y=x^((x^x))

Question

i can not solve the problem,please solve me the problem

Answer 1

y=xraesed tu x raesed tu x

It tern up very steep as x inkreez

zampel...1^1^1=1

2^2^2...2^4=16

3^3^3...3^27=7.62559748*10^12...very big

Answer 2

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)}