Derivate Question: Proving f'(0) Exists

[musamba]

Hi.
I have a problem with prooving this:
Q: Assume a function $$f(x): R --> R$$ is a continuous function such as $$f(x + y) = f(x)f(y)$$.
Proove that $$f$$ is deriveable if $$f'(0)$$ exists.

I've done this so far:
$$f(x) = f(x + 0) = f(x)f(0)$$

Then I use the product rule:
$$\\ f'(x) = f'(x)f(0) + f(x)f'(0)$$

And manipulate the equation:
$$\\ \\ f'(x) - f'(x)f(0) = f(x)f'(0)$$
$$\\ \\ f'(x)(1-f(0)) = f(x)f'(0)$$
$$\\ \\ (*) f'(x) = \frac{f(x)f'(0)}{1-f(0)}$$

But this must be wrong when:
$$\\ f(x) = f(x + 0) = f(x)f(0)$$
Wrtiting different gives:
$$\\ f(x) = f(x)f(0)$$
$$\\ \\ f(0) = \frac{f(x)}{f(x)}$$
$$\\ \\ f(0) = 1$$

When $$f(0) = 1 , (*)$$ only exist when $$f(0)$$ is not equal to 1.
Please tell me what went wrong in my calculation.

 

[HallsofIvy]

Hi.
I have a problem with prooving this:
Q: Assume a function $$f(x): R --> R$$ is a continuous function such as $$f(x + y) = f(x)f(y)$$.
Proove that $$f$$ is deriveable if $$f'(0)$$ exists.

I've done this so far:
$$f(x) = f(x + 0) = f(x)f(0)$$

Then I use the product rule:
$$f'(x) = f'(x)f(0) + f(x)f'(0) \\$$

This is your error. f'(0) means f'(x) evaluated at x=0, not "the derivative of f(0)". f(0) is a constant and necessarily has derivative 0. You are just saying that f'(x)= f'(x)f(0). (And so showing in a different way that f(0)= 1.)

I am also not happy with using the product rule to show that f ' does exist. Part of the hypotheses for the product rule is that the derivatives must exist.
Go back to the original definition:
$$f'(x)= lim_{h \rightarrow 0}\frac{f(x+h)- f(x)}{h}$$
using the fact that f(x+h)= f(x)f(h), that f '(0) exists, and that f(0)= 1.\

Your proof of that:

$$f(x) = f(x)f(0)$$
$$f(0) = \frac{f(x)}{f(x)}$$
$$f(0) = 1$$

is correct.

It can be shown, by the way, that the only continuous (and so only differentiable) functions satisfying f(x+y)= f(x)f(y) are the exponential functions: f(x)= a^x.

 

[musamba]

Thank you, HallsofIvy.
I will give that a shot.