A function satisfies the equation $f(x+y) = f(x) f(y)$. Show that $f(x)$ is differentiable at all $x$, and determine its $f'(x)$ in terms of $f(x)$. If $f(x)$ is differentiable at $x=0, \space f'(0) = c \space$ where $\space c \space$ is a constant and $\space f(x) \space$ is not equals to $0$.
Someone please help me. Im stuck on this question. I've no idea where to even start.
 A: First show that $f(0)$ must be $1$ if $f$ is not the zero function. Then one has
$$
\mathop {\lim }\limits_{y \to 0} \frac{{f(x + y) - f(x)}}{y} = \mathop {\lim }\limits_{y \to 0} \frac{{f(x)f(y) - f(x)}}{y} = f(x)\mathop {\lim }\limits_{y \to 0} \frac{{f(y) - 1}}{y} = f(x)f'(0) =cf(x).
$$
A: Use the definition of a derivative. Let $x \in \mathbb{R}$. Then, we need to study the limit:
$$\lim_{h \to 0} \frac{f(x+h)-f(x)}{h}$$
By the definition of the function, we know that $f(x+h) = f(x)f(h)$. So, we simplify the above:
$$\lim_{h \to 0} \frac{f(x)(f(h)-1)}{h}$$
You mentioned that $f(x)$ is differentiable at $x = 0$ and that $f'(0) = c$. We also know that $f(x) \neq 0$. Now, I claim that $f(0) = 1$. Why? Consider:
$$f(0) = f(0+0) = f(0) \cdot f(0)$$
Since $f(0) \neq 0$, $f(0) = 1$. Since $f(x)$ is differentiable at $x = 0$, it is the case that:
$$\lim_{h \to 0} \frac{f(x)(f(h)-1)}{h} = \lim_{h \to 0} f(x) \cdot \lim_{h \to 0} \frac{f(h)-f(0)}{h}$$
The second limit is just the derivative of $f(x)$ at $x = 0$ and the first one just evaluates to $f(x)$. So, we have:
$$f'(x) = f(x) \cdot c$$
And this proves that $f(x)$ is differentiable at any $x \in \mathbb{R}$.
I should mention that it's always a good idea to start from the definitions of words and work through those definitions to try and get somewhere. It might not end up really working but this sort of experimenting is a very natural part of learning how to solve problems.
