Whats the derivative of $ln((e^x)/(1+e^x))$ ?

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Answer 1 · 2016-06-21T19:51:48

$1/(1+e^x)$

$y = ln((e^x)/(1+e^x)) = ln (u), \qquad u = (e^x)/(1+e^x)$

and we use the chain rule $dy/dx = dy/(du) * (du)/dx$

$dy/(du) = 1/u = (1+ e^x)/(e^x)$

$(du)/(dx) = (e^x (1+ e^x) - e^x * e^x)/(1+e^x)^2$
$= (e^x )/(1+e^x)^2$

$\implies (dy)/(dx) = (1+ e^x)/(e^x) * (e^x )/(1+e^x)^2 = 1/(1+e^x)$

Whats the derivative of ln((e^x)/(1+e^x))ln((e^x)/(1+e^x))?