# Prove: f(x) = alnxProving Derivatives and ln Homework Statement

• JG89
In summary, Dick figured out that if a differentiable function f satisfies f(x+y) = f(x)f(y), then f(x) = 0 or f(x) = e^(ax) for some constant a. He also knows 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.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?f'(x) =
JG89

## Homework Statement

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

## 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.

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:
$$f'(x) = \lim_{h \rightarrow 0} \frac{f(x+h) - f(x)}{h} = \lim_{h \rightarrow 0} \frac{f(x+h) + (-f(x))}{h}$$.

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.

You should be able to show f(1/x)=-f(x) pretty easily.

0 = f(1) = f(x(1/x)) = f(x) + f(1/x) and so f(1/x) = -f(x).
So then I have $$f'(x) = \lim_{h \rightarrow 0} \frac{f(x+h) + f(1/x)}{h} = \lim_{h \rightarrow 0} \frac{f(1 + h/x)}{h}$$

But I can't see where to go from here to show that f'(x) = a/x

Isn't this a candidate for L'Hopital's now or is that a step in the wrong direction?

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.

No using L'Hopital's you have

$$\lim_{h\rightarrow 0} \frac{1}{x} f'\left(1 + \frac{h}{x} \right) = \frac{1}{x}$$

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)?

Chain rule? And you keep writing x/h when you mean h/x, be careful.

Sorry if I'm being difficult, but how did you use the chain rule there?..

Okay, I think I have it.

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

So we have $$f'(x) = \lim_{h \rightarrow 0} f'(1 + h\x)*(1/x) = a/x$$

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?

If anything I would let $$\phi(h) = 1 + \frac{h}{x}$$ but you have the right idea there.

Next, you wrote $$f'(1+h)$$, you probably meant $$f'\left(1 + \frac{h}{x} \right)$$?

JG89 said:
0 = f(1) = f(x(1/x)) = f(x) + f(1/x) and so f(1/x) = -f(x).
So then I have $$f'(x) = \lim_{h \rightarrow 0} \frac{f(x+h) + f(1/x)}{h} = \lim_{h \rightarrow 0} \frac{f(1 + h/x)}{h}$$

But I can't see where to go from here to show that f'(x) = a/x

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.

NoMoreExams said:
If anything I would let $$\phi(h) = 1 + \frac{h}{x}$$ but you have the right idea there.

Next, you wrote $$f'(1+h)$$, you probably meant $$f'\left(1 + \frac{h}{x} \right)$$?

Yup, that's what I meant. And of course, since f'(x) = a/x, integrating gives us f(x) = alnx.

Thanks guys

Dick said:
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.

How are you able to so easily spot these things?

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.

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.

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.

I will try redoing it with another proof. Thanks :)

## What is the formula for the derivative of f(x) = alnx?

The formula for the derivative of f(x) = alnx is f'(x) = a/x.

## How do you prove the derivative of f(x) = alnx?

To prove the derivative of f(x) = alnx, we use the definition of a derivative which states that the derivative of a function is the limit of the difference quotient as the change in x approaches 0. We substitute alnx into the definition and simplify to arrive at the formula f'(x) = a/x.

## What is the value of f'(x) when x = 1?

When x = 1, the value of f'(x) is a/1 = a.

## How does the value of a affect the graph of f(x) = alnx?

The value of a affects the steepness of the graph of f(x) = alnx. A larger value of a will result in a steeper graph, while a smaller value of a will result in a flatter graph.

## What is the domain and range of f(x) = alnx?

The domain of f(x) = alnx is all positive real numbers. The range of f(x) = alnx is all real numbers.

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