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Mathematics 16 Online
OpenStudy (anonymous):

Find the derivative and simplify your answer as much as possible.

OpenStudy (anonymous):

\[f(x) = \ln(2+lnx)\]

OpenStudy (anonymous):

Find \[f'(\frac{1}{e})\]

hartnn (hartnn):

f'(x) = .... ?

hartnn (hartnn):

chain rule

OpenStudy (anonymous):

I used it and got \[\frac{1}{x(2+lnx)}\]

hartnn (hartnn):

correct, just plug in x = 1/e and use the fact that ln e =1

OpenStudy (anonymous):

but look at this\[\frac{1}{(\frac{1}{e}*(2+\ln(\frac{1}{e}))}\]

hartnn (hartnn):

ln (1/A) = ln A^-1 = -ln A

hartnn (hartnn):

by the property that \(\Large \log y^x =x\log y \)

OpenStudy (anonymous):

So it all comes to become e

hartnn (hartnn):

correct :)

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