How do you differentiate #f(x) = (e^x-1)/(x-e^x)# using the quotient rule?

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Answer 1 · 2017-02-09T17:10:39

The answer is #f'(x)=(xe^x-2e^x+1)/(x-e^x)^2#

We use

#(u/v)'=(u'v-uv')/v^2#

Here,

#f(x)=(e^x-1)/(x-e^x)#

#u=e^x-1#, #=>#, #u'=e^x#

#v=x-e^x#, #=>#, #v'=1-e^x#

#f'(x)=(e^x(x-e^x)-(e^x-1)(1-e^x))/(x-e^x)^2#

#=(xe^x-(e^x)^2+(e^x)^2-e^x+1-e^x)/((x-e^x)^2)#

#=(xe^x-2e^x+1)/(x-e^x)^2#