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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2$\begingroup$ The derivative is defined recursively. Take logs of both sides and implicitly differentiate. $\endgroup$– DarylAug 17, 2012 at 8:34
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$\begingroup$ A google search for "differentiation" AND "tetration" brings up a lot of relevant items. $\endgroup$– Dave L. RenfroAug 17, 2012 at 15:08
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1 Answer
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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 $\frac{d}{dx}(^{n-1}x)$. $\endgroup$– JSCBAug 17, 2012 at 9:14
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2$\begingroup$ 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 $n$. $\endgroup$– DarylAug 17, 2012 at 9:26