real analysis - $f(x+y)=f(x)f(y)$ for all $x,y$ . Show that $f'(x)=f'(0)f(x)$ and determine the value of $f'(0)$

Problem is that Suppose f is differentiable everywhere and $f(x+y)=f(x)f(y)$ for all $x,y$ . Show that $f'(x)=f'(0)f(x)$ and determine the value of $f'(0)$.

I can show $f'(x)=f'(0)f(x)$

but i don't know how to determine the value of $f'(0)$.

Please help!

Answer

It's impossible to determine the value of $f'(0)$ from the information given - perhaps you left something out, perhaps the question was stated somewhat differently, or perhaps it's a bad question.

If $a$ is any real number and $f(t)=e^{at}$ then $f$ is differentiable, $f(x+y)=f(x)f(y)$, and $f'(0)=a$. So the information given literally says nothing at all about the value of $f'(0)$; that value can be anything.