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Math Help - Finding the derivative of a fraction

  1. #1
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    Finding the derivative of a fraction

    How do I find the derivative of (4x+(x^3)-(x^4))/x^2, where x does not equal 0

    I tried using the quotient rule but I can't seem to get the right answer. Can some please work this out step by step so I can see where I'm going wrong?
    Thanks in advance
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  2. #2
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    Re: Finding the derivative of a fraction

    Hello,


    Try to use the following rule:
    (\frac{f}{g})'=\frac{f'g-g'f}{g^2}
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  3. #3
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    Re: Finding the derivative of a fraction

    I have but Im still getting it wrong
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  4. #4
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    Re: Finding the derivative of a fraction

    It would help if you showed us your working... we're not going to give you the full answer with working if we see no effort on your part.
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    Re: Finding the derivative of a fraction

    Quote Originally Posted by JellyOnion View Post
    How do I find the derivative of (4x+(x^3)-(x^4))/x^2, where x does not equal 0

    I tried using the quotient rule but I can't seem to get the right answer. Can some please work this out step by step so I can see where I'm going wrong?
    Thanks in advance
    \displaystyle \frac{4x + x^3 - x^4}{x^2} = \frac{4x}{x^2} + \frac{x^3}{x^2} - \frac{x^4}{x^2} = \frac{4}{x} + x - x^2 = 4x^{-1} + x - x^2

    Finding the derivative should now be a piece of cake
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  6. #6
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    Re: Finding the derivative of a fraction

    But then wouldn't you get (-4x^-2)+1-2x?

    My textbook says the answer is (-2x^3+x^2-4)/x^2
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  7. #7
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    Re: Finding the derivative of a fraction

    Not sure how, this is my first time on this forum
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  8. #8
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    Re: Finding the derivative of a fraction

    y= \frac{4x+x^3-x^4}{x^2}=\frac{x(4+x^2-x^3)}{x^2}=\frac{4+x^2-x^3}{x}\Rightarrow \frac{dy}{dx}= can you get it from here using the quotient rule?
    Thanks from JellyOnion
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  9. #9
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    Re: Finding the derivative of a fraction

    Ohhhhhhh, yeah I got it thanks so much Joe
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  10. #10
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    Re: Finding the derivative of a fraction

    Quote Originally Posted by JellyOnion View Post
    But then wouldn't you get (-4x^-2)+1-2x?

    My textbook says the answer is (-2x^3+x^2-4)/x^2
    Aren't those the same thing?

    If you want to use the quotient rule directly, it's more work, but here's what it looks like:

    \left(\frac{4x+x^3-x^4}{x^2}\right)' =

    \frac{(x^2)(4x+x^3-x^4)'-(4x+x^3-x^4)(x^2)'}{x^4}=

    \frac{x^2(4+3x^2-4x^3)-(4x+x^3-x^4)(2x)}{x^4}=

    \frac{4x^2+3x^4-4x^5-8x^2-2x^4+2x^5}{x^4}=

    \frac{-4x^2+x^4-2x^5}{x^4}=

    -\frac{4}{x^2}+1-2x

    - Hollywood
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  11. #11
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    Re: Finding the derivative of a fraction

    Oh yeah your right, I didn't see that. Thanks Hollywood
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  12. #12
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    Re: Finding the derivative of a fraction

    私に大きな助けを共有していただきありがとうございます。私の家へようこそ!中国ドラマ宮廷女官ジャクギ
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  13. #13
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    Re: Finding the derivative of a fraction

    nice job!
    Last edited by mathnerd15; April 12th 2013 at 04:13 AM.
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  14. #14
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    Re: Finding the derivative of a fraction

    no problem JellyOnion good luck
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