Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Chain rule differentiation

  1. Nov 1, 2008 #1
    1. The problem statement, all variables and given/known data

    f(x)= x^2(x-2)^4 solve for f '(x)

    2. Relevant equations

    f(x) = x^2(x-2)^4

    3. The attempt at a solution


    The answer is given in the book as 2x(x-2)^3(3x-2)

    i'm not following any progression that gets me to that solution regardless of how many times I work through it.

    If someone could give me a step by step on this on I'd greatly appreciate it.

  2. jcsd
  3. Nov 1, 2008 #2


    User Avatar
    Homework Helper
    Gold Member

    You'll need to use the product rule since f(x) is the product of x^2 and (x-2)^4.
  4. Nov 1, 2008 #3
    If you have a function of two products:
    f(x)=g(x)h(x) where g and h are two functions with respect to x, you have to use the product rule so,
    f '(x)= g(x)h'(x) + h(x)g'(x)
    can you apply this to your function f(x)?
    (in your case, g(x)=x^2 and h(x)=(x-2)^4 but it could be the other way round too)
  5. Nov 1, 2008 #4
    ok, if i'm using the product rule correctly, that takes me to:

    x^2[4(x-2)^3(1)] + (x-2)^4(2x)

    for some reason I'm still not seeing how that will go to 2x(x-2)^3(3x-2) for a final solution

    Thanks for the help!
  6. Nov 1, 2008 #5
    sara's reply wasn't showing when I posted my last reply...let me work on it some more from what she posted and see if I can get anywhere yet.

  7. Nov 1, 2008 #6


    User Avatar
    Homework Helper
    Gold Member

    Your solution is correct, to get it into the same form as the one in your answer key, factor out a [itex]2x(x-2)^3[/itex], that gives [itex]2x(x-2)^3[2x+(x-2)][/itex] then just simplify [itex][2x+(x-2)][/itex]
  8. Nov 1, 2008 #7
    gotta love how algebra mistakes cause more errors than actual cal fundamentals.

    I was out of school many years, in fact the last math class I had was college algebra in 2001. So, i'm really struggling with stuff that isn't all that difficult in the grand scheme of things.

    Thank you both for being patient and helping me out!
  9. Nov 1, 2008 #8
    ok, one more problem here that i've tried to work. I think i'm ok on it except for the factoring out in the last step.

    the given problem is f(x)= x(3x-7)^3

    I got f '(x) = x[3(3x-7)^2] + (1)(3x-7)^3 via the product rule.

    after attempting to factor I ended up with f '(x) = (6x-7)(3x-7)^2, which I don't think is correctly factored.
  10. Nov 1, 2008 #9
    I think you made a small mistake, when you take (3x-7)^2 as common factor, you should get:
  11. Nov 1, 2008 #10
    once again, gotta love algebra.

    Thanks Sara!
  12. Nov 1, 2008 #11
    U're welcome.
  13. Nov 1, 2008 #12


    User Avatar
    Homework Helper
    Gold Member

    The derivative of 3x-7 is (3x), not just x. You should have;

    f '(x) = (3x)[3(3x-7)^2] + (1)(3x-7)^3=(12x-7)(3x-7)^2
  14. Nov 1, 2008 #13
    i meant -7
    sorry. (misprint)
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook