Question

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

Answer 1

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

Explanation:

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`