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Derivatives and ln

  1. Jan 25, 2009 #1
    1. The problem statement, all variables and given/known data

    If a differentiable function f(x) satisfies the equation f(xy) = f(x) + f(y), prove then that f(x) = alnx.

    2. Relevant equations

    3. The attempt at a solution

    I have proved that if f satisfies f(x + y) = f(x)f(y) then f(x) = 0 or f(x) = e^(ax)

    and I also know that if a function f satisfies f'(x) = af(x) for some constant a, then f(x) = ce^(ax) for some constant c and a.
  2. jcsd
  3. Jan 25, 2009 #2


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    Define f'(1)=a. Now write down f'(x) as a difference quotient and mess around with it. Maybe make a change of variables on the h. Can you show f'(x)=a/x?
    Last edited: Jan 25, 2009
  4. Jan 25, 2009 #3
    [tex] f'(x) = \lim_{h \rightarrow 0} \frac{f(x+h) - f(x)}{h} = \lim_{h \rightarrow 0} \frac{f(x+h) + (-f(x))}{h}[/tex].

    I think the negative sign in front of f(x) messes things up because I don't think I can write f(x+h) + (-f(x)) as f(x(x+h)).

    EDIT: I just saw your edit, I will continue to work on it.
  5. Jan 25, 2009 #4


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    You should be able to show f(1/x)=-f(x) pretty easily.
  6. Jan 26, 2009 #5
    0 = f(1) = f(x(1/x)) = f(x) + f(1/x) and so f(1/x) = -f(x).
    So then I have [tex]f'(x) = \lim_{h \rightarrow 0} \frac{f(x+h) + f(1/x)}{h} = \lim_{h \rightarrow 0} \frac{f(1 + h/x)}{h}[/tex]

    But I can't see where to go from here to show that f'(x) = a/x
  7. Jan 26, 2009 #6
    Isn't this a candidate for L'Hopital's now or is that a step in the wrong direction?
  8. Jan 26, 2009 #7
    I was thinking that, so using l'hospital I have f'(x) = f'(1+x/h), no? But to me this doesn't make sense because if I want to show that the derivative is a/x then this cannot be true.

    Plus in my textbook we haven;t covered l'hospital yet so I assume I'm expected to solve the question without using it.
  9. Jan 26, 2009 #8
    No using L'Hopital's you have

    [tex] \lim_{h\rightarrow 0} \frac{1}{x} f'\left(1 + \frac{h}{x} \right) = \frac{1}{x} [/tex]
  10. Jan 26, 2009 #9
    Isn't the derivative of f(1 + x/h) = f'(1 + x/h)? Plus where did the 1/x come from before the f'(1+ h/x)?
  11. Jan 26, 2009 #10
    Chain rule? And you keep writing x/h when you mean h/x, be careful.
  12. Jan 26, 2009 #11
    Sorry if I'm being difficult, but how did you use the chain rule there?..
  13. Jan 26, 2009 #12
    Okay, I think I have it.

    Letting [tex]\phi(h) = h/x [/tex], we have [tex]d\dx (f(1+\phi)) = f'(1 + \phi)*\phi'(h) = f'(1 + \phi) * (1/x)[/tex].

    So we have [tex] f'(x) = \lim_{h \rightarrow 0} f'(1 + h\x)*(1/x) = a/x [/tex]
  14. Jan 26, 2009 #13
    Ugh, I messed up some of the latex in the previous post and whenever I try to edit it, it won't save the changes, so I will just retype what I wanted to type..

    letting g(h) = h/x, we have d/dx(f(1+g)) = f'(1+g)*g' = f'(1 + h/x)*1/x.

    Now, since f'(x) = lim as h approaches 0 of f'(1 + h/x)*1/x, we have f'(x) = f'(1)/x = a/x.

    Dick, how did you see right away to define f'(1) = a?
  15. Jan 26, 2009 #14
    If anything I would let [tex] \phi(h) = 1 + \frac{h}{x} [/tex] but you have the right idea there.

    Next, you wrote [tex] f'(1+h) [/tex], you probably meant [tex] f'\left(1 + \frac{h}{x} \right) [/tex]?
  16. Jan 26, 2009 #15


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    You probably also showed that f(1)=0. So your difference quotient is
    (1/x)*((f(1+h/x)-f(1))/(h/x))). Since if h->0, so does h/x->0, the second factor is a difference quotient for f'(1). So it's (1/x)*f'(1)=a/x.
  17. Jan 26, 2009 #16
    Yup, that's what I meant. And of course, since f'(x) = a/x, integrating gives us f(x) = alnx.

    Thanks guys
  18. Jan 26, 2009 #17
    How are you able to so easily spot these things?
  19. Jan 26, 2009 #18


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    Practice, I guess. But it wasn't all THAT easy. I did have to think about it. I knew the answer was f(x)=a*ln(x), so f'(1)=a. Now you just have to figure out how to turn the difference quotient at a general point x into a difference quotient at x=1 times something using that f(xy)=f(x)+f(y). Forget the solution, get a blank piece of paper and try it. It's not super hard either.
  20. Jan 26, 2009 #19
    Do you know of any questions similar to the one that I posted? There aren't anymore in my textbook and I'd love to try another one.
  21. Jan 26, 2009 #20


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    Not exactly. There aren't a lot of functions that have such a simple relation as f(xy)=f(x)+f(y) or f(x+y)=f(x)f(y). Try redoing the proof that the latter is c*exp(ax) where a=f'(0) for the latter case, if you can forget how it was proved before. It's even easier.
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