Question

How do you find the derivative of `f(x)=(2x+6)/(3x^2+9)`?

Answer 1

`f'(x)=(-6(x^2+6x-3))/(3x^2+9)^2`

Explanation:

We should use quotient rule i.e if `f(x)=g(x)/(h(x))`, then

`f'(x)=(h(x)xxg'(x)-g(x)xxh'(x))/(h(x))^2`

As `g(x)=2x+6` and `h(x)=3x^2+9`

Hence `f'(x)=((3x^2+9)xx2-(2x+6)xx6x)/(3x^2+9)^2` or

`f'(x)=(6x^2+18-12x^2-36x)/(3x^2+9)^2` or

`f'(x)=((-6x^2-36x+18))/(3x^2+9)^2` or

`f'(x)=(-6(x^2+6x-3))/(3x^2+9)^2` or

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

Answer 2

I get`" " (dy)/(dx)=- (2(x^2+6x-3))/(3(x^2+3)^2)`

Explanation:

Quotient rule states that for `y=u/v` where `u=f(x)" and "v=g(x)`

Then `dy/dx= (v (du)/(dx) - u (dv)/(dx))/v^2`

Set `y=f(x)=(2x+6)/(3x^2+9) =u/v`

Then `(du)/(dx)=2" "(dv)/(dx)=6x`

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

`color(brown)(=>dy/dxcolor(green)(= (v (du)/(dx) - u (dv)/(dx))/v^2) -> ((3x^2+9)(2)-(2x+6)(6x))/((3x^2+9)^2))`

Not that :`(3x^2+9)^2 = (3(x^2+3))^2 = 9(x^2+3)^2`

`" "=(2cancel((3x^2+9)))/((3x^2+9)^(cancel(2))) -(2(x+3)(6x))/(9(x^2+3)^2)`

`" "=2/(3(x^3+3))- (4x(x+3))/(3(x^3+3)^2)`

`" "=(2(x^2+3)-4x(x+3))/(3(x^2+3)^2)`

`" "=(2x^2+6-4x^2-12x)/(3(x^2+3)^2)`

`" "=(-2x^2-12x+6)/(3(x^2+3)^2)`

`" "(dy)/(dx)= (-2(x^2+6x-3))/(3(x^2+3)^2)`

`" "color(blue)((dy)/(dx)=- (2(x^2+6x-3))/(3(x^2+3)^2))`