What is the derivative of #x^(lnx)#?

let #y =x^(lnx)#
There are no rules that we can apply to easily differentiate this equation, so we just have to mess with it until we find an answer.

If we take the natural log of both sides, we are changing the equation. We can do this as long as we take into account that this will be a completely new equation:
#lny=ln(x^(lnx))#
#lny=(lnx)(lnx)#
Differentiate both sides:
#((dy)/(dx))*(1/y)=(lnx)(1/x)+(1/x)(lnx)#
#((dy)/(dx))=(2*y*lnx)/x#

Okay, now we're done messing with that equation. Let's go back to the original problem:
#y =x^(lnx)#

We can rewrite this as #y=e^[ln(x^(lnx))]# because e to the power of a natural log of some number is that same number.
#y=e^[ln(x^(lnx))]#

Now, let's differentiate this using the exponent rule:
#(dy)/(dx) = d/dx[ln(x^(lnx))] * [e^[ln(x^(lnx))]]#

Conveniently, we already found the first term above, so we can easily simplify this.
#(dy)/(dx) = [(2*y*lnx)/x] * [x^(lnx)]#
#(dy)/(dx)=(2*y*(lnx)*(x^(lnx)))/x#

Let # y = x^(lnx) #

Take Natural logs of both sides
# ln y = ln x^(lnx) #
# :. ln y = (lnx)(lnx) #
# :. ln y = (ln^2x) #

Differentiate (implicitly) wrt #x# on the LHS and apply the chain rule to the RHS:
# :. 1/ydy/dx = (2ln x)1/x #
# :. dy/dx = (2yln x)/x #
# :. dy/dx = ((2x^(lnx))ln x)/x #
# :. dy/dx = ((2x^(lnx))ln x)x^-1 #
# :. dy/dx = (2x^((lnx)-1))ln x #