prove $f'(x)=f(x)$ 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 $x$?
 A: 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 $f'(x_0)=f(x_0)$.
A: 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
$$f'(x)=f(x)f'(0)$$
A: $$
\begin{align} f(x+h)-f(x) & = f(x)\cdot f(h) - f(x), \quad\text{from (a)} \\[8pt]
& = f(x)(f(h)-1) \\[8pt]
& = f(x)\cdot hg(h),\quad \text{from (b)}
\end{align}
$$
So $\displaystyle f'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}=\lim_{h \to 0}f(x)g(h)=f(x)$
A: $$f'(x) = \lim_{\Delta x \to 0} \frac{f(x+\Delta x)-f(x)}{\Delta x}= \lim_{\Delta x \to 0} \frac {f(x)(f(\Delta x)-1)}{\Delta x} = \lim_{\Delta x \to 0} \frac {f(x) \Delta xg(\Delta x)}{\Delta x}=\lim_{\Delta x\to 0}f(x)g(\Delta x)=\lim_{\Delta x\to 0}f(x)\lim_{\Delta x\to 0}g(\Delta x)=f(x)$$
