Where is the highest point of $f(x)=\sqrt[x]{x}$ in the $x$-axis? I mean, the highest point of the $f(x)=\sqrt[x]{x}$ is when $x=e$.
I'm trying to calculate how can I prove that or how can it be calculated.
 A: The typical proof is to use derivatives and find critical points etc which is a general method and you ought to have it in your toolbox. You already have multiple answers for that.
Here is a different approach.
A simple trick works for this problem:
Use the inequality
$$e^t \ge 1 + t$$
which is valid for all real $t$.
Let $x$ be any real $\gt 0$.
Then
$$ e^{(x/e) - 1} \ge 1 + (x/e) - 1 = x/e$$
Thus
$$e^{x/e} \ge x$$
and so
$$e^{1/e} \ge x^{1/x}$$
A: Write the function as $y = x^{1/x}$.  To take the derivative, use logarithmic differentiation.  That is, let
$$\ln y = \ln x^{1/x} = \frac{1}{x} \ln x.$$
The derivative of this is
$$\frac{y'}{y} = -x^{-2} \ln x + x^{-2} = \frac{1}{x^2} (1 - \ln x)$$
Since we know $y$, we can multiply both sides by $y$ to have a solution of $y'$:
$$y' = x^{1/x} \frac{1}{x^2} (1 - \ln x)$$
Now, the max will occur when this is 0.  The only solution is when $1 - \ln x = 0$, or when $x = e$.  Now, check to make sure this actually gives you the absolute max by seeing that the derivative is positive to the left of $e$ and negative to the right of $e$.  That is easy, since the only term that will ever change sign is the $1 - \ln x$.  Done.
A: Well, write $$f(x) = e^{\frac{1}{x}\ln(x)}$$ and differentiate and set equal to 0 to get:
$$\dfrac{d}{dx}(e^{\frac{1}{x}\ln(x)})=\bigg(-\frac{1}{x^2}\ln(x)+\frac{1}{x^2}\bigg)e^{\frac{1}{x}\ln(x)}=0$$
Which implies (after dividing by the exponential term) that
$$\frac{1}{x^2}(1-\ln(x))=0$$
Whence $1=\ln(x)$ or $x=e$.
Now you just need to check whether this gives you a local minimum or maximum via the second derivative.
A: It really only makes sense for $x\gt 0$, at least if you stick to real numbers. 
On $(0,\infty)$ can rewrite the function as
$$f(x) = x^{1/x} = e^{(\ln x)/x}.$$
Note that as $x\to\infty$,
$$\lim_{x\to\infty}\frac{\ln x}{x} = 0,$$
so $\lim\limits_{x\to\infty}f(x) = e^0 = 1$ and as $x\to 0^+$, we have
$$\lim_{x\to 0^+}\frac{\ln x}{x} = -\infty$$
so $\lim\limits_{x\to 0^+}f(x) = \lim\limits_{t\to-\infty}e^t = 0$. 
So that means that the function is bounded. We find its critical points by taking the derivative:
$$\begin{align*}
\frac{d}{dx}f(x) &= \frac{d}{dx} e^{(\ln x)/x}\\
&= e^{(\ln x)/x}\left(\frac{d}{dx}\frac{\ln x}{x}\right)\\
&= e^{(\ln x)/x}\left(\frac{x\frac{1}{x} - \ln x}{x^2}\right)\\
&= e^{(\ln x)/x}\left(\frac{1-\ln x}{x^2}\right).
\end{align*}$$
This is zero if and only if $1-\ln x=0$, if and only if $\ln x = 1$, if and only if $x=e$. So the only critical point is at $x=e$.
If $0\lt x \lt e$, then $f'(x)\gt 0$ (since $\ln x \lt 1$), so the function is increasing on $(0,e)$, and if $e\lt x$, then $f'(x)\lt 0$, so the function is decreasing on $(e,\infty)$. Thus, $f$ has a local maximum at $x=e$, and since it is the only local extreme of the function, which is continuous, $f(x)$ has a global extreme at $x=e$.
