Question

How do you differentiate `f(x)=(x-3)/(x+3)^2` using the quotient rule?

Answer 1

`f'(x)=-(x-9)/(x+3)^3`

Explanation:

We use the quotient rule to find `f'(x)`:

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

So

`d/dx(x-3)/(x+3)^2=((x+3)^2-2(x+3)(x-3))/(x+3)^4=(x+3-2x+6)/(x+3)^3=-(x-9)/(x+3)^3`

Answer 2

`d/dx(f(x))=-(x-9)/(x+3)^3`

Explanation:

`f(x)=(x-3)/(x+3)^2`
let
`y=f(x)`
`y=(x-3)/(x+3)^2`
Let
`u=x-3`
Then
`du/dx=1`
Let
`v=(x+3)^2`
`(dv)/dx=2(x+3)`
and
`v^2=(x+3)^4`
We have by quotient rule in diferentiation

`d/dx(u/v)=(v(du)/dx-u(dv)/dx)/v^2`
Substituting
`d/dx((x-3)/((x+3)^2))=((x+3)^2(1)-(x-3)(2(x+3)))/(x+3)^4`
Simplifying
`d/dx(y)=((x+3)(x+3-2(x-3)))/(x+3)^4`

`d/dx(f(x))=((x+3)(x+3-2x+6))/(x+3)^4`

`=(-x+9)/(x+3)^3`
`d/dx(f(x))=-(x-9)/(x+3)^3`