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Help with the derivative: (x+1)(x-3)^3

  1. Mar 7, 2009 #1
    1. The problem statement, all variables and given/known data
    find f'x
    f(x) = (x+1)(x-3)^3






    3. The attempt at a solution

    i know its the product and chain rules.
    (x+1)*3(x-3)^2 + (x-3)^3 *1
    but then i get all confused with the numbers.
    could someone walk me through this?
    thanks!
     
  2. jcsd
  3. Mar 7, 2009 #2

    HallsofIvy

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    Okay, you have the derivative! What more do you want? I see no reason to multiply it out because the original problem has it factored. I might suggest factoring out (x-3)2: 3(x+1)(x-3)2+ (x-3)3= (x-3)2(3(x+1)+ x-3)= (x-3)2(3x+ 3+ x- 3)= 4x(x-3)2. Is that better?
     
  4. Mar 7, 2009 #3
    Maybe you need a refresher in Algebra?
     
  5. Mar 7, 2009 #4

    Hurkyl

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    I should point out to the original poster that this isn't an insult -- calculus is a subject where you make use of

    1. Calculus
    2. Arithmetic
    3. Algebra
    4. Trigonometry
    5. (Whatever I forgot to put in this list)

    It seems far too common for people to think "I'm having trouble in my calculus class, therefore I need to work on my calculus skills" -- but to the contrary, I assert that the vast majority of difficulties people have in calculus is for reasons completely unrelated to calculus.
     
  6. Mar 7, 2009 #5
    Yea I didn't mean it as an insult either sorry if it came off that way, being able to do algebraic manipulation with ease is a key to success in higher maths. So if you are having troubles with that sort of stuff don't let it slide, it will only get worse.
     
  7. Mar 7, 2009 #6

    Hurkyl

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    I never know if such comments really do come off that way, but I always worry. :frown: Maybe the opening poster can tell us what his impression was when he read that comment, that way he can educate me. :smile:
     
  8. Mar 7, 2009 #7
    An safer approach than the product rule is to expand it completely then look for the derivative. This method is reasonable since the expression you have there isn't too complicated for expansion.
     
  9. Mar 7, 2009 #8

    Hurkyl

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    That's debatable. I think I'm less likely to make a typographical error when using the product rule and factoring than I am of expanding that out.
     
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