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Math Help - derivative of x/x-1-including h!!!

  1. #1
    Boo
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    derivative of x/x-1-including h!!!

    Hello!
    Now, I have to count derivative from :

    (\frac{x}{x-1}
    it would of course not be a problem IF the result would not be:
    -\frac{1}{(x-1)(x-1+h)}
    So, obviously it is the whole procedure including the proof of the rule for the derivative of the rational function and that is what I am asking about.

    (Of course I am avare h is almost 0, but I am interested to the whoel procedure how didi they get it!)
    I tried like this:
    = \frac{x+h}{x+h-1}-\frac{x}{x-1}
    and got of course:

    =\frac{(x-1)(x+h)-x(x+h-1)}{(x+h-1)(x-1)}= \frac{-h}{(x+h-1)(x-1)}


    can someone help?
    Did I make a mistake?
    Many thanks!
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  2. #2
    MHF Contributor

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    Re: derivative of x/x-1-including h!!!

    Quote Originally Posted by Boo View Post
    Hello!
    Now, I have to count derivative from :
    I tried like this:
    = \frac{x+h}{x+h-1}-\frac{x}{x-1}
    and got of course:

    =\frac{(x-1)(x+h)-x(x+h-1)}{(x+h-1)(x-1)}= \frac{-h}{(x+h-1)(x-1)}

    You forgot that it is:
    =\frac{ \frac{x+h}{x+h-1}-\frac{x}{x-1}}{h}=\frac{-1}{(x+h-1)(x-1)}
    Thanks from Boo
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  3. #3
    Boo
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    Re: derivative of x/x-1-including h!!!

    O !
    Thanks, Plato!!!
    So, we obvioulsy get the rule for the derivative of the rational function out of it...
    so:
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    Re: derivative of x/x-1-including h!!!

    Quote Originally Posted by Boo View Post
    O !
    Thanks, Plato!!!
    So, we obvioulsy get the rule for the derivative of the rational function out of it...
    so:
    WHAT? What is the limit as h goes to 0?
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  5. #5
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    Re: derivative of x/x-1-including h!!!

    Quote Originally Posted by Boo View Post
    Hello!
    Now, I have to count derivative from :

    (\frac{x}{x-1}
    it would of course not be a problem IF the result would not be:
    -\frac{1}{(x-1)(x-1+h)}
    So, obviously it is the whole procedure including the proof of the rule for the derivative of the rational function and that is what I am asking about.

    (Of course I am avare h is almost 0, but I am interested to the whoel procedure how didi they get it!)
    I tried like this:
    = \frac{x+h}{x+h-1}-\frac{x}{x-1}
    and got of course:

    =\frac{(x-1)(x+h)-x(x+h-1)}{(x+h-1)(x-1)}= \frac{-h}{(x+h-1)(x-1)}


    can someone help?
    Did I make a mistake?
    Many thanks!
    A first principles approach for this problem may be easier if you can see that \displaystyle \begin{align*} f(x) = \frac{x}{x - 1} \equiv 1 + \frac{1}{x - 1} \end{align*}.

    \displaystyle \begin{align*} f'(x) &= \lim_{h \to 0}\frac{f(x+h) - f(x)}{h} \\ &= \lim_{h \to 0}\frac{\left( 1 + \frac{1}{x + h - 1} \right) - \left( 1 + \frac{1}{x - 1} \right) }{ h } \\ &= \lim_{h \to 0}\frac{\frac{1}{x + h - 1} - \frac{1}{x - 1}}{h} \\ &= \lim_{h \to 0}\frac{ \frac{x - 1}{\left(x - 1 \right) \left( x + h - 1 \right) } - \frac{x + h - 1}{\left( x - 1 \right) \left( x + h - 1 \right) } }{h} \\ &= \lim_{h \to 0}\frac{ -h}{h \left( x - 1 \right) \left( x + h - 1 \right) } \\ &= \lim_{h \to 0}\frac{-1}{\left( x - 1 \right) \left( x + h - 1 \right) } \\ &= -\frac{1}{ \left( x - 1 \right) \left( x + 0 - 1 \right) } \\ &= -\frac{1}{ \left( x - 1 \right) ^2 } \end{align*}
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