Question

How do you find the derivative of `y=x^(ln(x))`?

Answer 1

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

Explanation:

Write the function as:

`y= x^(lnx) = (e^lnx)^lnx = e^(ln^2x)`

then using the chain rule:

`d/dx(x^lnx) = d/dx (e^(ln^2x)) = e^(ln^2x) d/dx (ln^2x) = (2e^(ln^2x)lnx)/x`

and simplifying:

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