How do you differentiate #y=e^x/(1+x)#?

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Answer 1 · 2016-12-25T08:21:18

The answer is #=(xe^x)/(1+x)^2#

We need the differentiation of a quotient

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

Here, #u=e^x#, #=>#, #u'=e^x#

#v=1+x#, #=>#, #v'=1#

Therefore,

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

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

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