Question

How do you differentiate `(lnx)^(x)`?

Answer 1

`d/dx(lnx)^x = (lnx)^x{1/lnx + ln((lnx))}`

Explanation:

Let `y=(lnx)^x`

Take (Natural) logarithms of both sided:

`" " lny = ln((lnx)^x )`
`:. lny = xln((lnx) )`

Differentiate Implicitly (LHS) and apply product rule and chain rule (RHS).

`\ \ \ \ \ \ 1/ydy/dx = (x)(1/lnx*1/x) + (1)(ln((lnx))`
`:. \ 1/ydy/dx = 1/lnx + ln((lnx))`
`:. " " dy/dx = y{1/lnx + ln((lnx))}`
`:. " " dy/dx = (lnx)^x{1/lnx + ln((lnx))}`