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Old 2004-08-16, 03:08   #5
jinydu
 
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Dec 2003
Hopefully Near M48

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I do know how to differentiate f(x) = x^x. Maybe that can help:

f(x) = x^x = (e^ln x)^x = e^(x*ln x)

f '(x) = x*ln x * e^(x*ln x) = x*ln x * x^x = (x^x) * x * ln x

Last fiddled with by jinydu on 2004-08-16 at 03:09
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