system
Given the following post, respond as if you're commenting on Reddit:
human
Why does "e" come out as "x" for the maximum "y" value for the function of f(y) = x^(1/x)?

This question is tied to a fact that, while I had a little fun with tetration (in the x\^x\^x\^x... form), I noticed that there is a region that does not go to infinity, while the number of exponents go to infinity. So I started searching for the max "x" value where this happens. 

Long story short, I arrived at  y = x\^(1/x) as the function I need the maximum "y" value of.  I was lazy, so I homed in on the "x" any "y" values simply by using a calculator and trying numbers. After about 20 steps, I got  1.44466786 as "y" and 2.71815 as "x". 

I was really surprised that I homed in on "e". This also means, that for infinitley many tetration of the same value as in x\^x\^x\^x..., the maximum finite value is "e", and the "x" value where this happens is e\^(1/e). Which is very interesting.

So my question question again: why does f(y) = x\^(1/x) has it's maximum at x = e? 

I'm a third year physics bachelor student, so I can handle higher level math if it's needed for the answer.
gpt
In general, a^b = e^[b\*ln(a)]. You can prove this yourself by setting a^b = e^(X), and solving by logging both sides.

This is your first hint, because any power can be expressed quite naturally as something with e and a natural log in it, so it shouldn't be a surprise that an e might drop out at the end.

So we have:

f(x) = x^(1/x) = e^[ln(x)/x]

To find the extremal points (including the maximum), we want df/dx=0. So we differentiate e^[ln(x)/x], set that to zero, and solve. Using the chain rule and the quotient rule, this gives us:

[1-ln(x)]/x^2 \* e^[ln(x)/x] = 0

This is zero when 1-ln(x)=0, i.e. x=e.

You can go further to prove that this is indeed a maximum etc.