Derivatives using Logarithmic Differentiation

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Homework Help Overview

The discussion revolves around finding the derivative of the function y = e^(x^x) using logarithmic differentiation. Participants are exploring the application of logarithmic properties to simplify the differentiation process.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to apply logarithmic differentiation but expresses uncertainty about the next steps after reaching a certain point. Another participant suggests taking the logarithm twice, while a different participant questions the transition from x ln(x) to ln(x) + 1.

Discussion Status

The discussion is ongoing, with participants sharing different approaches and questioning specific steps in the differentiation process. There is no explicit consensus, but various lines of reasoning are being explored.

Contextual Notes

Participants are navigating the complexities of logarithmic differentiation and its implications, with some expressing confusion about the manipulation of logarithmic expressions. No additional constraints or rules are noted in the discussion.

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Homework Statement



Using logarithmic differentiation calculate the derivative of y=e^(x^x)



The Attempt at a Solution



y=e^x^x
LNy=LNe^x^x
LNy=x^xLNe
...
Stuck!

This seems to be the only way you can do it, but once I get to that part I'm not sure what else there is to do. I know lne=1, so does that mean the answer is y'=e^(x^x)*x^x
 
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[tex]\ln(\ln(y))=\ln(x^{x}\ln(e))=\ln(x^{x})=x\ln(x)[/tex]

And:
[tex](\ln(\ln(y))'=\frac{y'}{y\ln(y)}=\ln(x)+1[/tex]
whereby follows:
[tex]y'=y\ln(y)(\ln(x)+1)=e^{x^{x}}x^{x}(\ln(x)+1)[/tex]
 
Maybe you need to take the logarithm two times...
 
I don't see how it goes from xLNx to LNx+1
 

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