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Homework Help: Seems I can't differentiate properly or I need some help with it.

  1. May 14, 2008 #1


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    1. The problem statement, all variables and given/known data
    Part of a question here...

    Show that

    [tex]\frac{d}{dx} \left[ \frac{x}{(4-x^2)^n} \right ] = \frac{1-2n}{(4-x^2)^n} - \frac{8n}{(4-x^2)^{n+1}}[/tex]

    2. Relevant equations

    [tex]\frac{d}{dx}(\frac{u}{v}) = \frac{v\frac{du}{dx}-u\frac{dv}{dx}}{v^2} [/tex]

    3. The attempt at a solution

    So, using the quotient law I get

    [tex] \frac{(4-x^2)^n \times 1 -(x) \times n(4-x^2)^{n-1} \times -2x}{(4-x^2)^{2n}}[/tex]


    [tex] =\frac{1}{(4-x^2)^n} - \frac{2nx^2}{(4-x^2)^{n+1}}[/tex]

    I seem to have the denominators correct but not the numerators. Did I do it wrong or are there more ways to simplify?
    Last edited: May 14, 2008
  2. jcsd
  3. May 14, 2008 #2
    I can't get the result either... I get the same as you, only a + instead of a - between the fractions...

    Maple backs me up... According to maple the result is:
    [tex]\frac{1}{(4-x^2)^n} + \frac{2nx^2}{(4-x^2)^{n+1}}[/tex]

    And no matter what I do I cannot get it to simplify / expand / factor, whatever, into what you have to show... I don't think they are equal.
  4. May 14, 2008 #3


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    I checked it with Mathematica, Nick89 is correct. The thread starter is off by a sign but
    \frac{d}{dx} \left[ \frac{x}{(4-x^2)^n} \right ] = \frac{1-2n}{(4-x^2)^n} - \frac{8n}{(4-x^2)^{n+1}}
    definitely doesn't hold.
  5. May 14, 2008 #4


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    If you are really feeling unconfident that you might be missing a simplification, try putting say x=0 and n=2 and comparing the results of the two. They are different, so your's in right (once you fix that sign error).
  6. May 14, 2008 #5


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    Well I guess I should post the entire question:

    [tex]I_n =\int _{0} ^{1} \frac{1}{4-x^2}[/tex]

    for n=1,2,3,...
    Verify that

    \frac{d}{dx} \left[ \frac{x}{(4-x^2)^n} \right ] = \frac{1-2n}{(4-x^2)^n} - \frac{8n}{(4-x^2)^{n+1}}[/tex]

    and hence prove that

    [tex]8I_{n+1}=(2n-1)I_n +\frac{1}{3^n}[/tex]

    EDIT: Ahh...-ve*-ve=+ve...Will fix now... but the question being wrong is really odd.
  7. May 14, 2008 #6
    Rockfreak, look at this, and now take the [tex]x^2[/tex] term on the numerator of the second term and put in


    and simplify the result.
  8. May 14, 2008 #7


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    :frown: I hate questions where you always need to remember that 1-1=0
  9. May 14, 2008 #8
    My students absolutely hated those 1/1=1 and 1-1=0 tricks you use to simplify algebraic expressions! There is reason behind them of course, but if done with no explanation they look like magic.

    In this one you know that you want to have 4-x^2 on the top of the 2nd term to split it up into the two types of terms that appear in the answer. So you're like I need to express the numerator [tex]x^2[/tex] in the form [tex]a(4-x^2)+b[/tex] for some numbers a and b to get the expression into the form that I want.
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