Question

How do you find the derivative of `G(x) = (4-cos(x))/(4+cos(x))`?

Answer 1

`(8sinx)/(4+cosx)^2`

Explanation:

The derivative of the quotient is defined as follows:
`(u/v)'=(u'v-v'u)/v^2`

Let `u=4-cosx` and `v=4+cosx`

Knowing that `color(blue)((d(cosx))/dx=-sinx)`

Let us find `u'` and `v'`

`u'=(4-cosx)'=0-color(blue)((-sinx))=sinx`
`v'=(4+cosx)'=0+color(blue)((-sinx))=-sinx`

`G'(x)=(u'v-v'u)/v^2`
`G'(x)= (sinx(4+cosx)-(-sinx)(4-cosx))/(4+cosx)^2`
`G'(x)=(4sinx+sinxcosx+4sinx-sinxcosx)/(4+cosx)^2`
`G'(x)=(8sinx)/(4+cosx)^2`