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

Find the maximum of the function $$f(x)=x^{1/x}$$ and the value of $x$ which gives the maximum value?

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Like every other maximum problem: differentiate, set the derivative to 0, solve for $x$, check whether any of the solutions happen to be maxima. –  Henning Makholm Mar 3 '12 at 21:32
It's a bit easier if you consider $\ln f(x)$ instead. –  Harald Hanche-Olsen Mar 3 '12 at 21:34
@HaraldHanche-Olsen: How does taking the logarithm make it easier? –  celtschk Jul 9 '12 at 10:40
@celtschk: Just take a look at J.D.'s answer. –  Harald Hanche-Olsen Jul 10 '12 at 19:28
I'm looking. But I still don't see how it's easier than just deriving $x^{1/x}$ directly: $(x^{1/x})' = (1/x)x^{1/x-1} + x^{1/x}\ln x \cdot (-1/x^2) = x^{1/x-2}(1-\ln x)$ –  celtschk Jul 10 '12 at 19:43

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. (Hint: maximum occurs at $x = e.$)

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What is y'/y? i don't get it –  Victor Mar 3 '12 at 21:57
I differentiated both sides of the equation $\ln y = (1/x) \ln x$. $\dfrac{d}{dx} \ln y = (\dfrac{d}{dx} y)/y$ by the rules of derivatives of logarithmic functions, and $y'$ is a shorthand for $\dfrac{d}{dx} y$. –  user2468 Mar 3 '12 at 22:00

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!

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+1 for presentation! I watch them magic tricks close though, I do. –  Eugene Shvarts Jul 9 '12 at 8:28