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How can one find the dervative of a function

  1. Oct 14, 2008 #1
    How can I find the derivative of a function say, f(n) = ln(n), where n is a natural number?

    In other words, how can find g(x) in the following equation:
    f(x) + g(x) = f(x+1)

    I know there isn't a general formula for this, but are there some techniques to find g(x) for a specific equation f(x)?
  2. jcsd
  3. Oct 14, 2008 #2


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    If you mean the domain of f is the set of natural numbers, then there is no derivative. The domain has to be a continuous set in order to define a derivative.

    As far as finding g(x) in f(x)+ g(x)= f(x+1) is concerned, what's wrong with the obvious g(x)= f(x+1)- f(x)? Unless you have more information on f(x) I don't know what else you want.

    Are you, perhaps, talking about "finite differences"? The finite difference [itex]\Delta f[/itex] is defined as f(n+1)- f(n) for functions defined only on the natural numbers or, more generally, as (f(x+h)- f(x))/h for h a fixed, non-zero, real number. Still the only way to calculste f(n+1)- f(n) is to evaluate f(n+1) and f(n) and actually do that algebra.
    Last edited by a moderator: Oct 15, 2008
  4. Oct 16, 2008 #3
    I guess technically I was talking about "finite differences," but I wanted to find it in terms of the actual derivative of the function. Here's an easy example:

    f(n) = n^2, n ∈ ℕ

    The finite difference here is expressed through the derivative of the function. I realize that this would readily cancel out, just like f(n) - f(n) + f(n+1) = f(n+1) would, as it would with most polynomial functions, but I'm wondering that for a function like ln(n), there exists the finite difference in terms of f'(n) where it wouldn't readily cancel out.

    If we take for example, f(n) = ln(n), a simple guess could be made about the finite difference in terms of f'(n):


    though probably not true.
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