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

1 Answer
Mar 31, 2017

see below

Explanation:

Use the Product Rule and the Quotient Rule

(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))