in backpropogation intuition of week 3 , it is written as dz=da*g’(z) but in the previous videos dz = a-y , how did dz=da*g’(z) come up?? sir has told that, it has come by chain rule, but i dont understand how… for dz[1] [image] for dz[2] [image] how is dz[1] = da*g’(z) , but dz[2]=a[2]-y ?

[image] Rohan_Devaki: dz[1] = w[1]T * dz[2] * da[1] * g’(z[1]). Yes, as Saif has pointed out, the first operand on the RHS there is incorrect and should be W^{[2]T}. There’s another important error there as well: the operation between W^{[2]T} and dZ^{[2]} is not * (elementwise multiply), but…

I still have a doubt at the end , what would be the derivative of da[1]/dz[1] ? if it is just g’( z[1] ) then ,how is dz[1] = w[1]T * dz[2] * da[1] * g’(z[1]) ? even after checking the chain rule, i am not able to get to this as sir mentioned, please give a elaborate solution dL/dz[1] = dL/da[2]…

[image] Rohan_Devaki: how is dz[1] = w[1]T * dz[2] * da[1] * g’(z[1]) ? How do you get this? Derivative of dZ1 is: \frac{dL}{dZ1} = dZ^{[1]} = W^{[2][T]}.dZ^{[2]}* g^{[1]} {’} (Z^{[1]}) You already saw the chain rule.

I am asking you to elaborate the chain rule, dz[1] = w[1]T * dz[2] * da[1] * g’(z[1]) is written from the sirs classes, as i have written my steps above, i am not able to get this as a conclusion these are my steps: dL/dz[1] = dL/da[2] * da[2]/dz[2] * dz[2]/da[1]/ *da[1]/dz[1] dL/a[2] * da[2]/d…

[image] Rohan_Devaki: dz[1] = w[1]T * dz[2] * da[1] * g’(z[1]) is written from the sirs classes, as i have written my steps above, i am not able to get this as a conclusion From which class do you get this? Can you share a link with us? Regarding the explanation of the Chain Rule, it just n…

[image] Backpropagation Intuition (Optional) - Shallow Neural Networks | Coursera Video created by DeepLearning.AI for the course "Neural Networks and Deep Learning". Build a neural network with one hidden layer, using forward propagation and backpropagation. …

I am glad your doubts are cleared now. However, a link which you gave never say that dz[1] = w[1]T * dz[2] * da[1] * g’(z[1]). It is a better idea to watch the video again…

dz[1] = w[1]T * dz[2] * da[1] * g’(z[1]). it said the same, i have already attached the screenshot in the question itself , please check the beginning of the thread

[image] Rohan_Devaki: dz[1] = w[1]T * dz[2] * da[1] * g’(z[1]) I’ve checked the screenshot, it said dz[1] = w[2]T * dz[2] * g[1]’(z[1]). This is different from the one you mentioned. Maybe it is a good time for you to watch that video again, or at least the screenshot which you shared.

Derivation of dz=da* g'(z) ? or dz= a- y? how is derivation of dz[1] and dz[2] different?

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

saifkhanengr May 31, 2023, 2:59pm 2

Check this, and this, and this.

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Derivation of dz=da* g'(z) ? or dz= a- y? how is derivation of dz[1] and dz[2] different?

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