Find the derive of x^^x

[Nasser]

I am looking for the first derive of

f(x) = x^^x

I searched the web but no results found.
I also searched in tetration forum but nothing new.

please help.

[Klitzing]

Don't know exactly what you mean by "^^".
If that's the power, I'd usually would write a single "^". Then:
y = x^x, y' = ?
y = x^x = e^(ln(x^x)) = e^(x*ln(x))
y' = e^(x*ln(x))*(ln(x)+x*1/x) = x^x*(ln(x)+1)

--- rk

[Nasser]

"^^" is different thing
it is not "^"

[wendy]

Ye need to spread the function out further. I use a^^x to refer to a particur kind of isomorphic power, loosely related to 2 cosh(x acosh(a/2)), or a^^x = cho(x acho a). But if ye be using it, ye need to spell it out more clearly.

[Keiji]

Since he mentions the tetration forum and many people use ↑↑ to mean tetration, I assume that's what he's talking about.

I'm not sure if you can find a derivative. Wikipedia's Tetration article mentions a piecewise differentiable approximation, though.

[Nasser]

Since he mentions the tetration forum and many people use ↑↑ to mean tetration, I assume that's what he's talking about.

yes you are abslutely right.

It seems there is no derive formula for x^^x.

thank you Keiji.

[quickfur]

Since he mentions the tetration forum and many people use ↑↑ to mean tetration, I assume that's what he's talking about.

yes you are abslutely right.

It seems there is no derive formula for x^^x.

thank you Keiji.

There may not be a formula, but one thing for sure is that the derivative will grow faster than x^^x itself.

This is because for any function f that grows slower than e^x, the derivative of f grows slower than x (e.g., (x^n)' = x^(n-1) = (1/x)*f(x)). When f=e^x, f'=e^x, so e^x is the fixed point of the derivative operation. When f grows faster than e^x, say e^(x^2), the derivative grows faster: the derivative of e^(x^2) = x*e^(x^2) = x*f(x).

In fact, the further you get away from e^x, the faster the growth rate of the derivative will be relative to the original function. For example, the derivative of e^e^x = e^x*e^e^x = (e^x)*f(x). And the derivative of e^e^e^x = e^x*e^e^x*e^e^e^x = (e^x*e^e^x)*f(x).

So when f(x) = x^^x, the derivative of f(x) must have a vastly higher rate of growth than f(x) itself.

In fact, let's say g1(x) = e^x), g2(x) = e^e^x, g3(x) = e^e^e^x, etc.. The derivative of gn(x) for some n, is of the form (e^x * e^e^x * ... * g{n-1}(x)) * gn(x) > gn(x)^m for some number m that increases as n increases. Now since h=e^^x is the limit of g1, g2, g3, ..., it follows that the derivative of h must be some limit of gn(x)^m, which we can say is approximately gn(x)^^x, or, IOW, the derivative of e^^x is approximately (e^^x)^^x.

Now, since x^^x > e^^x, that means the derivative of x^^x must grow even faster than (x^^x)^^x. How much faster, I have no idea, but this should give you a little glimpse into how incredibly fast it grows.