Prove that $f′(x)=f′(0)f(x)$ derivatives Let $f:I \to R$ be differrentiable on an open interval $I \subseteq R$ with
$$f(a + b) = f(a)f(b) \quad \forall a, b \in R$$
Suppose that $f(0) = 1$ and that $f'(0)$ exists. Show that:
$$f'(x) = f'(0)f(x) \quad \forall x \in R$$
So far, this is what I got:
$$  f(x+0)= f(x)f(0)$$
 $$f(x) = f(x)f(0)$$
For all $x$ , $f(0)$ has to be $1$.
$$0=f(x)f(0) - f(x)$$
$$0 = f(x)(f(0) -1)$$
$f(x)=0$ for all $x$ or $f(0) = 1$.
I'm stuck here. Could you please give a hint?
 A: I am expanding on my hint in the comment. The question is simpler than it appears. Moreover as others have pointed out the question gives more hypotheses than necessary to get to the conclusion.
The following is the corrected version without any redundant hypotheses:
Let $f: \mathbb{R} \to \mathbb{R}$ be a function such that $f(a + b) = f(a)f(b)$ for all $a, b \in \mathbb{R}$. If $f'(0)$ exists then $f'(x) = f'(0)f(x)$ for all $x \in \mathbb{R}$.
Since $f'(0)$ exists we have by definition of derivative $$\lim_{h \to 0}\frac{f(h) - f(0)}{h} = f'(0)\tag{1}$$ Now it is easily seen (again via definition of derivative) that
\begin{align}
f'(x) &= \lim_{h \to 0}\frac{f(x + h) - f(x)}{h}\notag\\
&= \lim_{h \to 0}\frac{f(x + h) - f(x + 0)}{h}\notag\\
&= \lim_{h \to 0}\frac{f(x)f(h) - f(x)f(0)}{h}\text{ (because }f(a + b) = f(a)f(b))\notag\\
&= \lim_{h \to 0}f(x)\cdot\frac{f(h) - f(0)}{h}\notag\\
&= f(x)f'(0)\text{ (using }(1))\notag
\end{align}
As you can see we don't need to know the value of $f(0)$. It is immaterial. However there are only two choices for $f(0)$. Either $f(0) = 0$ or $f(0) = 1$ and both are valid choices.
A: Consider the map $g_y$, for a given $y$, $x\mapsto f(x+y)$.
Then, $g_y'(x) = f'(x+y)$.
But $g_y(x) = f(x)f(y)$, so $g_y'(x) = f'(x)f(y)$.
Thus, for $x=0$, $g_y'(0) = f'(y) = f'(0)f(y)$.
