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Derivate question

  1. Oct 26, 2005 #1
    I have a problem with prooving this:
    Q: Assume a function [tex] f(x): R --> R [/tex] is a continous function such as [tex] f(x + y) = f(x)f(y) [/tex].
    Proove that [tex] f [/tex] is deriveable if [tex] f'(0) [/tex] exists.

    I've done this so far:
    [tex] f(x) = f(x + 0) = f(x)f(0)[/tex]

    Then I use the product rule:
    [tex]\\ f'(x) = f'(x)f(0) + f(x)f'(0)[/tex]

    And manipulate the equation:
    [tex]\\ \\ f'(x) - f'(x)f(0) = f(x)f'(0)[/tex]
    [tex]\\ \\ f'(x)(1-f(0)) = f(x)f'(0)[/tex]
    [tex]\\ \\ (*) f'(x) = \frac{f(x)f'(0)}{1-f(0)}[/tex]

    But this must be wrong when:
    [tex]\\ f(x) = f(x + 0) = f(x)f(0)[/tex]
    Wrtiting different gives:
    [tex]\\ f(x) = f(x)f(0)[/tex]
    [tex]\\ \\ f(0) = \frac{f(x)}{f(x)}[/tex]
    [tex]\\ \\ f(0) = 1[/tex]

    When [tex] f(0) = 1 , (*)[/tex] only exist when [tex]f(0)[/tex] is not equal to 1.
    Please tell me what went wrong in my calculation.
    Last edited: Oct 26, 2005
  2. jcsd
  3. Oct 26, 2005 #2


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    Science Advisor

    This is your error. f'(0) means f'(x) evaluated at x=0, not "the derivative of f(0)". f(0) is a constant and necessarily has derivative 0. You are just saying that f'(x)= f'(x)f(0). (And so showing in a different way that f(0)= 1.)

    I am also not happy with using the product rule to show that f ' does exist. Part of the hypotheses for the product rule is that the derivatives must exist.
    Go back to the original definition:
    [tex]f'(x)= lim_{h \rightarrow 0}\frac{f(x+h)- f(x)}{h}[/tex]
    using the fact that f(x+h)= f(x)f(h), that f '(0) exists, and that f(0)= 1.\

    Your proof of that:
    is correct.

    It can be shown, by the way, that the only continuous (and so only differentiable) functions satisfying f(x+y)= f(x)f(y) are the exponential functions: f(x)= ax.
    Last edited by a moderator: Oct 26, 2005
  4. Oct 26, 2005 #3
    Thank you, HallsofIvy.
    I will give that a shot.
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