What is the derivative of a sexp()

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d(x^^n)/dx = ?
 
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moriheru said:
What is (x^^n)?

It's tetration which represents a double up arrow in Knuth's up arrow notation. Basically a power tower of x's n high:
[tex]x^{x^{.^{.^{.^x}}}}[/tex]

And I don't see any simple solution to this problem.
 
Mentallic said:
It's tetration which represents a double up arrow in Knuth's up arrow notation. Basically a power tower of x's n high:
[tex]x^{x^{.^{.^{.^x}}}}[/tex]

And I don't see any simple solution to this problem.
Im only looking for the formula for integer values of x, will that make it simpler?
 
HallsofIvy said:
That function is not even defined for integer values of x.
Sorry, what I meant was positive integers.
 
You can rewrite ## x^x = \exp(\ln(x)\cdot x) ##. similar ##x^{x^x}=\exp(\exp(\ln(x)\cdot x)\cdot \ln(x)) ##. But you'll have to derivate yourself, I'm too lazy right now. o0)
 
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ok I got it

d(x ^^ n) / dx = x ^^ n * d(x ^^ ( n -1) * ln x ) / dx

Follow up question:
lim i→ ∞ { di(x ^^ n) / dxi } converge?
 
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japplepie said:
ok I got it

d(x ^^ n) / dx = x ^^ n * d(x ^^ ( n -1) * ln x ) / dx

Follow up question:
lim i→ ∞ { di(x ^^ n) / dxi } converge?
It would be better if you type in latex.