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

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Answer 1 · 2016-04-24T14:08:44

#f'(x)=e^x/(1+e^x)#

The chain rule says that if

#f(x)=g(h(x))#

then

#f'(x)=h'(x)g'(h)#

Substituting in the functions from the question,

#h(x)=1+e^x#

#g(h)=ln(h)#

Then the derivatives are

#h'(x)=e^x#

#g'(h)=1/h#

According to the chain rule,

#f'(x)=(e^x)*(1/(h(x)))#

Expanding out and substituting in that #h(x)=1+e^x#,

#f'(x)=e^x/(1+e^x)#