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We know that $(x^{n})' = nx^{n - 1}$ and $(n^{x})' = n^{x}\ln n$.

My question is: how can calculate the formula of $x^x$?

What about:

$$\left(x^{x^{{ \begin{array}{ccc} &\;&.\cdot^.\\ &n\; times & \\ .\cdot^. & & \; \end{array} }^x}}\right)' = ? $$

Is there any way to find a general formula?

When $n$ is 2, we will have $x^x$. If n is 3, we will have $x^{x^x}$, and so on.

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    $\begingroup$ Let $f_n(x) = x^{x^{\cdots^x}} = x^{f_{n-1}(x)}$ with $f_0(x) = 1$. Then $f_n'(x) = x^{f_{n-1}(x)-1}(f_{n-1}(x)+f_{n-1}'(x)\cdot x\log x)$, which you can compute recursively. $\endgroup$
    – user856
    Oct 21, 2013 at 19:05
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    $\begingroup$ Being a programming hobbist, this notation kind of reminds me function pointers and lambda functions. It pleases me. $\endgroup$
    – chubakueno
    Oct 21, 2013 at 23:15
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    $\begingroup$ A hobbist is sort of like a hobbit. $\endgroup$
    – GEdgar
    Oct 22, 2013 at 0:34

8 Answers 8

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I don't mean to bring this question back from the dead, but Elias' answer is quite wrong, and Anders seems to have been to only one to notice, but he was ignored. Testing any of $n=1,2,3$, for example, all give wrong answers, so I'll give my own answer, which is a similar approach.

A power tower with $n$ $x$s total can be described for $n\in\mathbb{N}^0$ as $$\varphi_{n}(x)= \left\{ \begin{array}{rcl} 1, & \mbox{ if } & n=0\\ x^{\varphi_{n-1}(x)}, & \mbox{ if } & n> 0 \end{array} \right.$$ Then $$\log{\varphi_n(x)}=\varphi_{n-1}(x) \log(x) $$$$ \implies\varphi_n'(x)=\frac{\varphi_n(x)\varphi_{n-1}(x)}{x}+\varphi_n(x)\varphi_{n-1}'(x)\log(x)$$ Recursively using this identity, $$\varphi_n'(x)=\frac{\varphi_n(x)\varphi_{n-1}(x)}{x}+\varphi_n(x)\left(\frac{\varphi_{n-1}(x)\varphi_{n-2}(x)}{x}+\varphi_{n-1}(x)\varphi_{n-2}'(x)\log(x)\right)\log(x)$$ $$=\frac{\varphi_n(x)\varphi_{n-1}(x)}{x}+\frac{\varphi_n(x)\varphi_{n-1}(x)\varphi_{n-2}(x)\log(x)}{x}+\varphi_n(x)\varphi_{n-1}(x)\varphi_{n-2}'(x)\log^2(x)$$ $$=\frac{\varphi_n(x)\varphi_{n-1}(x)}{x}+\frac{\varphi_n(x)\varphi_{n-1}(x)\varphi_{n-2}(x)\log(x)}{x}+\frac{\varphi_n(x)\varphi_{n-1}(x)\varphi_{n-2}(x)\varphi_{n-3}(x)\log^2(x)}{x}+\varphi_n(x)\varphi_{n-1}(x)\varphi_{n-2}(x)\varphi_{n-3}'(x)\log^3(x)$$ Continuing this pattern, we have $$\varphi_n'(x)=\frac1x\sum_{k=1}^{n}\left(\prod_{i=0}^k\varphi_{n-i}(x)\right)\log^{k-1}(x)$$

