show by calculation that the derivative of the fermi function (logistic function) can be expressed by the function itself

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I'm taking a course on Neural Networks.

one of the questions on our exam will (likely) be:

Show by calculation that the derivative of the Fermi function (logistic function) can be expressed by the function itself.

Accordingly I've consulted with my friend, he proposed me this as a solution:

Step 1

$ f(x) = 1/(1+e^{-x}) $

$ f'(x) = \frac{d}{dx} (1 + e^{-x})^{-1}$

To interject here, the $\frac{d}{dx}$ term is basically meaningless- it just indicates that we're taking a derivative- isn't it?


Step 2

$ f'(x) = \frac{d}{dx} (1 + e^{-x})^{-1}=$

$ = -1 ( (1 + e^{-x})^{-2} ) ( (0 + e^{-x}) (-1) )=$


Step 3

$= -1 ( (1 + e^{-x})^{-2} ) ( (0 + e^{-x}) (-1) )=$

$ = e^{-x} / (1 + e^{-x})^2 =$


Step 4

$ = e^{-x} / (1 + e^{-x})^2 =$

$ = (1+ e^{-x} -1) / (1 + e^{-x})^2 =$


Step 5

$ = (1+ e^{-x} -1) / (1 + e^{-x})^2 =$

$ = ( 1 / ( 1 + e^{-x} ) ) - ( 1 / (1 + e^{-x})^2 )= $


Step 6

$ = ( 1 / ( 1 + e^{-x} ) ) - ( 1 / (1 + e^{-x})^2 ) =$

$ = f(x) - ( f(x) )^2 =$


Step 7

$ = f(x) - ( f(x) )^2 =$

$= f(x) ( 1 - f(x) )$


So essentially what I would like to know is: is that correct?

But also- I would like to know how we arrived to that solution, is it possible that someone could explain this to me in English as if I were a 5 years old child?