Question

How do you differentiate `e^x/ln x`?

Answer 1

`y'=(e^x(xlnx + 1))/(xln^2x)`

Explanation:

`y=f/g`

`y'=(f'*g-f*g')/g^2`

`y=e^x/lnx => f=e^x, g=lnx`

`f'=e^x, g'=1/x`

`y'=(e^xlnx - e^x/x)/ln^2x`

`y'=(e^x(xlnx - 1))/(xln^2x)`