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How can we differentiate $(x^{-1})^{({x^{-1})^{x^{-1}}}}$ with respect to $x$?

AakashM
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David Holden
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  • -1 for adding useless text instead of actually making your post higher quality (i.e. adding relevant information, such as what you have tried so far) – Doorknob Dec 23 '13 at 14:59
  • @doorknob thank you for the advice. I'm sorry I thought it might be a "be kind to noobs week" in view of the imminent seasonal festivities. another of my (many) erroneous assumptions. – David Holden Dec 23 '13 at 15:02
  • 75% of your question is now completely unrelated to the topic. If you would make the question text relevant to the question, I would remove my downvote. – Doorknob Dec 23 '13 at 15:04
  • please, you may keep your downvote intact, I actually treasure them more than the (occasional) upvote. but note that, for the reason stated, without the added text I could not have posted the question at all, so I think the text more relevant, in a meta- sense, than your explication allows. however I appreciate your efforts to keep the site in focus, and accept your argument for that reason. – David Holden Dec 23 '13 at 15:08
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    wolframalpha says: $$\frac{d}{dx}\left( \left(\frac{1}{x} \right)^{\left({1 \over x}\right)^{1 \over x}}\right)=-\left[\left(\frac{1}{x}\right)^{\left({1 \over x}\right)^{1 \over x}+{1 \over x }+2}\right]\left(x+\log^2 \left({1 \over x}\right)+\log \left({1 \over x}\right)\right)$$ – K. Rmth Dec 23 '13 at 15:09
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    Write your function as $\exp\Bigl( (1/x)^{1/x} \cdot\ln(1/x) \Bigr)$. Use the chain and product rules (and a similar trick to find the derivative of $(1/x)^{1/x}$). – David Mitra Dec 23 '13 at 15:13
  • It often helps to log the function and then use L'Hospital's rule. Did you try this? – Alex Dec 23 '13 at 15:14
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    @DavidMitra the details of the automatic quality filter are not disclosed to us, the general public; however, I suspect that the most significant part of my edit, as far as the filter was concerned, was starting the sentence with a capital letter (see also this tag on the pan-network meta site, and links therefrom) – AakashM Dec 23 '13 at 15:28
  • ;-) I have read about these "capital letters", and am eager to learn more. could anyone suggest a (fairly brief) text suitable for someone at my beginner's level of knowledge of the English tongue? – David Holden Dec 23 '13 at 15:32
  • @AakashM btw thanks for the edit, and the possibly useful advice re initial capitals – David Holden Dec 23 '13 at 15:43

3 Answers3

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Well, for positive real $x,$ we can use the fact that $\alpha^\beta=\exp(\beta\ln\alpha)$ for any positive real $\alpha$ and any real $\beta,$ together with the power rule of logarithms, to see that $$\begin{align}\left(x^{-1}\right)^{\left(x^{-1}\right)^{x^{-1}}} &= \exp\left(\left(x^{-1}\right)^{x^{-1}}\ln\left(x^{-1}\right)\right)\\ &= \exp\left(-\left(x^{-1}\right)^{x^{-1}}\ln(x)\right)\\ &= \exp\left(-\exp\left(x^{-1}\ln\left(x^{-1}\right)\right)\ln(x)\right)\\ &= \exp\left(-\exp\left(-x^{-1}\ln(x)\right)\ln(x)\right).\end{align}$$

After that, it simply becomes an exercise in repeated use of the chain rule, together with the product rule.

Cameron Buie
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$$(x^{-1})^{({x^{-1})^{x^{-1}}}}=(x^{-1})^{({x^{-1/x})}}=x^{-x^{-1/x}}=y$$ $$\ln y=-x^{-1/x}\ln x=z\ln x$$ $$y'=y(z'\ln x+\frac{z}{x})$$ where $$-x^{-1/x}=z$$ $$\ln z=\frac{1}{x}\ln x\Rightarrow z'=z\frac{1-\ln x}{x^2}$$

Adi Dani
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spurred on by Cameron's excellent approach, I'm wondering about:

let $z=\frac1x$, so $$f(x) = z^{z^z}$$ and $$f'(x) = \frac{d}{dz} z^{z^z} \frac{dz}{dx} = -x^{-2}\frac{d}{dz} z^{z^z} $$ as a possible alternative route?

David Holden
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    This is certainly a good idea, but you should not post non-answers as answers. – Carsten S Dec 23 '13 at 16:00
  • the question of differentiating $z^{z^z}$ has recently been very adequately dealt with here: http://math.stackexchange.com/questions/616014/nth-derivative-of-a-tetration-function/616170#616170 and in the posts linked to by OP there – David Holden Dec 23 '13 at 16:14
  • What then was the purpose of your question? – Carsten S Dec 23 '13 at 16:27
  • the substitution I just mentioned had not occurred to me when I asked the question. my purpose is to learn mathematics, though from a few weeks experience I would add that it is extremely curious (in Dodgson's sense of the term) how many amateur moral philosophers one encounters in the process. – David Holden Dec 23 '13 at 16:32