Question

How do you differentiate `y = (3x - 5) / (3x^2 + 7)`?

Answer 1

Use the quotient rule.

Answer:

`f'(x)=(-9x^2+30x+21)/(3x^2+7)^2`

Explanation:

Quotient rule

For function

`f(x)=u/v`

`f'(x)=(vu'-uv')/v^2`

So:

For `f(x)=(3x-5)/(3x^2+7)`

let
`u=3x-5`
so
`u'=3`
let
`v=3x^2+7`
so
`v'=6x`

`f'(x)=((3x^2+7)*3-(3x-5)*6x)/(3x^2+7)^2`

Expand and simplify:

`f'(x)=(9x^2+21-18x^2+30x)/(3x^2+7)^2`

`f'(x)=(-9x^2+30x+21)/(3x^2+7)^2`