Use the function `f(x)=x^(1/x),x >0,` to determine the bigger of the two numbers `e^(pi)a n dpi^edot`

Explanation

`y=x^((1)/(x))`
Taking logarithm of both sides
In `y=(1)/(x) In x = (In x)/(x)`
Differentiating ` w.r.t` x , we have
`(1)/(y)(dy)/(dx)=(x.(1)/(x)-Inx)/(x^(2))=(1-Inx)/(x^(2))`
`:. (dy)/(dx)=(y(1-Inx))/(x^(2))=(x^((1)/(x))(1-Inx))/(x^(2))=(x^((1)/(2)))/(x^(2))(1-Inx)` ....(i)
`(dy)/(dx)=0 implies 1- Inx =0 implies Inx=1 " ":. x =e `
From (i)
`(d^(2)y)/(dx^(2))=(x^((1)/(x)))/(x^(2))(-(1)/(x))+(1-Inx)""(d)/(dx)((x^((1)/(x)))/(x^(2)))`
`(d^(2)y)/(dx^(2))|_(x=e) =(e^((1)/(e)))/(e^(2))(-(1)/(e)) lt 0`
Thus y is maximum at x=e .
But ` x=e` is the only extreme value , hence y is greatest at x=e , i.e., ` e^((1)/(3)) ge x^((1)/(x)) AA x gt 0` .
Thus ` e^((1)/(e)) gt pi^((1)/(pi))`
Raising both sides to power `epi` , we have
`(e^(1//e))^(epi) gt (pi^(1//pi))^(epi`
`implies e^(pi) gt pi^(e)`