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While differentiating ${x}^{{x}^{x}}$ using power rule, what should be the base and exponent, i.e. base=$x$, exponent=$x^x$ or base=$x^x$, exponent=$x$. Any WHY?

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  • $\begingroup$ There is no inbuilt power rule to differentiate functions which have both base and exponent as variable. The usual formulas apply for the case when one of the base and exponent is constant. $\endgroup$ – Paramanand Singh Dec 3 '17 at 7:21
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    $\begingroup$ And the expression in question means $x^{(x^{x})} $ and not $(x^{x}) ^{x}$. You can use technique of logarithmic differentiation here. $\endgroup$ – Paramanand Singh Dec 3 '17 at 7:22
  • $\begingroup$ @ParamanandSingh: A CAS needs a rule. It goes upto last power to say topmost is exponent and $x^x$ is base which is by all means true relatively. $\endgroup$ – Rorschach Dec 3 '17 at 7:32
  • $\begingroup$ Ok, got your point. Using CAS does require you to adapt to their ways (whereas it should have been the other way round). $\endgroup$ – Paramanand Singh Dec 3 '17 at 7:41
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Think of $x^{x^x}$ as $$f(x)^{g(x)}\tag 1$$ where $f(x)=x$ and $g(x)=x^x$. I you consider $$F(x)^{G(x)}\tag 2$$ with $F(x)=x^x$ and $G(x)=x$ then you will have $(x^x)^x=x^{x^2}$ which is different from $(1)$.

So, stick to $(1)$. There is a generalized power rule to be applied to $(1)$: $$\frac d{dx}f(x)^{g(x)}=g(x)f(x)^{g(x)-1}\frac {df}{dx}+f(x)^{g(x)}\ln(f(x))\frac {dg}{dx}$$

if $f(x)>0$. We have $\frac{df}{dx}=1$ and $\frac{dg}{dx}=x^x(1+\ln(x)).$ The final result can be given now. Note that this is true if $f(x)=x>0$.

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  • $\begingroup$ I enforced a rule in CAS that as (base^exp)^exp to base^(exp^exp) and it seems to work fine now. I hope it doesn't collide with other expression structures. $\endgroup$ – Rorschach Dec 3 '17 at 9:04

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