Is there an everywhere-defined function that satisfies $f(x+y)=\frac{f(x)+f(y)}{1-f(x)f(y)}$ Is there a function $f:\mathbb{R}\to\mathbb{R}$ which is differentiable and satisfies the following:
(1) $f(x+y)=\frac{f(x)+f(y)}{1-f(x)f(y)}$
(2) $f'(0)=1$
(1) is the functional equation for $\tan$ function, but $\tan x$ is not defined at $\frac{\pi}{2}+n\pi$, so $f$ cannot be the $\tan$ function. Then does such $f$ exist?
 A: Here is the outline of a proof.


*

*Setting $x = 0$ and $y = 0$ in (1), we find that $f(0) = 0$.

*By the second condition, we see that $f(a) > 0$ for some small $a > 0$.

*Letting $x = y$ in (1), we see that $1 - f(x)^2 \ne 0$, so $f(x) \ne \pm 1$. Since also $f(0) = 0$, by the intermediate value theorem $f(x) \in (-1,1)$ for all $x$.

*Letting $x = y$ in (1), we note first that $0 < 1 - f(x)^2 \leq 1$ by point 3. Thus f(x) > 0 implies $f(2x) > 2f(x)$.

*By induction from point 4, we find $f(2^n a) > 2^n f(a)$.

*Point 5 is obviously incompatible with point 3.
So there is no such function $f$.
A: It is not possible. It is straightforward to show that $f$ satisfies
$f'(x) = 1+ f(x)^2$, with $f(0) = 0$ (because $f'(0) = 1$). The unique solution on $(-{\pi \over 2}, {\pi \over 2})$ is $f(x) = \tan x$.
A: Note that by setting $y=0$ we get
$$
f(x)  =\frac{f(x)+f(0)}{1-f(0)f(x)} \Rightarrow (1+f(x)^2) f(0) = 0 ~\forall x$$
Hence $f(0)=0$.
Differentiating both sides of 1) with respect to $x$ you ge
$$
f'(x+y) = \frac{1+f(y)^2}{(1-f(x) f(y))^2} f'(x)
$$
similarly, differentiating with respect to $y$
$$
f'(x+y) = \frac{1+f(x)^2}{(1-f(x) f(y))} f'(y)
$$
Hence
$$ (1+f(x)^2) f'(y) = (1+f(y)^2) f'(x)$$
Substitute $y=0$ to get
$$
f'(x) = 1+f(x)^2
$$
We also have $f(0)=0$. Hence $f(x)=\tan(x)$
Added in response to comment
if $f$ exists, it has to be $\tan(x)$ and since you want a function other than $\tan(x)$, no other function exists.
A: No, there is not; the proof is based on three observations:
First, suppose that $f$ passes through $\pm 1$. Near such an $x$, the numerator of the functional equation is near $2$ and the denominator is near $0$, so the function is unbounded (hence discontinuous) at $2x$. 
Second, note that $f(0) = 0$. This is immediate from setting $x = y = 0$ in the functional equation.
Third, suppose that $f(x) > 0$ for some small positive $x$. Then $$f(2x) = \frac{2 f(x)}{1 - f(x)^2} > 2f(x)$$
Inducting on this, we find that
$$f(2^n x) > 2^n f(x)$$ for all $n = 1, 2, ...$. Choose an $n$ large enough, and this exceeds $1$, contradicting the first observation. A similar argument shows that if $f(x) < 0$ we arrive at a contradiction; thus $f \equiv 0$ is the only differentiable solution.
A: Differentiability is not needed to show that all solutions of the functional equation are $\tan Ct$ for constant $C$. 
The complex function $g(x) = \frac{1+if(x)}{1-if(x)}$ has $|g|=1$ (for $f$ real-valued) and, from the functional equation on $f$, satisfies the addition formula 

$g(x)g(y)=g(x+y)$ for all $x,y \in \mathbb{R}$ 

The only measurable solutions of that equation with real $f$ (that is, $|g|=1$) are $g(t)=e^{i\alpha t}$ with real $\alpha$, and no other solutions can be proved to exist without some form of the Axiom of Choice for uncountable sets.  In that generality, the solution is to take $g(t)=e^{i\alpha t}$, solve for $f(t)$ from $g(t)$ (each is a linear fractional function of the other), and find that $f(t) = \tan \frac{\alpha}{2} t$.
Specializing to differentiable solutions, the only way to have $f'(0)=1$ is $f(t)=\tan t$. 
