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I have a problem with prooving this:

Q: Assume a function [tex] f(x): R --> R [/tex] is a continous function such as [tex] f(x + y) = f(x)f(y) [/tex].

Proove that [tex] f [/tex] is deriveable if [tex] f'(0) [/tex] exists.

I've done this so far:

[tex] f(x) = f(x + 0) = f(x)f(0)[/tex]

Then I use the product rule:

[tex]\\ f'(x) = f'(x)f(0) + f(x)f'(0)[/tex]

And manipulate the equation:

[tex]\\ \\ f'(x) - f'(x)f(0) = f(x)f'(0)[/tex]

[tex]\\ \\ f'(x)(1-f(0)) = f(x)f'(0)[/tex]

[tex]\\ \\ (*) f'(x) = \frac{f(x)f'(0)}{1-f(0)}[/tex]

But this must be wrong when:

[tex]\\ f(x) = f(x + 0) = f(x)f(0)[/tex]

Wrtiting different gives:

[tex]\\ f(x) = f(x)f(0)[/tex]

[tex]\\ \\ f(0) = \frac{f(x)}{f(x)}[/tex]

[tex]\\ \\ f(0) = 1[/tex]

When [tex] f(0) = 1 , (*)[/tex] only exist when [tex]f(0)[/tex] is not equal to 1.

Please tell me what went wrong in my calculation.

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# Derivate question

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