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Given a function $f$ satisfying the following two conditions for all $x$ and $y$:
(a) $f(x+y)=f(x)\cdot f(y)$,
(b) $f(x)=1+xg(x)$, where $\displaystyle \lim_{x\rightarrow 0}g(x)=1$.
Prove that $f'(x)=f(x)$.

The only thing I know is that $f'(x)=f(x)$ is true for $x=\{0,1\}$ , but how do we know that it's true for all $x$?

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Hint: $\frac{f(x+h) - f(x)}h = f(x) \cdot \frac{f(h) - 1}h$ (now use what you now about $f'(0)$). –  martini May 22 '12 at 15:10
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Where you wrote $x=\{0,1\}$, did you mean $x\in\{0,1\}$? You can write $x=0,1$ (with no braces) and it means the same thing. But "$x=\{0,1\}$" means $x$ is a set with two members: $0$ and $1$. –  Michael Hardy May 22 '12 at 15:20
    
Note that $ f'=f $ then implies $ f(x)=A\exp(x) $, but from (a) $A=A*A$ and (b) rules out $A=0$ so $f$ is exactly $\exp $. –  Mark Hurd May 23 '12 at 2:32
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3 Answers

$$ \begin{align} f(x+h)-f(x) & = f(x)\cdot f(h) - f(x), \quad\text{from (a)} \\[8pt] & = f(x)(f(h)-1) \\[8pt] & = f(x)\cdot hg(h),\quad \text{from (b)} \end{align} $$

So $\displaystyle f'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}=\lim_{h \to 0}f(x)g(h)=f(x)$

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Let $x_0$ be an arbitrary real number.

Note that you have $f(0)=1$ from the second condition.

You have $$f'(x_0)=\lim\limits_{y\to 0} \frac{f(x_0+y)-f(x_0)}y = \lim\limits_{y\to 0} \frac{f(x_0)(f(y)-1)}y = f(x_0) \lim\limits_{y\to 0} \frac{f(y)-1}y = f(x_0)f'(0).$$

Since you wrote you have already proved this for $0$ and $1$, you know that $f'(0)=f(0)=1$. Thus the above equation is the same as $f'(x_0)=f(x_0)$.

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In fact, what happens at $0$ isn't even necessarily to comment separately since $\frac{f(y)-1}{y}=\frac{1+y\,g(y)-1}{y}=g(y)\to 1$ as $y\to 0$. –  Thomas E. May 22 '12 at 15:20
    
That's even better, nice observation. (And it is essentially TonyK's solutions.) –  Martin Sleziak May 22 '12 at 15:22
    
Martini's hint was enough. To get derivate of f at 0 =1 he needs to make use of (b) differentiating f and evaluating that derivative at zero incorporating the limit property given for g. –  Michael Chernick May 22 '12 at 15:26
    
@Michael I agree that martini's hint was enough. I did not see it when I start typing my answer. –  Martin Sleziak May 22 '12 at 15:46
    
@MartinSleziak My comment was not meant to be critical of you putting out the solution. I was just wondering whether Mathematics treats homework problems differently from CrossValidated. This problem was not designated as homework but our members and moderators tend to ask the OP if it is homework when they suspect that it might be. If the answer is yes the OP is asked to label it as such and then we just give hints and leave some of the problem as a exercise for the student to do. –  Michael Chernick May 22 '12 at 17:45
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Assuming $f$ is differentiable:

You have that

$$f(x+y)=f(x)f(y)$$

Fix $x$ and differentiate wrt $y$.

$$f'(x+y)=f(x)f'(y)$$

Let $y=0$, and you get

$$f'(x)=f(x)f'(0)$$

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Maybe it is worth noticing that in this solution we have to assume that $f$ is differentiable. Or, we should read $f'(x+y)=f(x)f'(y)$ as: If $f'(y)$ exists then $f'(x+y)$ exists and it is equal to $f(x)f'(y)$. Similarly for the last equation. –  Martin Sleziak May 22 '12 at 16:50
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@MartinSleziak That is true. Since many aldready dealt with the quotient, I wanted to give another look to the problem. –  Pedro Tamaroff May 22 '12 at 16:56
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