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Question #cd7fb

1 Answer

Jun 16, 2017

#f'(x)=x^x(lnx+1)+2x#

Explanation:

#f(x)=e^(xlnx)+x^2#
#f'(x)=(e^(xlnx))'+(x^2)'#
By the power rule, #(x^2)'=2x#
Using the chain rule, #(e^(xlnx))'=e^(xlnx)*(xlnx)'#

#=e^ln(x^x)*[lnx+x(1/x)]#
#=x^x(lnx+1)#

Therefore, #f'(x)=x^x(lnx+1)+2x#

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