move forward; an upstream loss derivative moves backward and is multiplied by each operation’s local derivative.

02 · Experiment

Test the prediction

Exact computation graph · one ReLU hidden unit

The activity declares every weight and activation, then applies local derivatives and the chain rule to all parameters.

Loading the learning experiment…

03 · Reason from evidence

The backward pass reports nonzero gradients, yet a snapshot shows that every weight is unchanged after the batch.

Your brief: Trace the training contract from loss to stored parameters. Name the checks that would locate the first missing state transition.

Evidence your answer should use

  • Whether the optimizer step ran and was skipped by any guard
  • Whether the intended parameters belong to the optimizer
  • Gradient, update, and parameter values for one named weight
  • Mixed-precision scale, clipping, and zeroing events around the step

There is no reveal button here. A strong response names the claim, the evidence that would test it, and what the evidence cannot establish.

Your learning artifact

Write the argument you would defend

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04 · Check your understanding

What does backpropagation produce directly?

Optional referenceDeep dive / reference chapterOpen the complete essay, diagrams, mathematics, exercises, glossary, and sources when you want more depth.

Next: Efficient gradients let hidden layers learn features. Next we see how one new representation solves XOR.