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

Question

Emilia Aguilar · Accepted Answer

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

Christopher Agresta · Answer

undefined To find the derivative of ln((e^x)/(1+e^x)), you can use the quotient rule combined with the chain rule. The quotient rule states that for functions u(x) and v(x), the derivative of u(x)/v(x) is (u'(x)v(x) - u(x)v'(x))/(v(x))^2. Applying this rule and the chain rule to ln((e^x)/(1+e^x)), we get:

[1/(e^x/(1+e^x)) * ((1+e^x)*(e^x)' - e^x*(1+e^x)')]/(e^x/(1+e^x))^2

Simplify and compute derivatives:

= [1/(e^x/(1+e^x)) * ((1+e^x)*e^x - e^x*(1+e^x))/(1+e^x)^2]

= [1/(e^x/(1+e^x)) * (e^x + e^(2x) - e^x - e^(2x))]/(1+e^x)^2

= [1/(e^x/(1+e^x)) * (e^(2x) - e^(2x))]/(1+e^x)^2

= 0/(1+e^x)^2

= 0

So, the derivative of ln((e^x)/(1+e^x)) is 0.