how about a derivative this time?

Assuming a,b≠0 then:

y=[x^(ax)]^(bx)

lny = ln[[x^(ax)]^(bx)]

lny = bx*ln[x^(ax)]

lny = bx*ax*lnx

lny = abx²lnx

y'/y = 2abx*lnx + ab*x

y' = y[2abx*lnx +ab*x]

y' = [x^(ax)]^(bx)*[2abx*lnx +ab*x] =>

y' = x^(abx²)*x*[2ab*lnx+ab] =>

y' = ab*x^(abx² +1)*[2lnx +1]

d/dx (x^ax)^bx

=bx(x^ax)^(bx - 1)(d/dx x^ax)

=bx(x^ax)^(bx - 1)(ax)(x)^(ax - 1)(1)

=abx²(x^ax)^(bx - 1)(x)^(ax - 1)

=abx²{x^[ax^(bx - 1)(ax^1 - 1)]}

=abx²{x^[ax^(bx - 1 + 1) - 1]}

=abx²{x^[ax^(bx) - 1]}

=abx^[ax^(bx) -1 + 2]

=abx^[ax^(bx) + 1]

I know this isn't the standard method for solving f(x)^(g(x)) derivatives; it makes use of simpler differentiation methods, for one thing. But I don't see why the answer wouldn't still be correct.

@ Adrian

Think you might be incorrect on your interpretation. There's an exponent, ax, with an exponent of its own, bx. Not a single exponent (ax)(bx).

OK Sparky--

Wouldn't this be the same as x^(abx^2)??

let y = x^(abx^2).

ln (y)= abx^2*ln(x) now take derivatives of both sides and use the product rule

y'/y = abx^2*(1/x) + ln(x)*2abx

Now y' = y*[abx +2abx*ln(x)].

substituting back in for y and factoring out abx you have

y' = abx*x^(abx^2)*[1+2ln(x)] which can be simplified further to give

y' = ab*x^(abx^2+1)*[1+2ln(x)]