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Math Help - Differentiation

  1. #1
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    Differentiation

    I need to prove from the definition of defferntiability that the function

    f(x)= 2x-1/x-2

    is differentiable at the point 1 and find f'(1)

    So far I have

    The difference quotient for f at 1 is

    Q(h) = f(h)-f(1)/h

    = 2h-1/h-2 + 1 / h

    = ?

    can anyone help?
    Last edited by Arron; July 22nd 2011 at 03:42 AM.
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  2. #2
    MHF Contributor Siron's Avatar
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    Re: Differentiation

    When is a function differentiable? ...
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  3. #3
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    Re: Differentiation

    find a common denominator, cancel any like terms and then find the limit of h->0, if you get (some number)/0 then the function is not differentiable, however if you get 0/0 you get an indeterminate so then you must try another technique (the second derivative).
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  4. #4
    MHF Contributor Siron's Avatar
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    Re: Differentiation

    I'm sorry, maybe my question was unclear, but you have to calculate (like abdulsafan said). A function is differentiable in a point of her domain if she has a derivative.

    Notice you have to calculate
    lim_(h->0) (f(1+h)-f(1))/h
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  5. #5
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    Re: Differentiation

    would that = h/1+h?
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  6. #6
    MHF Contributor Siron's Avatar
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    Re: Differentiation

    You have to calculate:
    \lim_{h \to 0}\frac{f(1+h)-f(1)}{h}

    and
    f(1+h)=\frac{2(1+h)-1}{(1+h)-2}=\frac{2h+1}{h-1}
    f(1)=-1
    so
    \lim_{h \to 0} \frac{\frac{2h+1}{h-1}+1}{h}=\lim_{h \to 0}\frac{\frac{2h+1+(h-1)}{h-1}}{h}
    =\lim_{h \to 0} \frac{3h}{(h-1)h}=-3

    Therefore f'(1)=-3

    Check:
    D\left(\frac{2x-1}{x-2}\right)=\frac{D(2x-1)\cdot (x-2)-(2x-1)\cdot D(x-2)}{(x-2)^2}=\frac{2\cdot (x-2)-(2x-1)}{(x-2)^2}=\frac{-3}{(x-2)^2}
    f'(1)=\frac{-3}{(1-2)^2}=-3

    Therefore f'(1)=-3

    Do you understand now? ...
    Last edited by Siron; July 22nd 2011 at 05:05 AM.
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  7. #7
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    Re: Differentiation

    Sorry I should have said this before: a function is differentiable if
    A) f(a) exist
    B) limit as x approaches a exist
    C) f'(a) exist
    D) and f(a) equals the limit as x approaches a

    I think that ur function does all of the above...hope this Helps!
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  8. #8
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    Re: Differentiation

    definition of a derivative at a point ...

    f'(a) = \lim_{x \to a} \frac{f(x) - f(a)}{x-a}


    if it exists ...

    f'(1) = \lim_{x \to 1} \frac{f(x) - f(1)}{x-1}

    f'(1) = \lim_{x \to 1} \frac{\frac{2x-1}{x-2} + 1}{x-1}

    f'(1) = \lim_{x \to 1} \frac{(2x-1) + (x-2)}{(x-1)(x-2)}

    f'(1) = \lim_{x \to 1} \frac{3x-3}{(x-1)(x-2)}

    f'(1) = \lim_{x \to 1} \frac{3(x-1)}{(x-1)(x-2)}

    f'(1) = \lim_{x \to 1} \frac{3}{x-2} = -3
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