Question

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

Answer 1

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

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

Explanation:

Write the function as:

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

then using the chain rule:

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

and using the product rule:

`(d^2)/dx^2 x^(lnx) = (2lnx)/x d/dx x^(lnx) + x^(lnx) d/dx (2lnx)/x`

`(d^2)/dx^2 x^(lnx) = (4ln^2x)/x^2 x^(lnx) + 2x^(lnx) (1-lnx)/x^2`

`(d^2)/dx^2 x^(lnx) = x^(lnx) (4ln^2x -2lnx + 2)/x^2`

Note now that:

`x^(lnx)/x = x^(lnx)*x^-1 = x^(lnx-1)`

so we can simplify as:

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

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