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