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How is 1/f(x) differentiable at 0.

  1. Dec 1, 2007 #1
    I can prove derivative of 1/f(x) = -f'(x)/(f^2).
    But how can this be differentiable at x=0 (if f(x) is not equal to 0) as my class notes claim

    Thanks
     
  2. jcsd
  3. Dec 1, 2007 #2

    EnumaElish

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    Say f(0) = 2. This implies 1/f(0) = 1/2 and [1/f'(0)]' = -f'(0)/4.
     
    Last edited: Dec 1, 2007
  4. Dec 1, 2007 #3
    True but let me tell you my confusion ( should have posted in detail)

    I proved 1/f'(x) as follows

    - given a seq {an} -> a,

    1/f(an) - 1/f(a) / (an -a) -> a then this function is differentiable.

    Manipulating this:

    ( f(a) - f(an) ) /( f(a)f(an) (an-a) =

    -f'(a) / ( f(a) f(an) ).

    Since an-> a, f(an)-> f(a) and thus I get -f'(a)/f^2.

    But this is my problem- I say an-> a. But 'a' can be any value including 0 but yet it is give that f(x) is not equal to 0.

    Hope I am making sense

    Thanks

    Asif
     
  5. Dec 1, 2007 #4
    f(x) is not equal to zero, that in no way implies that a may not be zero.
     
  6. Dec 1, 2007 #5

    ktm

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    A function is differentiable at a point if the derivative exists at a point. From the derivative you got, the derivative exists at 0, as seen from just plugging in 0 and knowing that [tex]f(x) \neq 0[/tex]. Thus it is differentiable at 0. That is all.
     
  7. Dec 1, 2007 #6

    EnumaElish

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    Example: f(x) = 2 + x. In particular, f(0) = 2 > 0. And, f'(x) = 1 for all x.

    Then, derivative of 1/f(0) = -f'(0)/f(0)^2 = -1/4.
     
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