Does the argmax of $x^{1/x} = e$, and why?

Question

The function $x^{1/x}$ seems to be maximized when $x=e$. I don't know how to take the derivative. How do I prove this, if it's true?

Answer 1

$\log(y)=\frac{\log(x)}{x}$ differentiating, $\frac{y'}{y}=\frac{1-\log(x)}{x^2}$ $\implies$ $y'=\frac{x^{\frac{1}{x}}(1-\log(x))}{x^2}.$

Answer 2

As already said, logarithmic differentiation is the key $$y=x^{\frac 1x}\implies \log(y)={\frac 1x}\log(x)\implies \frac{y'}{y}=\frac{1-\log(x)}{x^2}$$ So, there is an extremum for $\log(x)=1$ that is to say $x=e$.

But you need to prove that this is a maximum and, for that, you need the second derivative test $$y'=y \left(\frac{y'}{y}\right)\implies y''=y'\left(\frac{y'}{y}\right)+y\left(\frac{y'}{y}\right)'$$ Since $y'$ is already $0$, at this point $(x=e)$ $$y''=y\left(\frac{y'}{y}\right)'=x^{\frac 1x}\left(\frac{1-\log(x)}{x^2}\right)'=x^{\frac 1x}\left(\frac{2 \log (x)-3}{x^3}\right)$$ and for $x=e$ $$y''=-e^{\frac{1}{e}-3}$$ which is negative. So, $x=e$ corresponds to a maximum.