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Evaluate $\frac{d}{dx}ⁿx$ which $ⁿx=x^{x^{x^{...}}}$, total $n$ $x$'s.

I have tried to observe when $n=1,2,3,4,5$, but it's difficult to see the pattern. Can anyone get it? Thank you.

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As I said in my comment above, define $y= ^n\!\!x$. Then $$\log y=\log ^nx=\log x^{^{n-1}x}=\, ^{n-1}x\log x.$$

Differentiating implicitly, we get $$\frac{d y}{d x}\frac{1}{y}=\frac{d}{dx}(\, ^{n-1}x)\log x+\, ^{n-1}x\frac{1}{x},$$ from which we see $$\frac{d}{dx}(\, ^{n}x)=\frac{\, ^{n}x}{x}\left(x\log x\frac{d}{dx}(\, ^{n-1}x)+\, ^{n-1}x\right).$$

For the first few values of $n$, we get

$n=1:\quad\frac{d}{dx}(\, ^{1}x)=\frac{d}{dx}(x)=1$

$n=2:\quad\frac{d}{dx}(\, ^{2}x)=\frac{\, ^{2}x}{x}\left(x\log x\frac{d}{dx}(\, ^{1}x)+\, ^{1}x\right)=x^x(\log x+1)$

$n=3:\quad\frac{d}{dx}(\, ^{3}x)=\frac{\, ^{3}x}{x}\left(x\log x\frac{d}{dx}(\, ^{2}x)+\, ^{2}x\right)=x^{x^x}\left(x^x(\log x+1)\log x+x^{x-1}\right)$

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  • $\begingroup$ But your answer still have $\frac{d}{dx}(^{n-1}x)$. $\endgroup$ – JSCB Aug 17 '12 at 9:14
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    $\begingroup$ You have to build up the definitions recursively. This is the only way to get an explicit formulas, like I was doing above. The implicit definition is the only solution for general $n$. $\endgroup$ – Daryl Aug 17 '12 at 9:26

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