stuck on f '(x)

[PTstudent]

Hello friends. I think I have most of this problem worked out. original f(x) is x(3x - 9)^3 so u = 3x - 9 and u' = 3
f '(x)= x(3)(3x-9)^2(3) + (3x-9)(1) then f '(x)= 9x(3x-9)^2 + (3x-9)^3 I am positive what I have so far is correct. But I don't know what to do with the (3x-9)^3 any help is appreciated.

 

[wjm11]

Hello friends. I think I have most of this problem worked out. original f(x) is x(3x - 9)^3 so u = 3x - 9 and u' = 3
f '(x)= x(3)(3x-9)^2(3) + (3x-9)(1) then f '(x)= 9x(3x-9)^2 + (3x-9)^3 I am positive what I have so far is correct. But I don't know what to do with the (3x-9)^3

What you have done looks fine (except for the missing "^3" in "(3x-9)(1)"). f '(x)= 9x(3x-9)^2 + (3x-9)^3 is correct. Why do you think you need to do something more?

 

[PTstudent]

oh ok that's what I thought. I wasn't sure if it could be factored somehow or something. Thank you for your help.