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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.2012-03-03
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    It's a bit easier if you consider $\ln f(x)$ instead.2012-03-03
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    @HaraldHanche-Olsen: How does taking the logarithm make it easier?2012-07-09
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    @celtschk: Just take a look at J.D.'s answer.2012-07-10
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    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)$2012-07-10
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    @HaraldHanche-Olsen: Sorry, forgot the at-addressing.2012-07-10
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    @celtschk: No, if you know the rule you just used, it's not a great improvement. But unless you have learned a specific rule for differentiationg $f(x)^{g(x)}$, this relies on a modest bit of multivariable calculus, using the partial derivatives of $x^y$ and the multivariable chain rule. But it's not really a big deal (which is why I put the words “a bit” in my comment).2012-07-11

2 Answers 2

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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.

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

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    What is y'/y? i don't get it2012-03-03
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    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 $\dfrac{d}{dx} y$.2012-03-03
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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.2012-07-09
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    I like this! It is so smooth just like exp itself.2018-04-09