Infinite tetration and superroot of infinitesimal

[andydude]

But what about dx^(1/dx)?

Simple calculus would state that:
\( \frac{d}{dx}\left(x^{1/x}\right) = x^{1/x}(1+\ln(x))\frac{1}{x^2} \)
so moving the dx part, we obtain:
\( d\left(x^{1/x}\right) = x^{1/x}(1+\ln(x))\frac{1}{x^2}dx \)
is this what you are talking about?

If not, then I suppose that \( d\left(x^{1/x}\right) \ne dx^{1/{dx}} \) which means we are no closer to knowing the answer, but since you said \( (1+dx)^{1/dx} = e \) then that means that:
\( dx^{1/dx} = \sum_{k=0}^{1/dx}\left({1/dx \atop k}\right)(-1)^{1/dx} (1+dx)^k = \sum_{k=0}^{1/dx}\frac{1}{k!}(-1)^{1/dx} \)
which means \( dx^{1/dx} \) is indeterminate, since \( (-1)^{\infty} \) is indeterminate.

Andrew Robbins