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Let y=lnx/e^x. Find the first and second derivative,
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u = lnx --> u' = ...? v = e^x --> v' = ....? Using quotient rule
\[\frac{ \frac{ 1 }{ x } e^x-e^x(\ln(x))}{ e^2x }\]
divided by \[e^{2x}\]
=\[\frac{ 1/x-\ln(x) }{ e^2 }\]
second derivative \[\frac{ (\frac{ -1 }{ x^2 }-\frac{ 1 }{ x })(e^2) -0}{ e^4 }\]
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=\[ \frac{ -x-x^2 }{ x^3e^2 } \]
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