Prove that $f$ is differentiable on $\mathbb R$ Suppose that $f'(a)$ exists for some $a$ which is a member of $\mathbb R$ 
$f(x+y) = f(x+a)f(y)$, for all values of $x$, where $y$ is a member of $\mathbb R$. Prove that the function $f$ is differentiable on $\mathbb R$.
I have interpreted the question such that I have evaluated the right hand side of the equation to see whats the differentiated form and I have come up with the following :
$f(x+a)f(y)$ when differentiated is $f (x+a)f '(y) + f (y)f '(x+a)$ I am not sure how to prove that $f$ is differentiable in relation to getting a result and am I not sure if what I have done so far if it has made any sense. can anyone help me?
 A: I agree with Reveillark that $y$ should be a variable. First substitute $0$ for $x$ to get
$$f(y)=f(a)f(y).$$
So $f(y)=0$ or $f(a)=1$.  If $f(y)=0$ for some $y\in \Bbb R$, say when $y=b$, then $$f(x+b)=f(x+a)f(b)=0$$ for all $x$, so $f(x)=0$ for all $x$, which makes $f$ a differentiable function.  
From now on, assume $f(y)\ne 0$ for all $y\in \Bbb R$.
  Then $f(a)=1$.  Thus
$$f(x+y)-f(y)=f(x+a)f(y)-f(y)=f(x+a)f(y)-f(a)f(y).$$
So
$$\frac{f(x+y)-f(y)}{x}=\frac{f(x+a)-f(a)}{x}f(y).$$
Since $f'(a)$ exists, the RHS tends to $f'(a)f(y)$ as $x\to 0$.  Therefore, $f'(y)$ exists and equals $f'(a)f(y)$.  Thus, $$f'(y)=f'(a)f(y)$$ for all $y\in\Bbb R$.  In particular, this shows that $$f(y)=A\exp\big(f'(a)y\big)$$ for some constant $A$.  Since $f(a)=1$, we get $$f(y)=\exp\big(f'(a)(y-a)\big)$$ for all $y\in \Bbb R$.
A: Suppose that for all $x, y$, we have 
$$
f(x+y) = f(x+a)f(y),
$$
where $a$ is a constant in $\Bbb R$, and that $f$ is differentiable at $a$. Let's try to compute
$$
f'(u)
$$
for some arbitrary $u$. To do so, we have to look at 
$$
\lim_{h \to 0} \frac{f(u+h) - f(u)}{h}
$$
Letting $x = h$ and $y = u$ in the main formula, we get
$$
f(u+h) = f(h + u) = f(h+a) f(u)
$$
Letting $x = 0$ and $y = u$ in that formula gives us
$$
f(u) = f(a) f(u)
$$
So the difference-quotient becomes
\begin{align}
\lim_{h \to 0} \frac{f(u+h) - f(u)}{h} 
&= \lim_{h \to 0} \frac{ f(h+a) f(u) - f(a)f(u)}{h}\\ 
&= f(u) \lim_{h \to 0} \frac{ f(h+a) - f(a)}{h}\\ 
&= f(u) f'(a) 
\end{align}
which shows the limit exists, and hence that $f$ is differentiable at $u$ for an arbitrary $u$. 
