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Math Help - [SOLVED] Derivative

  1. #1
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    [SOLVED] Derivative

    f(x)= <br />
\frac  {x^3-3x^2+4}{x^2}<br />


    well i used the power rule but came up with
    the wrong answer

    then i used the quotient rule and came out with the
    wrong answer

    am i doing my algebra wrong is is there something im not seeing
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by john doe View Post
    f(x)= x^3-3x^2+4/x^2

    well i used the power rule but came up with
    the wrong answer

    then i used the quotient rule and came out with the
    wrong answer

    am i doing my algebra wrong is is there something im not seeing
    why use the quotient rule (one thing you should have used though, was parentheses!)

    \frac {x^3 - 3x^2 + 4}{x^2} = x - 3 + 4x^{-2} .........(was that what you got?)

    use the power rule
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  3. #3
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    o i see what you did now

    you distributed the  x^-2

    <br />
\frac  {x^3-8}{x^3} <br />
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by john doe View Post
    o i see what you did now

    you distributed the  x^{-2}
    yes

    (if your power has more than one term, use {}. you should have typed x^{-2} in your LaTeX code)

    <br />
\frac {x^3-8}{x^3} <br />
    \frac {x^3 - 8}{x^3} = 1 - 8x^{-3}

    no more free rides you should be telling me this now
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  5. #5
    Member javax's Avatar
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    Like Jhevon said...much shorter way

     <br />
(x-3+4x^{-2})' = 1-0+4(-2)x^{-3} = 1-\frac{8}{x^3} = \frac{x^3-8}{3}<br />
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  6. #6
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    I think you made a mistake

    and i did solve it that way

    at first i tryed to do all this other stuff though
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  7. #7
    Super Member 11rdc11's Avatar
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    You could factor out the numerator if you wanted since it is the the form of a^3-b^3
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