Proof that given f'(v) is in fact the derivate of f(v)

Question

Given the logistic activation function for an artificial neuron:

$f(v) = \frac{1}{(1+e^{-\alpha v})}$

Prove that:

$f'(v) = \alpha[1-f(v)][1+f(v)]$

Answer 1

This is not correct. The correct formula (if $f(v) = \frac{1}{1 + e^{-v}}$) is that $f'(v) = f(v) (1 - f(v))$. To show this, you can try and use the quotient rule.

Answer 2

$f(v) = \frac {1}{1+e^{-v}}\\ e^{-v} = \frac {1 - f(v)}{f(v)}\\ \frac {d}{dv}e^{-v} = \frac {d}{dv}(\frac {1}{f(v)} - 1)\\ -e^{-v} = -\frac {1}{f^2(v)} f'(v)\\ \frac {1-f(v)}{f(v)} f^2(v) = f'(v)\\ f'(v) = f(v)(1-f(v))$

Answer 3

We take: $$p=1\to p'=0$$ $$q=1+e^{-v}\to q'=-e^{-v}$$

Then $$\bigg(\frac pq\bigg)'=\frac{p'q-pq'}{q^2}$$

Is the quotient rule. Here this is $\frac{q'}{q^2}$

Now use that $f(v)=\frac 1q$ and $q'=1-q$, which is apparent by looking at the first two lines.