Question

How do you differentiate `y=(2e^x-1)/(5e^x+9)`?

Answer 1

`(23e^x)/(5e^x+9)^2`

Explanation:

`y=(2e^x-1)/(5e^x+9)=u/v`

`(dy)/(dx)=(vu'-uv')/v^2`

`u'=2e^x`

`v'=5e^x`

`(dy)/(dx)=(2e^x(5e^x+9)-5e^x(2e^x-1))/(5e^x+9)^2`
`=(10e^(2x)+18e^x-10e^(2x)+5e^x)/(5e^x+9)^2`
`=(23e^x)/(5e^x+9)^2`