The Chain Rule and Credit Assignment

The output loss can compare a prediction with a target. An internal activation has no supplied “correct” value of its own. Training has to work out for itself how changing that activation would alter the final loss.

The problem has a name and a date. Marvin Minsky set it out in 1961, in a section of Steps Toward Artificial Intelligence headed “The Credit-Assignment Problem for Learning Systems”. He put it as a question rather than a definition: “But in the course of play, each ultimate success (or failure) is associated with a vast number of internal decisions. If the run is successful, how can we assign credit for the success among the multitude of decisions?” It reached the journal earlier still; the paper's own header records “Received by the IRE, October 24, 1960”. Schmidhuber's 2014 survey opens on the same difficulty, calling it “the fundamental credit assignment problem (Minsky, 1963)” and defining it as “Which modifiable components of a learning system are responsible for its success or failure?” Half a century apart, the same question.

Hidden units acquire useful structure precisely because nobody supplies them with a target. The 1986 paper in Nature that made the method famous claims exactly that in its abstract: “As a result of the weight adjustments, internal ‘hidden’ units which are not part of the input or output come to represent important features of the task domain, and the regularities in the task are captured by the interactions of these units.” The Nobel Committee for Physics restated the point in 2024: “A key advance was the demonstration by David Rumelhart, Hinton and Ronald Williams in 1986 of how architectures with one or more hidden layers could be trained for classification using an algorithm known as backpropagation”.

The chain rule supplies the missing link. It composes local derivatives along every dependency path.

Suppose x becomes z, z becomes a, and a contributes to loss L. The three local derivatives can be composed starting from either end. Which end you start from is the whole practical question.

There is a number on that choice. For a function with n inputs and m outputs, building the full m×n Jacobian costs n·c·ops(f) in forward mode but m·c·ops(f) in reverse mode, “where c is a constant guaranteed to be c < 6 and typically c ∼ [2, 3]”. That is the survey of automatic differentiation by Baydin and colleagues in the Journal of Machine Learning Research. A loss is a single number, so m is 1. One reverse sweep returns the sensitivity of L to every upstream quantity, at a small multiple of the cost of the forward pass. Forward mode would need a separate sweep per input.

Griewank fixes the ratio exactly at ω = 3 when only multiplications are counted, and notes the price: MEM{F′(x)} ∼ OPS{F(x)}. Reverse mode buys its speed by storing the forward values. His summary is why the arrows in the diagram below run right to left: “In other words, as Wolfe observed, gradients can ‘always’ be computed at a small multiple of the cost of computing the underlying function, irrespective of n the number of independent variables, which may be huge.”

Let z=2x+1, a=z², and L=3a at x=2.

If one value feeds two downstream paths, changing it affects the loss through both. The total derivative is the sum of those path contributions.

That is why gradients accumulate at residual additions, reused embeddings, tied weights, and any shared activation. The sum is not a bookkeeping nicety. Inside a deep residual network an identity skip splits ∂E/∂x_l into two additive terms, and one of them reaches the shallow unit without passing through any weight layer. He and colleagues drew the consequence in 2016: “This implies that the gradient of a layer does not vanish even when the weights are arbitrarily small.”

Veit and colleagues at Cornell, an independent group, then measured how much each of those summed paths actually carries. In a 110-layer ResNet, almost all of the gradient arrives along paths only 5-17 modules long — 0.45% of all the paths in the network. A network retrained on those effective paths alone reached 5.96% error against 6.10% for the full model. Failing to accumulate would omit valid influence. The measurement says which summands the influence was hiding in.

Two simple rules explain most reverse-mode bookkeeping. A stack of plain layers is the sequential case; the residual block measured by Veit and colleagues is the parallel one.

Use a small graph before trusting automatic differentiation.