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Implications of varying the definition of the derivative?

  1. Jan 10, 2016 #1
    I have been playing around with calculus for a while and I wondered what would it be like to make some changes to the definition of derivatives.

    I'd like to look at the original definition of derivatives in this way (everything is in lim Δx→0):
    F(x+Δx) - F(x) = F'(x) * Δx
    The Δx factor ensures that the RHS remains 0 when the LHS is 0.
    And we get the familiar
    (F(x+Δx) - F(x)) / Δx = F'(x)

    I'd like to think of it as taking infinitesimal differences over some infinitesimal interval, and I wondered what would it look like if I were to take infinitesimal ratios instead. And it came out looking like this:
    F(x+Δx) / F(x) = F*(x) ^ Δx
    The Δx exponent ensures that the RHS remains 1 when the LHS is 1.
    Then it comes out as
    (F(x+Δx) / F(x)) ^ (1/Δx) = F*(x)

    I've tried to use this on one sample function, F(x) = x^2; the results are F'(x) = 2x and F*(x)=e^(2/x)

    If F'(x) means the rate of change of between two things (like velocity when F(time) is position), then what does F*(x) mean?
     
  2. jcsd
  3. Jan 10, 2016 #2

    micromass

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    Take logarithms to reduce it to the concept of the derivative.
     
  4. Jan 10, 2016 #3
    If you take the logarithm, the denominator will be the same Δx, but the numerator will have logs wrapped around it
    log(F(x+Δx)) - log(F(x)) instead of F(x+Δx) - F(x).
     
  5. Jan 10, 2016 #4

    micromass

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    Right, so what function are you taking the derivative of?
     
  6. Jan 10, 2016 #5
    Ohh, the function log(F(x))
    -----------------------------
    So this means that
    log(F*(x)) = F'( log(F(x)) ) right?

    What else does it imply or what else could it be applied to?
     
    Last edited: Jan 10, 2016
  7. Jan 10, 2016 #6

    Erland

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    Hmm, not F'(log(F(x)), but G'(x), where G(x)=log(F(x)). But that is perhaps what you meant?
    Then what is G'(x), if you use the Chain Rule?
     
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