Remark: If we take an infinite power tower, we essentially have $\varphi_n(x)=\varphi_{n-1}(x)$, so that the derivative of an infinite power tower is $$\varphi_\infty'(x)=\frac1x\sum_{k=1}^{\infty}\left(\prod_{i=0}^k\varphi_{\infty}(x)\right)\log^{k-1}(x)=\frac1x\sum_{k=1}^{\infty}\varphi_{\infty}^{k+1}(x)\log^{k-1}(x)$$ $$\frac{\varphi_{\infty}^2(x)}{x}\sum_{k=0}^\infty\left(\varphi_\infty(x)\log(x)\right)^k=\frac{\varphi_{\infty}^2(x)}{x\left(1-\varphi_{\infty}(x)\log(x)\right)}$$ This is valid for all $x$ such that $\varphi_\infty(x)$ is finite (which is the interval $[e^{-e},e^{\frac1e}]$), because in that interval, $\left| \varphi_\infty(x) \log(x)\right|=\left|W(-\log(x))\right|\leq 1$, with equality only on the endpoints, so the series above converges.

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  • $\begingroup$ This is even better and cleaner. Thanks! $\endgroup$ Nov 21, 2014 at 7:46
  • $\begingroup$ All of the \varphi's make this look elegant. $\endgroup$ Dec 25, 2015 at 1:16
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You simply combine the two formulas you already mentioned to get:

$$(x^x)'=(x^n)'_{n=x}+(a^x)'_{a=x}=(nx^{n-1})_{n=x}+(a^x\ln a)_{a=x}=x\cdot x^{x-1}+x^x\ln x=x^x(1+\ln x)$$

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    $\begingroup$ This technique is also known as iterative differentiation and is much more efficient than log-differentiation. $\endgroup$
    – Squirtle
    Nov 11, 2013 at 19:11
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$$x>0\implies x^x=e^{x\log x}\implies (x^x)'=e^{x\log x}\left(\log x+1\right)=x^x(\log x+1)$$

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  • $\begingroup$ Yes, exactly. And for the second part of the question? How can we calculate x^x^x...n times ... ^x? $\endgroup$ Oct 21, 2013 at 18:49
  • $\begingroup$ see my answer.. $\endgroup$
    – ILoveMath
    Oct 21, 2013 at 19:02
  • $\begingroup$ @DanielRust Sorry, I didn't know that $log x$ is also acceptable. Learning something new every day. Thanks $\endgroup$ Oct 22, 2013 at 18:52
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LEt $f, g$ be any functions. Let $y = f^g \implies \ln y = g \ln f $

$$ \therefore \frac{y'}{y} = g' \ln f + g\frac{f'}{f} \implies \frac{ df^g}{dx}= y' = f^g ( f' \ln f + \frac{g f'}{f} )$$

Using this formula with $f = x = g $ gives desired resuld.

In general, if $y = x^{x^{..^{x^{x^x}}}} \implies y = x^y \implies \ln y = y \ln x$

$$ \therefore \frac{y'}{y} = y' \ln x + \frac{y}{x} \implies y' ( \frac{1}{y} - \ln x) = \frac{y}{x} \implies y' = \frac{y^2}{x(1 -( \ln x )y )}$$

In other words,

$$ ( x^{x^{..^{x^{x^x}}}} )' = \frac{ ( x^{x^{..^{x^{x^x}}}})^2}{x( 1 - ( \ln x) x^{x^{..^{x^{x^x}}}} ) } $$

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    $\begingroup$ It seems you've differentiated an infinite power tower, whereas the one in the question has finite height $n$. $\endgroup$
    – user856
    Oct 21, 2013 at 19:04
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    $\begingroup$ Interesting, but I think that $n$ must appear in the final general formula... Why did it disappear? $\endgroup$ Oct 22, 2013 at 3:42
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Here is my approach:

