1. ## Calculating With Derivatives

I have done 2 of the 3 parts of the following problem but now I'm stuck. Any help would be appreciated:

Let $\displaystyle \mathbb{R}^+=(0,\infty)$. Suppose $\displaystyle f:\mathbb{R}^+\rightarrow \mathbb{R}$ is differentiable at $\displaystyle x=1$ and $\displaystyle f(xy)=f(x)+f(y)$ for all $\displaystyle x,y\in \mathbb{R}^+$.

1)Prove $\displaystyle f(1)=0$
2)Prove $\displaystyle f(1/x)=-f(x)$
3)Prove $\displaystyle f$ is differentiable on $\displaystyle \mathbb{R}^+$, and $\displaystyle f'(x)=f'(1)/x$.

I have already proven parts 1 and 2. For part 3 I'm guessing you set it up something like this... $\displaystyle (f(xy)-f(a))/(xy-a)$ and then you use the fact that $\displaystyle f(xy)=f(x)+f(y)$ to rearrange to find the derivative but I'm struggling to see how. Any help would be great.

2. Originally Posted by zebra2147
I have done 2 of the 3 parts of the following problem but now I'm stuck. Any help would be appreciated:

Let $\displaystyle \mathbb{R}^+=(0,\infty)$. Suppose $\displaystyle f:\mathbb{R}^+\rightarrow \mathbb{R}$ is differentiable at $\displaystyle x=1$ and $\displaystyle f(xy)=f(x)+f(y)$ for all $\displaystyle x,y\in \mathbb{R}^+$.

1)Prove $\displaystyle f(1)=0$
2)Prove $\displaystyle f(1/x)=-f(x)$
3)Prove $\displaystyle f$ is differentiable on $\displaystyle \mathbb{R}^+$, and $\displaystyle f'(x)=f'(1)/x$.

I have already proven parts 1 and 2. For part 3 I'm guessing you set it up something like this... $\displaystyle (f(xy)-f(a))/(xy-a)$ and then you use the fact that $\displaystyle f(xy)=f(x)+f(y)$ to rearrange to find the derivative but I'm struggling to see how. Any help would be great.
See if this gives you a kick start
$\displaystyle \displaystyle \lim_{x \to a}\frac{f(x)-f(a)}{x-a}=\lim_{x \to a}\frac{f\left( \frac{x}{a}\right)}{a\left[\left( \frac{x}{a}\right)-1\right]}$
Now let $\displaystyle \displaystyle t=\frac{x}{a}$ then you get

$\displaystyle \lim_{x \to a}\frac{f\left( \frac{x}{a}\right)}{a\left[\left( \frac{x}{a}\right)-1\right]}=\lim_{t \to 1}\cdots$ and don't forget that $\displaystyle f(1)=0$

3. [duplicate]

The main idea is that no matter where you differentiate, you can scale the parameter of the function, so that the new parameter becomes near 1 (which you know is differentiable).

Algebraically:

$\displaystyle \displaystyle \frac{f(y) - f(x)}{y - x} = \frac{f(xy/x) - f(x)}{xy/x - x} = \frac{f(x) + f(y/x) - f(x)}{x(y/x - 1)} = \frac{f(y/x)}{x(y/x - 1)}$
$\displaystyle \displaystyle = \frac{1}{x}\frac{f(y/x) - f(1)}{y/x - 1}$