Question

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

Answer 1

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))`