$${ x }^{ x }={ e }^{ y }\\ x\ln { x } =y\\ { \left( { x }^{ x } \right) }'=y'{ e }^{ y }=\left( 1+\ln { x } \right) { e }^{ y }=\left( 1+\ln { x } \right) { x }^{ x }\\ $$$$\\ { x }^{ { x }^{ x } }={ e }^{ z }\\ { x }^{ x }\ln { x } =z\\ { \left( { x }^{ { x }^{ x } } \right) }'=z'{ e }^{ z }=\left( { \left( { x }^{ x } \right) }'\ln { x } +{ x }^{ x-1 } \right) { e }^{ z }=\left( \left( 1+\ln { x } \right){ x }^{ x } \ln { x } +{ x }^{ x-1 } \right) { x }^{ { x }^{ x } }\\ $$ $${ x }^{ { x }^{ { x }^{ x } } }={ e }^{ w }\\ { x }^{ { x }^{ x } }\ln { x } =w\\ { \left( { x }^{ { x }^{ { x }^{ x } } } \right) }'=w'{ e }^{ w }=\left( { \left( { x }^{ { x }^{ x } } \right) }'\ln { x } +{ x }^{ { x }^{ x }-1 } \right) { e }^{ w }=\left( \left( \left( \left( 1+\ln { x } \right) \ln { x } { x }^{ x }+{ x }^{ x-1 } \right) \right) { x }^{ { x }^{ x } }\ln { x } +{ x }^{ { x }^{ x }-1 } \right) { x }^{ { x }^{ { x }^{ x } } }$$

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Well for starters, if

$y = x^x, \tag{1}$

then we can use the technique of the logarithmic derivative. We have

$\ln y = x \ln x, \tag{2}$

from which

$y' / y = \ln x + 1, \tag{3}$

so

$y'(x) = y(\ln x + 1) = x^x(\ln x + 1). \tag{4}$

The other derivatives take a little more time, I'll try and get back to y'all!

Hope this little bit helps! Cheerio,

and as always,

Fiat Lux!!!

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  • $\begingroup$ It seems the solutions for the other derivatives, $x^{x^x}$ etc, have been adequately covered by some of the other answers, so I'll recuse myself from the task! Thanks to one and all! $\endgroup$ Oct 21, 2013 at 20:56
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Obviously $$\frac{d}{dx} [ x \%\% n ] =\sum_{k=1}^n (\ln (x))^{n-k} x^{ \sum_{i= k-2}^{n-1} [ x \%\% i ] - 1 } $$

where $a \%\% b = a^{a^{a^{a^{\dots {}^a}}}}$ ($b$ number of times)

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First derivative of $x^x$

The first derivative of $x^x$ is a rather elegant one:

$$\left(x^x\right)' = \left(e^{\ln x^x}\right)' = \left(e^{x\ln x}\right)'$$

Using the chain rule this becomes,

$$ e^{x\ln x} (x \ln x)' = e^{x \ln x} \left(\ln x + \frac{x}{x}\right) $$

We got $e^{x \ln x}$ by transforming $x^x$ so we use that substitution,

$$ \left(x^x\right)' = x^x(\ln x + 1) $$

First derivative of ${}^nx$

Your second question is about the first derivative of tetration. There are many notations used but one of the most popular is Rudy Rucker notation. Here $^0x = 1$ and ${}^nx = x^{^{n - 1}x}$, for example $x^{x^x} = {}^3x$.

My attempt at finding $\left({}^nx\right)'$:

If $n = 0$

$$ \left({}^0x\right)' = (1)' = 0 $$

If $n > 0$, we get using the definition,

$$ \left({}^nx\right)' = \left( x^{^{n - 1}x} \right)' $$

Using a similar approach as used above we get,

$$ \left( e^{^{n - 1}x \ln x} \right)' = e^{^{n - 1}x \ln x} \left(\left(^{n - 1}x\right)' \ln x + \frac{^{n - 1}x}{x} \right) $$

Which after substitution gives us,

$$ \left({}^nx\right)' = {}^nx \left( \left({}^{n - 1}x \right)' \ln x + \frac{^{n - 1}x}{x} \right)$$

I'm a CS student in the first bachelor so my solution is probably naive. It also seem too simple compared to the other results, so there could be/probably is something wrong. If anyone has any comments they would be highly appreciated!

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