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Trouble using the chain rule product rule and quotient rule

  1. Apr 2, 2007 #1
    im having a lot of trouble using the chain rule product rule and quotient rule..i can do them fine seperatly but when they're put together i cant get them like if you have (x^2-1)^4 (2-3x) i would start with
    4(x^2-1)^3(2x)(2-3x)+(x^2-1)^4(-3)
    have i done something wrong here because i never get the right answer with these ones i dont know if i messed up here or later on
     
  2. jcsd
  3. Apr 2, 2007 #2

    cristo

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    You evaluation of the derivative is correct.
     
  4. Apr 2, 2007 #3
    ok so the next thing i would do would be to factor out the (x^2-1)to get
    (8x)(x^2-1)^3(2-3x)+(x^2-1)(-3)
    is this still right?
     
  5. Apr 2, 2007 #4
    i should have a 1 in there before the plus sign too
     
  6. Apr 2, 2007 #5

    cristo

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    No. you've not factored out the (x^2-1), since you have (x^2-1)^3 in the first term. You're second term is incorrect since it should still be (x^2-1)^4. If you take out (x^2-1)^3 from both terms, and put it outside a bracket at the front, what do the remaining terms in the bracket look like?
     
  7. Apr 2, 2007 #6
    would it be
    ((x^2-1)^3)(1)(8X)(2-3X)+(x^2-1)(-3) ?
     
  8. Apr 2, 2007 #7

    cristo

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    You're missing a bracket.. write it like this, omitting the unnecessary ones:
    (x^2-1)^3(8x(2-3x)-3(x^2-1)).

    Now, can you simplify this?
     
  9. Apr 2, 2007 #8
    my first thought would be to multiple the 8x and -3 through their brackets then combine like terms?
     
  10. Apr 2, 2007 #9

    cristo

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    Good idea. Give it a go and see what you come up with.
     
  11. Apr 2, 2007 #10
    ok that workedi got the right answer thanks for your help
     
  12. Apr 2, 2007 #11

    cristo

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    You're very welcome!
     
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