Find the maximum of $f(x)=x^{1/x}$

This is not original with me.

If we know that $e^x \ge 1+x$ with equality only when $x = 0$, $e^{(x-e)/e} \ge 1 + (x-e)/e = x/e$ or $e^{x/e} \ge x$ or $e^{1/e} \ge x^{1/x}$ with equality only if $x = e$.

Ta-dah!

At no time do the fingers leave the hands!

Let $y = x^{1/x}.$ So $$ \ln y = (1/x)\ln x. $$ Differentiate both sides w.r.t $x$, we get $$ y'/y = (1/x)(1/x) + (-1/x^2) \ln x. $$ Rearranging, we have $$\dfrac{d}{dx}(x^{1/x}) = (1-\ln x)x^{1/x - 2} $$ Set $\dfrac{d}{dx}(x^{1/x}) = 0$ and work from there to get the maximum.

$$\text{(Hint: maximum occurs at x = e.)}$$