How do you find the derivative of #ln((e^x)/(1+e^x))#?

1 Answer
Nov 6, 2016

Answer:

#d/dx[ln((e^x)/(e^x+1))] =1/(e^x+1)#

Explanation:

Use the derivative of natural logs rule:
#d/dx[ln u] =(u')/u#

We want to simplify the following:
#d/dx[ln((e^x)/(e^x+1))]#

Dividing by the inner term is the same as multiplying by the reciprocal:
#= (d/dx[(e^x)/(e^x+1)])*((e^x+1)/(e^x))#

Use quotient rule to differentiate first term:
#=([(e^x+1)(e^x)-(e^x)(e^x)]/[(e^x+1)^2])*((e^x+1)/(e^x))#

Simplify:
#=([(e^x+1)cancel((e^x))-cancel((e^x))(e^x)]/[(e^x+1)^2])*((e^x+1)/cancel(e^x))#
#= [(cancel(e^x)+1-cancel(e^x))cancel((e^x+1))]/[(e^x+1)cancel((e^x+1)^2)]#
#= 1/(e^x+1)#