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

1 Answer
Eddie
Jun 21, 2016

# 1/(1+e^x)#

Explanation:

#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)#

from quotient rule:

#(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)#

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