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Another product rule question

  1. Sep 11, 2007 #1
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution



    I thought i was done here. The book takes a few more steps. it adds:


    to this:

    2(1-x)^1/2 (1-x)^1/2
    ------------ x -----------
    1 (1-x)^1/2

    Why does the last step exist?
  2. jcsd
  3. Sep 11, 2007 #2
    and one more question. i just got to a problem with sqrt(x+1) in the denominator. my origional thought was to put it in the numerator to the power of -1/2 and use the product rule.

    my book uses the quotient rule instead. how do i know which one is the best one to use?
    Last edited: Sep 11, 2007
  4. Sep 11, 2007 #3
    Are the dotted lines supposed to indicate that ir is a fraction?

    And have you used the chain rule yet?

    What is the original function now?
    Is it [tex]y=2x(1-x)^2[/tex] ?
    Last edited: Sep 11, 2007
  5. Sep 11, 2007 #4
    Did you do the product rule right?

    f'(x)g(x) + f(x)g'(x)?

    Or secondly, you can follow the definition of the limit and right

    lim as x approaches deltax 2(x+dx)(1-(x+dx)^2 all over dx

    where dx = delta x

    Use that to confirm if yo did the product rule right, I can't understand the way you wrote your work. Looking at your original function, I don't understand how you got a fraction.
    Last edited: Sep 11, 2007
  6. Sep 11, 2007 #5
    PowerIso, can you understand what he's writing? If the original function is just [tex]y=2x(1-x)^2[/tex]

    I see no need for the product rule as this is just a polynomial of the third degree.....Power Rule.


    Edit: I saw your edit....the fraction is messing me up too..that is why I belive he maybe copied the original function wrong...maybe he meant something else.:confused:
    Last edited: Sep 11, 2007
  7. Sep 11, 2007 #6
    Yes I can understand that, but his question regarded the product rule, so I tackled this problem with that view.
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