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Math Help - really simple multiplication prob

  1. #1
    Red
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    really simple multiplication prob

    got some variables in an example that are being combined, as follows:

    2x * x^2 + 4 - x^2 * 2x = 8x

    does this really equal 8x?

    (2x*x^2) + 4 -(x^2*2x)

    both left and right cancel out leaving the 4 no?

    i'm sure i'm missing something simple.
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  2. #2
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    that lot simplifies to 4=8x

    i think that was your question
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  3. #3
    Red
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    it was more along the lines that all of the problem "2x * x^2 + 4 - x^2 * 2x" simplifying to 8x. sorry if i wasn't clear. but yeah, i'm getting 4. this is part of a derivative problem but this part doesn't involve calculus....
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  4. #4
    MHF Contributor Unknown008's Avatar
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    What is the exact question?

    Because for you to get 8x, you need to cancel out the x^2 and have (4 * 2x) on the right...
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  5. #5
    Red
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    Find the value of the derivative of the function \frac{x^2}{x^2 + 4}, (if it exists) at the origin.
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  6. #6
    MHF Contributor Unknown008's Avatar
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    Ok, so, you will need to use the quotient rule, which is if

    y = \frac{u}{v}, then y' = \frac{v \frac{du}{dv} - u\frac{dv}{du}}{v^2

    This gives:

    \frac{d(\frac{x^2}{x^2 + 4})}{dx} = \frac{(x^2 + 4).2x - x^2(2x)}{(x^2 +4)^2}

     = \frac{2x^3 + 8x - 2x^3}{x^4 + 8x^2 + 16}

    Now, you get 8x.
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  7. #7
    MHF Contributor Unknown008's Avatar
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    EDIT: I think a bug occured as I typed my answer... ok, it seems to have been corrected. Thanks!
    Last edited by Unknown008; June 24th 2010 at 10:39 AM. Reason: Double post
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  8. #8
    MHF Contributor Unknown008's Avatar
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    Ok, so, you will need to use the quotient rule, which is if

    y = \frac{u}{v}, then y' = \frac{v \frac{du}{dv} - u\frac{dv}{du}}{v^2}

    This gives:

    \frac{d(\frac{x^2}{x^2 + 4})}{dx} = \frac{(x^2 + 4).2x - x^2(2x)}{(x^2 +4)^2}

     = \frac{2x^3 + 8x - 2x^3}{x^4 + 8x^2 + 16}

    Now, you get 8x.
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  9. #9
    Red
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    appreciate it, that cleared it up.
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