I said, using the quotient rule,

f'(x)[1(lnx) - x(1/x)]/(lnx)^2 = lnx - 1

so f'(e) = (lne - 1) / (lne)^2 = 1-1/1 = 0/1 = 0

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- Apr 4th 2010, 09:38 AM #1

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- Apr 4th 2010, 09:44 AM #2

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