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:
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.
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.
You simply combine the two formulas you already mentioned to get:
$$(x^x)'=(x^n)'_{n=x}+(a^x)'_{a=x}=(nx^{n-1})_{n=x}+(a^x\ln a)_{a=x}=x\cdot x^{x-1}+x^x\ln x=x^x(1+\ln x)$$
$$x>0\implies x^x=e^{x\log x}\implies (x^x)'=e^{x\log x}\left(\log x+1\right)=x^x(\log x+1)$$
x^x^x...n times ... ^x?
– Ionică Bizău
Oct 21 '13 at 18:49
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,
$$ ( x^{x^{..^{x^{x^x}}}} )' = \frac{ ( x^{x^{..^{x^{x^x}}}})^2}{x( 1 - ( \ln x) x^{x^{..^{x^{x^x}}}} ) } $$
Here is my approach:
$${ x }^{ x }={ e }^{ y }\\ x\ln { x } =y\\ { \left( { x }^{ x } \right) }'=y'{ e }^{ y }=\left( 1+\ln { x } \right) { e }^{ y }=\left( 1+\ln { x } \right) { x }^{ x }\\ $$$$\\ { x }^{ { x }^{ x } }={ e }^{ z }\\ { x }^{ x }\ln { x } =z\\ { \left( { x }^{ { x }^{ x } } \right) }'=z'{ e }^{ z }=\left( { \left( { x }^{ x } \right) }'\ln { x } +{ x }^{ x-1 } \right) { e }^{ z }=\left( \left( 1+\ln { x } \right){ x }^{ x } \ln { x } +{ x }^{ x-1 } \right) { x }^{ { x }^{ x } }\\ $$ $${ x }^{ { x }^{ { x }^{ x } } }={ e }^{ w }\\ { x }^{ { x }^{ x } }\ln { x } =w\\ { \left( { x }^{ { x }^{ { x }^{ x } } } \right) }'=w'{ e }^{ w }=\left( { \left( { x }^{ { x }^{ x } } \right) }'\ln { x } +{ x }^{ { x }^{ x }-1 } \right) { e }^{ w }=\left( \left( \left( \left( 1+\ln { x } \right) \ln { x } { x }^{ x }+{ x }^{ x-1 } \right) \right) { x }^{ { x }^{ x } }\ln { x } +{ x }^{ { x }^{ x }-1 } \right) { x }^{ { x }^{ { x }^{ x } } }$$
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!!!
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)
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,
$$ \left(x^x\right)' = x^x(\ln x + 1) $$
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!
