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Math Help - [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
    <br />
f(x) =  - \frac{{6 + 3x^2 }}{x}<br />

    if you choose to use the quotient rule:

    \begin{array}{l}<br />
 f(x) = \frac{{g(x)}}{{w(x)}} \\ <br />
 f' (x) = \frac{{g' (x)w(x) - g(x)w' (x)}}{{w^2 (x)}} \\ <br />
 \end{array}<br /> <br />

    in our case:


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

    thus:

    <br />
f(x) = \frac{{ - 6x^2  + 6 + 3x^2 }}{{x^2 }} =  - 3 + \frac{6}{{x^2 }}<br />

    -----------------------------------------------------------------------
    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.


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


    <br />
\begin{array}{l}<br />
 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) =  \\ <br />
  = \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) \\ <br />
 \end{array}<br />

    <br />
\begin{array}{l}<br />
  = \frac{1}{{(x + 5)x^2 }} + \frac{{6x + 1}}{{(x + 5)^2 x}} \\ <br />
 {\rm common}\;{\rm denominator} \\ <br />
 {\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 }} \\ <br />
 \end{array}<br />
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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

    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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