Hi, it ist stated in Week 3 , Video 1 that logistic gradient descents looks the same linke linear gradient descent, only with a different function f: dJ/dw = Sum[(f(x)-y)*x]
For the linear Cost Function J(w,b) = Sum[((w*x+b)-y)^2], I understand that J’ =2*((wx+b)-y)

*x.*

(This is chain rule with outer function u(x)^2 → 2u(x) and inner function v(x) wx+b → x)

(This is chain rule with outer function u(x)^2 → 2

Please help me to understand: How can the new logistic J-function get derivated to the same result (f(x)-y)*x?
J for logistic regression is Sum[y*log(1/(1+e^(wx+b)))+(1-y)log(1-(1/(1+e^(wx+b))))] and when I derivate it, there’s a couple of e^… in several fractions, sums and with different exponent.

→ I can’t reduce that derivation to the simple term of (f(x)-y)*x.

→ Is there a link where I can see that transformation? Or am I completely misunderstanding the topic?

Thank you a lot in advance!