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Derivative problem

  1. Oct 19, 2014 #1
    1. The problem statement, all variables and given/known data
    Show that f'(x) = k/x

    2. Relevant equations
    f is defined from zero to infinity
    f(xy) = f(x) + f(y)
    f'(1) = k
    f(1) = 0
    f(x+h) = f(x) + f (1+h/x)


    3. The attempt at a solution

    I know i can write f'(x) = f'(1)/x but thats all I've got so far...
     
  2. jcsd
  3. Oct 19, 2014 #2

    LCKurtz

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    If you have that and you are given f'(1)=k aren't you done?
     
  4. Oct 19, 2014 #3
    no, when i say "I know i can write f'(x) = f'(1)/x but thats all I've got so far..." i am using the result I'm supposed to find... i was trying to go backwards... don't really have a clue how to start...
     
  5. Oct 19, 2014 #4

    Ray Vickson

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    You are GIVEN ##f'(1) = k## as part of the problem input---so you would not using a result you are supposed to find.
     
  6. Oct 19, 2014 #5
    I thought i could go backwards, i.e from the result i was supposed to find to some equation i already knew... but this is probably not the way to go...
     
  7. Oct 19, 2014 #6
    Clever partial differentiation will be the tool to do this problem.
     
  8. Oct 19, 2014 #7
    I don't think ##f'(x) = \frac {f'(1)}{x}## one of his premises.
    I think he's trying to manipulate the data given into the form ##\frac {f'(1)}{x}## so that he can substitute ##f'(1)=k##.
     
  9. Oct 19, 2014 #8

    pasmith

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    You have an equation which gives [itex]f(x + h) = f(x) + f(1 + \frac hx)[/itex]. Try substituting that into the definition [tex]
    f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}.[/tex]
     
  10. Oct 19, 2014 #9
    then i get

    lim f(1+h/x)/h as h goes to zero

    if i then use l hopital and the chainrule, i get f'(1)/x with solves the problem! thanks !
     
  11. Oct 19, 2014 #10

    pasmith

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    Using l'Hopital requires that [itex]g(h) = f(1 + \frac hx)[/itex] be differentiable in some open neighbourhood of [itex]h = 0[/itex] so that the limit of [itex]g'(h)[/itex] as [itex]h \to 0[/itex] can be taken; all you know (from the chain rule, which you can use) is that [itex]g'(0) = f'(1)/x[/itex] exists, but you don't yet know whether [itex]g'(h)[/itex] exists for [itex]h \neq 0[/itex], still less whether [itex]\lim_{h \to 0} g'(h) = g'(0)[/itex].

    Instead, you should start again from [tex]
    f'(x) = \lim_{h \to 0} \frac{f(1 + \frac hx)}{h} = \lim_{h \to 0} \frac{f(1 + \frac hx) - f(1)}{h}[/tex] since [itex]f(1) = 0[/itex]. The right hand side is almost [tex]f'(1) = \lim_{p \to 0} \frac{f(1 + p) - f(1)}p,[/tex] but you need to do some further manipulation before taking the limit.
     
  12. Oct 19, 2014 #11
    edit: nevermind!
     
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