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Math Help - Simple derivative question.

  1. #1
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    Simple derivative question.

    F(x) = (x+1)/(x-1) a=3

    Compute derivative at x = a using the limit definition and find an equation of the tangent line.

    I get up to:

    [(3+h)+1/(3+h)-1] - 2
    ____________________
    h

    but I don't know where to go from there, help!
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  2. #2
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    Quote Originally Posted by coach2uf View Post
    F(x) = (x+1)/(x-1) a=3

    Compute derivative at x = a using the limit definition and find an equation of the tangent line.

    I get up to:

    [(3+h)+1/(3+h)-1] - 2
    ____________________
    h

    but I don't know where to go from there, help!
    I'll start at the start. First of all, you should make life as easy for yourself as possible.

    So write \frac{x + 1}{x - 1} as \frac{(x - 1) + 2}{x - 1} = 1 + \frac{2}{x - 1}.

    I'll do it for general a. If you prefer, you can replace all the a's with 3 .......


    Note that if f(x) = 1, then \frac{f(a + h) - f(a)}{h} = 0 since f(x) = 1 for all values of x including x = a + h and x = a.

    So you only have to worry about getting \lim_{h \rightarrow 0}\frac{f(a + h) - f(a)}{h} for f(x) = \frac{2}{x - 1}:

    f(a + h) - f(a) = \frac{2}{a + h - 1} - \frac{2}{a - 1}


    Put over a common denominator:


    = \frac{2(a - 1)}{(a + h - 1)(a - 1)} - \frac{2(a + h - 1)}{(a - 1)(a + h - 1)} = \frac{2(a - 1) - 2 (a + h - 1)}{(a + h - 1)(a - 1)}


    Simplify:


     = \frac{-2h}{(a + h - 1)(a - 1)}.


    So \frac{f(a + h) - f(a)}{h} = \frac{-2}{(a + h - 1)(a - 1)}.


    Now take the limit


    \lim_{h \rightarrow 0}\frac{f(a + h) - f(a)}{h} = \lim_{h \rightarrow} \frac{-2}{(a + h - 1)(a - 1)} = \frac{-2}{(a - 1)^2}.

    When a = 3 the value this limit is equal to -\frac{1}{2}.

    This means that the gradient of the tangent line is -\frac{1}{2}.
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