deltaL/deltaWij

#1

Didn’t calculate the differentiation of sigmoid(z)

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#2

didn’t calculate delA/delZ

#3

Hello Kush,
‘A’ here signifies the activation function which we have assumed to be the sigmoid function in this case. We have written the value of del(A)/del(Z) in terms of the differentiation of the sigmoid function directly, we are not concerned with the arithmetic value it returns.

I hope this helps.

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#4

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