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August 27, 2014

August 27, 2014

Posted by **Ifi** on Sunday, October 28, 2012 at 8:48am.

I get f'(x)=e^(xlnx) (1+lnx) while at wolfram is x^x(ln x+x)...sorry for my miscalculation / typo error...

- Math -
**Steve**, Sunday, October 28, 2012 at 11:59amYour answer is correct, since e^(x lnx) = (e^(lnx))^x = x^x

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