# $F({x+y\over 1+xy}) = F(x)\cdot F(y)$ for all real $x, y$ except $-1$

If $$F(x)$$ is a differentiable function that satisfies the above functional equation for all real x and y except -1. If $$F(0) \ne 0$$ and $$F'(0) = 1$$ then $$F(x) = ?$$

• I assume "!=" means "not equals"? – Ethan Dlugie Apr 23 at 16:52
• Yes. Sorry, I did not realise that it could be misinterpreted as factorial. Will edit. – Mamta Kumari Apr 23 at 17:06

## 1 Answer

Hint

Let $$x=\tan x_1$$ and $$y=\tan y_1$$ with $$|x_1|<{\pi\over 2}$$ and $$|y_1|<{\pi\over 2}$$. Then by defining $$g(x)=e^{f(\tan x)}$$ we have $$g(x_1+y_1)=g(x_1)+g(y_1)$$

• I don't think that's working! But I did find the answer by exploiting the fact that the function is said to differentiable. F(x) turned out to be ((1+x)/(1-x))^(0.5) – Mamta Kumari Apr 23 at 17:34