Derivative of sigmoid

Question

I'm creating a neural network using the backpropagation technique for learning.

I understand we need to find the derivative of the activation function used. I'm using the standard sigmoid function

f(x) = 1 / (1 + e^(-x))

and I've seen that its derivative is

dy/dx = f(x)' = f(x) * (1 - f(x))

This may be a daft question, but does this mean that we have to pass x through the sigmoid function twice during the equation, so it would expand to

dy/dx = f(x)' = 1 / (1 + e^(-x)) * (1 - (1 / (1 + e^(-x))))

or is it simply a matter of taking the already calculated output of f(x), which is the output of the neuron, and replace that value for f(x)?

Answer 1

The two ways of doing it are equivalent (since mathematical functions don't have side-effects and always return the same input for a given output), so you might as well do it the (faster) second way.

Answer 2

Dougal is correct. Just do

f = 1/(1+exp(-x))
df = f * (1 - f)

Answer 3

A little algebra can simplify this so that you don't have to have df call f.
df = exp(-x)/(1+exp(-x))^2

derivation:

df = 1/(1+e^-x) * (1 - (1/(1+e^-x)))
df = 1/(1+e^-x) * (1+e^-x - 1)/(1+e^-x)
df = 1/(1+e^-x) * (e^-x)/(1+e^-x)
df = (e^-x)/(1+e^-x)^2

Answer 4

You can use the output of your sigmoid function and pass it to your SigmoidDerivative function to be used as the f(x) in the following:

dy/dx = f(x)' = f(x) * (1 - f(x))