Derivative problem

  • Thread starter johann1301
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  • #1
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Homework Statement


Show that f'(x) = k/x

Homework 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)


The Attempt at a Solution


[/B]
I know i can write f'(x) = f'(1)/x but thats all I've got so far...
 

Answers and Replies

  • #2
LCKurtz
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Homework Statement


Show that f'(x) = k/x

Homework 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)


The Attempt at a Solution


[/B]
I know i can write f'(x) = f'(1)/x but thats all I've got so far...

If you have that and you are given f'(1)=k aren't you done?
 
  • #3
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1
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...
 
  • #4
Ray Vickson
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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...

You are GIVEN ##f'(1) = k## as part of the problem input---so you would not using a result you are supposed to find.
 
  • #5
217
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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.

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...
 
  • #6
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Clever partial differentiation will be the tool to do this problem.
 
  • #7
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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.
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##.
 
  • #8
pasmith
Homework Helper
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Homework Statement


Show that f'(x) = k/x

Homework 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)


The Attempt at a Solution


[/B]
I know i can write f'(x) = f'(1)/x but thats all I've got so far...

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]
 
  • #9
217
1
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 !
 
  • #10
pasmith
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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 !

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.
 
  • #11
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3
edit: nevermind!
 

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