Question

What is the derivative of `g(w)= 1/(2^w+e^w)`?

Answer 1

`-(2^wln(2)+e^w)/(2^w+e^w)^2`

Explanation:

Writing

`g(w)=(2^w+e^w)^(-1)`
so we get

`g'(w)=(-1)(2^w+e^w)^(-2)(2^wln(2)+e^w)`