Hello, Thank you for this amazing course. In the notes, the backpropagation for the entire training set of length m is given as (L = #output layer): dz[L] =a[L]-y dW[L] = 1/m*dz[L].a[L-1].T db[L] = 1/m*sum(dz[L], …) etc… I worked out the backpropagation manually, and I obtained the following: …

Remember that L is the vector output of the loss function and the cost J is the average of the L values across the samples. So it’s just a question of how you define things. Prof Ng’s notation is slightly ambiguous, but the key thing to realize is that only the dW and db expressions are partial deri…

Thank you a lot for your reply! I forgot L being defined as a vector, only having in mind L(a(i), y(i)) as a raw number. It makes total sense now, thank you!

Great! I guess I should have included this formula just to make everything complete: \displaystyle \frac {\partial J}{\partial W^{[l]}} = \frac {1}{m} \sum_{i = 1}^{m} \frac {\partial L(\hat{y_i},y_i)}{\partial W^{[l]}} Using the fact that the derivative of the average is the average of the deriva…

Hi Paulin, I’m still having a hard time understanding your explanations here. for the dZ2 notation, does that mean dL/dZ2, and not dJ/dZ2? If so, could you clarify why we are not taking the derivative of J but L for Z2? Thank you!

Yes, all the dZ values are Chain Rule factors at a given layer. It is literally only the dW^{[l]} and db^{[l]} where all the Chain Rule factors get multiplied together to form the full gradients w.r.t. J.

The partial derivatives you worked out are for a network with forward propagation like Z[l] = W[l] * a[l-1] + b[l], and cost function like J=1/m*Sum(A[L]) Andrew’s equations are actually for a network like the following, Z[i]=1/m*W[i]*a[i-1] + 1/m * b[i], and the cost function like J=Sum(A[L]) So,…

Sorry, but I disagree. What you are missing is that dW^{[l]} = \displaystyle \frac {\partial J}{\partial W^{[l]}}. There is no factor of \frac {1}{m} needed at the level of the individual layer forward prop functions in order to get the derivatives that Prof Ng shows. The \frac {1}{m} comes from the…

[image] Arnault_CAILLET: dW[L] = 1/m*dz[L].a[L-1].T What we are discussing is whether Andrew’s equation dW[L] = 1/m*dz[L].a[L-1].T is correct. If you look at one of the reading material under ** Week 4 Optional Reading: Feedforward Neural Networks in Depth** Feedforward Neural Net…

The derivation in the “I Deep Learning” material is incomplete, because it does not specify any particular loss function. It is focused only on specifics of individual layers, with no view of the total network or what task you are trying to solve. You cannot implement such a network. Nor can you c…

[image] luohan1992: It clearly shows Andrew’s equation is wrong Jonas does things a bit differently than Andrew does. Notice that he keeps the RHS as derivatives of J, so the \frac {1}{m} is still buried in the \displaystyle \frac {\partial J}{\partial Z^{[l]}}. Andrew’s formula is correct b…

Course 1 - Week 4 - 1/m in backpropagation

Course Q&A Deep Learning Specialization Neural Networks and Deep Learning

paulinpaloalto February 1, 2026, 8:34pm 20

I’m glad to hear that it helped. This particular issue has come up before. Here’s another historical thread on essentially the same topic that is worth a look.

Here’s one with a more general discussion of back prop and the Chain Rule and has links to reference info.

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Course 1 - Week 4 - 1/m in backpropagation

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