Question

Question #cd7fb

Answer 1

`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`