1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    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

    User Avatar
    Science Advisor
    Homework Helper

    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


    User Avatar
    Homework Helper

    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


    User Avatar
    Homework Helper

    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!
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted

Similar Discussions: Derivative problem
  1. Derivation Problem (Replies: 3)

  2. Derivative Problem (Replies: 5)

  3. Derivative problem (Replies: 3)