Why does the minimum value of $x^x$ equal $1/e$? The graph of $y=x^x$ looks like this:

As we can see, the graph has a minimum value at a turning point. According to WolframAlpha, this point is at $x=1/e$.
I know that $e$ is the number for exponential growth and $\frac{d}{dx}e^x=e^x$, but these ideas seem unrelated to the fact that the mimum value of $x^x$ is $1/e$. Is this just pure coincidence, or could someone provide an intuitive explanation (i.e. more than just a proof) of why this is?
 A: Let $f(x)=x^x$. Note that $f(x)$ is only defined for $x>0$.
Then
$$
\ln f(x)=x\cdot\ln x\tag{1}
$$
Differentiating (1) gives
$$
\frac{1}{f(x)}f^\prime(x)=x\frac{1}{x}+\ln x=1+\ln x
$$
Note that we have used the chain rule and the product rule.
Solving for $f^\prime(x)$ gives
$$
f^\prime(x)=f(x)(1+\ln x)=x^x(1+\ln x)
$$
Can you use this to locate the critical points of $f(x)$?
A: Hint: actually you are looking for a local/global minimum .. so look at the derivative of the function $f(x) = x^x$
$$f'(x) = x^x (\log (x)+1)$$
which equals $0 \iff x = \frac{1}{e}$ 
A: Try to minimize logarithm of  $ y= x^x,$ i.e., $y=x \, \log x$ 
Its derivative is
$$ 1 + \log(x) $$
When equated to zero, it solves to that the minimum of $y(x)$ ocuurs at 
$$ x= \dfrac{1}{e}= 0.36788$$
and the minimum value is
$$ y_{min}= \dfrac{1}{{e}^{\frac{1}{e}}} \approx 0.6922$$
The tiny red dot shown in your graph for minimum is:
$$ (x,y) = (0.36788, 0.6922) $$
Note that the minimum value is not $\dfrac{1}{e}\, ! \, $ ... but is the reciprocal of $e^{th} $root of $e.$
A: Hint: Note that $x^x = e^{x\log x}$.
So minimizing $x^x$ is the same as minimizing $x\log x$
