How do you differentiate $y=(x/e^x)Lnx$ ?

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Answer 1 · 2017-03-31T19:41:58

see below

$(fg)'=fg'+gf'$
$(f/g)' = (gf'-fg')/g^2$

$y=(x/e^x)lnx$

$color(red)(f=x/e^x), color(blue)( g=lnx$

$color(red)(f'=(e^x-xe^x)/e^(2x)), color(blue)( g'(x)=1/x$

$color(green)(y'=fg'+gf'$

$color(green)(y'=x/e^x*1/x+lnx*( e^x-xe^x)/e^(2x))$

$color(green)(y'=cancelx/e^x*1/cancelx+lnx*( e^x-xe^x)/e^(2x))$

$color(green)(y'=1/e^x+(lnx( e^x-xe^x))/e^(2x))$

$color(green)(y'=(e^x+lnx( e^x-xe^x))/e^(2x))$

How do you differentiate y=(x/e^x)Lnxy=(x/e^x)Lnx?