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Math Help - Prove g is differentiable at a

  1. #1
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    Prove g is differentiable at a

    Let
    f be a function that is differentiable at a a E Reals, with f(a) not 0.
    Define g(x)=1/f(x) for x near a. Prove that g is differentiable at a and give the formula for g'(a).

    F(x) is differentiable at a implying it is continuos at a. So g(x) tends to 1/f(a) as x tends to a. g'(a) = lim((g(x)- g(a))/x-a)) = (1/f(x) - 1/f(a))/(x-a)
    Not sure what the next steps are.
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  2. #2
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    Quote Originally Posted by poirot View Post
    Let
    f be a function that is differentiable at a a E Reals, with f(a) not 0.
    Define g(x)=1/f(x) for x near a. Prove that g is differentiable at a and give the formula for g'(a).

    F(x) is differentiable at a implying it is continuos at a. So g(x) tends to 1/f(a) as x tends to a. g'(a) = lim((g(x)- g(a))/x-a)) = (1/f(x) - 1/f(a))/(x-a)
    That is correct.
    Note that can be written as [(f(a)-f(x))/(x-a)][1/(f(x)f(a)]
    Now find \lim_x\to a.
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  3. #3
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    Quote Originally Posted by Plato View Post
    That is correct.
    Note that can be written as [(f(a)-f(x))/(x-a)][1/(f(x)f(a)]
    Now find \lim_x\to a.
    ok so g'(a)= f'(a)/f(x)f(a) but how do I show g is differentiable
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  4. #4
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    Quote Originally Posted by poirot View Post
    ok so g'(a)= f'(a)/f(x)f(a) but how do I show g is differentiable
    Not quite. You missed a minus sign.
    Note that [(f(a)-f(x))/(x-a)]= -[(f(x)-f(a))/(x-a)]

    Also f(x)->f(a) so you get [f(a)]^2 in the denominator.
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  5. #5
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    g'(a) = -F'(a)/(f(a))^2. Is that enough to show its differentiable since f'(a) and f(a) exsist
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  6. #6
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    Quote Originally Posted by poirot View Post
    g'(a) = -F'(a)/(f(a))^2. Is that enough to show its differentiable since f'(a) and f(a) exsist
    Yes.
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