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How do you differentiate # [(e^x)(lnx)]/x #?

1 Answer

Aug 19, 2016

#e^x/x^2(1+(x-1)lnx)#

Explanation:

#f(x) = (e^xlnx)/x#

#f'(x) = (x(e^x*1/x+e^xlnx)-e^xlnx*1)/x^2#
(Quotient rule and Produce rule)

#f'(x) = (e^x+xe^xlnx-e^xlnx)/x^2#

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

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