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I got $y'= x^2 + \ln({1\over x})\times ({1\over x})^x$

Am I correct?

So $y'=-x^{-x}(\ln{x} + 1)$

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    @ data: You $should$ use natural log.2012-10-17

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$y = x^{-x} \implies \log(y) = -x \log(x)$ Can you now do implicit differentiation?

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    Well, log is more widely used, but this unimportant.2012-10-17
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Some logarithms and the chain rule:

$y=\left(\frac{1}{x}\right)^x=e^{\frac{1}{x}\log x}\Longrightarrow y'=e^{\frac{1}{x}\log x}\left[-\frac{1}{x^2}\log x+\frac{1}{x^2}\right]=$

$=\frac{1}{x^2}\left(\frac{1}{x}\right)^x\left(1-\log x\right)=\frac{1}{x^{x+2}}(1-\log x)$