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Thread: [SOLVED] help with a derivative

  1. #1
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    [SOLVED] help with a derivative

    ok, I'm in calc. concepts and I have two questions. Help with either would be greatly appreciated.

    1)
    Find an equation for the tangent line to f(x) = [-(6+3x^2)/x)] at (2, -9)
    I know that the answer is 3x + 2y = -12 I just can't find it! I shouldn't have any problem finding the actual tan line, it's the derivatibe that I'm having trouble with.

    I tried using quotient rule and this is what I got:

    (-x)(-6x) - (-1)(6 + 3x^2)
    ------------------------- =
    x^2


    6x^2 + 6 + 3x^2
    ----------------------- =
    x^2

    9x^2 + 6
    ---------
    x^2

    which I don't think is right?



    2) Find the derivative of
    y = -6 - (1/x)
    ----------
    X + 5

    yikes! We don't have anything like this in the text, he just kind of snuck it in on the test and I'm trying to figure it out. The answer is:
    6x^2 + 2x + 5
    --------------
    x^2(x+5)^2
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  2. #2
    Senior Member Peritus's Avatar
    Joined
    Nov 2007
    Posts
    397
    $\displaystyle
    f(x) = - \frac{{6 + 3x^2 }}{x}
    $

    if you choose to use the quotient rule:

    $\displaystyle \begin{array}{l}
    f(x) = \frac{{g(x)}}{{w(x)}} \\
    f' (x) = \frac{{g' (x)w(x) - g(x)w' (x)}}{{w^2 (x)}} \\
    \end{array}

    $

    in our case:


    $\displaystyle \begin{array}{l}
    g(x) = - 6 - 3x^2 - > g'(x) = - 6x \\
    w(x) = x - > w'(x) = 1 \\
    \end{array}$

    thus:

    $\displaystyle
    f(x) = \frac{{ - 6x^2 + 6 + 3x^2 }}{{x^2 }} = - 3 + \frac{6}{{x^2 }}
    $

    -----------------------------------------------------------------------
    First I'd suggest that you don't get fixated on finding special formulas for everything. The trick is to express the function in a convinient form and use the product rule.


    $\displaystyle y(x) = \frac{{ - 6 - {\raise0.7ex\hbox{$1$} \!\mathord{\left/
    {\vphantom {1 x}}\right.\kern-\nulldelimiterspace}
    \!\lower0.7ex\hbox{$x$}}}}{{x + 5}} = - \left( {6 + \frac{1}{x}} \right) \cdot \left( {\frac{1}{{x + 5}}} \right)$


    $\displaystyle
    \begin{array}{l}
    y'(x) = - \left( {\frac{1}{{x + 5}}} \right) \cdot \frac{d}{{dx}}\left( {6 + \frac{1}{x}} \right) - \left( {6 + \frac{1}{x}} \right)\frac{d}{{dx}}\left( {\frac{1}{{x + 5}}} \right) = \\
    = \frac{1}{{x^2 }} \cdot \left( {\frac{1}{{x + 5}}} \right) + \left( {6 + \frac{1}{x}} \right) \cdot \left( {\frac{1}{{(x + 5)^2 }}} \right) \\
    \end{array}
    $

    $\displaystyle
    \begin{array}{l}
    = \frac{1}{{(x + 5)x^2 }} + \frac{{6x + 1}}{{(x + 5)^2 x}} \\
    {\rm common}\;{\rm denominator} \\
    {\rm = }\frac{{{\rm x + 5 + 6x}^{\rm 2} + x}}{{x^2 (x + 5)^2 }} = \frac{{{\rm 6x}^{\rm 2} + 2x + 5}}{{x^2 (x + 5)^2 }} \\
    \end{array}
    $
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  3. #3
    MHF Contributor kalagota's Avatar
    Joined
    Oct 2007
    From
    Taguig City, Philippines
    Posts
    1,026
    after getting the correct answer in the first one, which is

    $\displaystyle f'(x)=-3 + \frac{6}{x^2} $ = slope of the tangent line), all you have to do is to substitute x=2, and you'll get the slope of the line and just use the point-slope form to get the line..

    --------------------------------
    there are many ways to attack the problem..
    just do the one you think you are good at..
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