Backpropagation: Chain Rule to the Third Last Layer I'm trying to solve dLoss/dW1. The network is as in picture below with identity activation at all neurons:



Solving dLoss/dW7 is simple as there's only 1 way to output:

$Delta = Out-Y$

$Loss = abs(Delta)$

The case when Delta>=0, partial derivative of Loss over W7 is:

$\dfrac{dLoss}{dW_7} = \dfrac{dLoss}{dOut} \times \dfrac{dOut}{dH_4} \times \dfrac{dH_4}{dW_7} \\
= \dfrac{d(Out-Y)}{dOut} \times \dfrac{d(H_4W_{13} + H_5W_{14})}{dH_4} \times \dfrac{d(H_1W_7 + H_2W_8 + H_3W_9)}{dW_7} \\
= 1 \times W_{13} \times H_1$

However, when solving dLoss/dW1, the situation is very different, there are 2 chains to W1 through W7 and W10, and now, how should the chain for $\dfrac{dLoss}{dW_1}$ be?

Furthermore, at an arbitrary layer, with all outputs of all layers already calculated plus all gradients of weights on the right side also calculated, what should a formula for $\dfrac{dLoss}{dW}$ be?
I finally solved it out but it's long, it's not only chain rule, it includes quotient rule too. And, this is only the third last layer, once a DNN has more layers then it's more complex.

$\dfrac{dLoss}{dW_1} = \dfrac{d}{dW_1}(Out-Y) = \dfrac{d}{dW_1}Out = \dfrac{d}{dW_1}(H_4W_{13} + H_5W_{14}) \\= \dfrac{d}{dW_1}H_4W_{13} + \dfrac{d}{dW_1}H_5W_{14} \\= W_{13} \times \dfrac{d}{dW_1}H_4 + W_{14} \times \dfrac{d}{dW_1}H_5 \\= W_{13} \times \dfrac{d}{dW_1}(H_1W_7 + H_2W_8 + H_3W_9) + W_{14} \times \dfrac{d}{dW_1}(H_1W_{10} + H_2W_{11} + H_3W_{12}) \\= W_{13} \times \dfrac{d}{dW_1}H_1W_7 + W_{14} \times \dfrac{d}{dW_1}H_1W_{10} \\= W_{13}W_7 \times \dfrac{d}{dW_1}H_1 + W_{14}W_{10} \times \dfrac{d}{dW_1}H_1 \\= (W_{13}W_7 + W_{14}W_{10}) \times \dfrac{d}{dW_1}(X_1W_1 + X_2W_2) \\= (W_{13}W_7 + W_{14}W_{10}) \times X_1$