Derivative of $z(s) = \frac{1}{1 + e^{-s}}$

Question

My lecture notes state that, if $z(s) = \dfrac{1}{1 + e^{-s}}$, then $z'(s) = 1 - z^2$. Isn't this incorrect? Shouldn't this be $z'(s) = \dfrac{ e^{-s} }{ ( 1 + e^{-s} )^2}$?

Clarification/Confirmation would be appreciated.

Answer 1

It should be $z'=z-z^2$: $$ z' = \frac{e^{-s}}{(1+e^{-s})^2} = \frac{1+e^{-s}}{(1+e^{-s})^2} - \frac{1}{(1+e^{-s})^2} = \frac{1}{1+e^{-s}}-\frac{1}{(1+e^{-s})^2} = z-z^2. $$