# Derivative problem

1. Oct 19, 2014

### johann1301

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. Oct 19, 2014

### LCKurtz

If you have that and you are given f'(1)=k aren't you done?

3. Oct 19, 2014

### johann1301

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. Oct 19, 2014

### Ray Vickson

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. Oct 19, 2014

### johann1301

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. Oct 19, 2014

### sankalpmittal

Clever partial differentiation will be the tool to do this problem.

7. Oct 19, 2014

### GFauxPas

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. Oct 19, 2014

### pasmith

You have an equation which gives $f(x + h) = f(x) + f(1 + \frac hx)$. Try substituting that into the definition $$f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}.$$

9. Oct 19, 2014

### johann1301

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. Oct 19, 2014

### pasmith

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

Instead, you should start again from $$f'(x) = \lim_{h \to 0} \frac{f(1 + \frac hx)}{h} = \lim_{h \to 0} \frac{f(1 + \frac hx) - f(1)}{h}$$ since $f(1) = 0$. The right hand side is almost $$f'(1) = \lim_{p \to 0} \frac{f(1 + p) - f(1)}p,$$ but you need to do some further manipulation before taking the limit.

11. Oct 19, 2014

### GFauxPas

edit: nevermind!