How do you find the derivative of $x^lnx$ ?

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Answer 1 · 2016-06-25T05:42:28

$(dy)/(dx)=2x^((lnx-1))lnx$

Let $y=x^(lnx)$

hence $lny=lnx xx lnx=(lnx)^2$

hence $1/y(dy)/(dx)=2lnx xx 1/x$

and $(dy)/(dx)=2lnx xx 1/x xx x^(lnx)=2x^((lnx-1))lnx$

How do you find the derivative of x^lnxx^lnx?