# Differential of the exp function prove

1. Nov 9, 2009

### Sennap

1. The problem statement, all variables and given/known data
Let f: R->R be a function which satisfies the condition:
$$f(x+y) = f(x) \cdot f(y)$$
$$\forall x,y \in R$$

a)
Show that if f is everywhere differntial it satisfies the equation:
$$f'(x) = f'(0) \cdot f(x)$$

Conclude that either f is the zero function or there exists a number c so that
$$f(x) = exp(c \cdot x)$$
$$\forall x \in R$$

b)
Show that if f is differntial in 0, then f is the zero function or there exists a number c so that:
$$f(x) = exp(c \cdot x)$$
$$\forall x \in R$$

2. Relevant equations
$$\frac{d}{dx} ln|f(x)| = \frac{f'(x)}{f(x)}$$
$$\frac{(f \cdot g)'}{f \cdot g} = \frac{f'}{f} + \frac{g'}{g}$$

As well as we might need some ln and exp rules, the most important one being:
$$a^{x} = exp(x \cdot ln(a))$$

3. The attempt at a solution
Here's what i've done so far:

If we diffrientiate the ln (natural log) both sides we get.

Left side:
$$(ln|f(x+y)|)' = \frac{f'(x+y)}{f(x+y)}$$

Right side:
$$(ln|f(x) \cdot f(y)|)' = \frac{f'(x)}{f(x)} + \frac{f'(y)}{f(y)}$$

Next, I isolated f'(x) but I couldn't see that new equation helping me anything.

Could someone please point me in the right direction, I've really no idea how to approach this problem.

Sennap

Last edited: Nov 9, 2009
2. Nov 9, 2009

### clamtrox

Should this be $$f(x+y) = f(x) f(y)$$?

3. Nov 9, 2009

### Sennap

Yes, sorry about that. It's fixed now.

- Sennap

4. Nov 9, 2009

### HallsofIvy

Staff Emeritus
How about using the definition of derivative?
$$f'(x)= \lim_{h\to 0}\frac{f(x+h)- f(x)}{h}= \lim_{h\to 0}\frac{f(x)f(h)- f(x)}{h}= f(x)lim_{h\to 0}\frac{f(h)- 1}{h}$$
The fact that f is differentiable tells you that that limit exists. But
$$f'(0)= \lim_{h\to 0}\frac{f(h)- f(0)}{h}$$
and f(x+0)= f(x)f(0) tells you that f(0)= 1.