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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Originally Posted by poirot

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)

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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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Originally Posted by Plato

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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ok so g'(a)= f'(a)/f(x)f(a) but how do I show g is differentiable

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Originally Posted by poirot

ok so g'(a)= f'(a)/f(x)f(a) but how do I show g is differentiable

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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.

g'(a) = -F'(a)/(f(a))^2. Is that enough to show its differentiable since f'(a) and f(a) exsist

Quote:

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Originally Posted by poirot

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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Yes.