y=x^(x^x) ··· Eq.1 Here we assume that Eq.1 is asking the first derivative, and x>0.
Since x>0, both sides of Eq.1 are positive. Apply natural logarithm to both sides.
log y=log x^(x^x) We have: log y=(x^x)·(log x) ··· Eq.2
Defferentiate both sides of Eq.2 with respect to x, using identities of the first derivative such as,
(log x)'=1/x, (log y)'=d(log y)/dy·(dy/dx)=(1/y)·y'=y'/y, and (x^x)'=(x^x)(log x + 1)
Eq.2 can be restated using the product rule as follows: (log y)'=(x^x)'(log x)+(x^x)(log x)' That is:
y'/y={(x^x)(log x + 1)}(log x)+(x^x)(1/x) so that: y'/y={x^(x^1)}·{(log x)²+(log x)+(1/x)}
Multiply both sides by Eq.1. We have: y'={x^(x^1)}·{(log x)²+(log x)+(1/x)}·{x^(x^x)} That is:
y'={x^(x^(x+1))}·{(log x)²+(log x)+(1/x)}
Therefore, the first derivative of given equation is: y={x^(x^(x+1))}·{(log x)²+(log x)+(1/x)}