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

Jun 10, 2018

$- \frac{{2}^{w} \ln \left(2\right) + {e}^{w}}{{2}^{w} + {e}^{w}} ^ 2$

Explanation:

Writing

$g \left(w\right) = {\left({2}^{w} + {e}^{w}\right)}^{- 1}$
so we get

$g ' \left(w\right) = \left(- 1\right) {\left({2}^{w} + {e}^{w}\right)}^{- 2} \left({2}^{w} \ln \left(2\right) + {e}^{w}\right)$