Question

How do you differentiate `x^lnx`?

Answer 1

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

Explanation:

Let `y = x^(lnx)`

Take Natural logarithms of both sides:
`lny = ln{x^(lnx)}`
`:. lny = (lnx)(ln{x)` as `ln a^b = blna`
`:. lny = (lnx)^2`

Differentiate Implicitly (LHS), and apply chain rule to RHS:
`:. 1/y dy/dx = 2(lnx)(1/x)`
`:. 1/(x^(lnx)) dy/dx = 2(lnx)x^-1`
`:. dy/dx = (x^(lnx))2(lnx)x^-1`
`:. dy/dx = 2x^(ln(x)-1)lnx`