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let $f(0)=1$ and $f'(0)=1$.

and $f(x+y)=f(x)f(y)$ for $x,y\in R$.

How can I found $f(x)$ by using a definition of derivative?

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  • $\begingroup$ What have you tried? And how could you use the definition of the derivative here to learn more about $f$? $\endgroup$ – TMM Sep 15 '11 at 13:34
  • $\begingroup$ $f(x+h)=f(x)f(h)$, $f(x+h)-f(h)=f(x)(f(h)-1)$, $\frac{f(x+h)-f(h)}{h}=f(x)\frac{f(h)-1}{h}$, $\lim_{h\to 0}$, then $f'(x)=f(x)$ $\endgroup$ – Xiang Sep 15 '11 at 13:39
  • $\begingroup$ If you get $f'(x) = f(x)$ you're done? $\endgroup$ – TMM Sep 15 '11 at 13:42
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    $\begingroup$ Well you can already show $f'=f$ and you know $f(0)=1 $so now can you prove that $ e^x $ is the unique solution to that Initial value problem? Hint: Assume $y=e^x g(x)$ solves the IVP and plug it in to see if you can say anything about what $g$ must be. $\endgroup$ – Ragib Zaman Sep 15 '11 at 13:46
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    $\begingroup$ If $f'(x) = f(x)$, then $f'(x) / f(x) = 1$. But $f'(x)/f(x) = \frac{d}{dx} \ln(f(x)) = 1$ implies $\ln(f(x)) = x + C$ and $f(x) = D e^x$. If of course you are allowed to use all that. $\endgroup$ – TMM Sep 15 '11 at 13:47
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Let me just turn that last comment into an answer. As you already pointed out in the comments, by the definition of the derivative and using $f(x+y) = f(x)f(y)$, $f(0) = 1$ and $f'(0) = 1$ we have

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

Now you're almost done, as $f'(x) = f(x)$ is one of the essential properties of the exponential function. From $f'(x) = f(x)$ it follows that

$$\frac{f'(x)}{f(x)} = \frac{d}{dx}\left(\ln f(x)\right) = 1$$

Integrating both sides, we get $\ln f(x) = x + C$ for some constant $C$. This implies $f(x) = D e^x$ for some constant $D$ (i.e. $D = e^C$). We know that $f(0) = 1$, so it follows that $D = 1$ and $f(x) = e^x$.

Alternatively, as pointed out by Ragib, after noticing $f'(x) = f(x)$ you could write $f(x) = e^x g(x)$ for some function $g(x)$. Then $f'(x) = e^x g'(x) + f(x) = f(x)$, so $e^x g'(x) = 0$ for all $x$. Since $e^x \neq 0$, it follows that $g'(x) = 0$ and $g(x) = D$ is constant. Using $f(0) = 1$ it then follows that $D = 1$ and $f(x) = e^x$.

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