What is derivative a of a continuous function such that $f(x+y) = f(x) +f(y)$? Let $f: \mathbb{R} \to \mathbb{R} $ be a continuous function with $f(x+y) = f(x) +f(y),\forall x,y\in \mathbb{R}.$ 
Find $\frac {df} {dx} $, if it is exist?
 A: This is called Cauchy's functional equation. You can show that $f(0)=0$, and if you let $c=f(1)$ then you can deduce that $$f(2)=f(1+1)=f(1)+f(1)=2c\qquad \text{ and } \qquad f\left(\frac{1}{2}\right)=\frac{f(1)}{2}= \frac{c}{2} $$
This extends similarly to
$$f(x)=cx$$ for all rational $x$, but because $f$ is continuous, $f(x)=cx$ for all real $x$. You should be able to show that the derivative of this linear function exists.
A: If a function $f$ fulfills $f(x+y)=f(x)+f(y)$ for any $x,y\in\mathbb R$, then by differentiating by $x$ you get
$$f'(x+y)=f'(x).$$
(Here we use the assumption that $f$ is differentiable.)
This equality is true, again, for any $x,y\in\mathbb R$. So we see that $f'(x)$ is constant.
Let $f'(x)=a$. Then we get $f(x)=ax+b$. 
It is relatively easy to show that $f(0)=0$, from which we get $b=0$.
Therefore $f(x)=ax$ for some $a\in\mathbb R$. We get that $f'(x)=a=f(1)$.

It is worth mentioning that the equation $f(x+y)=f(x)+f(y)$ is called Cauchy's functional equation. Solutions of this equation are called additive functions.
We saw that if $f$ is differentiable, then $f(x)=ax$. In fact, we get only solutions of this form even with much weaker conditions on $f$. But if we do not have any restrictions on $f$, then there are also different solutions.
See Overview of basic facts about Cauchy functional equation.
