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Find the derivative and simplify your answer as much as possible.
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\[f(x) = \ln(2+lnx)\]
Find \[f'(\frac{1}{e})\]
f'(x) = .... ?
chain rule
I used it and got \[\frac{1}{x(2+lnx)}\]
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correct, just plug in x = 1/e and use the fact that ln e =1
but look at this\[\frac{1}{(\frac{1}{e}*(2+\ln(\frac{1}{e}))}\]
ln (1/A) = ln A^-1 = -ln A
by the property that \(\Large \log y^x =x\log y \)
So it all comes to become e
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correct :)
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