Differentiability of $f(x+y) = \frac{f(x)+f(y)}{1-f(x)f(y)}$ let $f$ be a function defined on the interval $(-1, 1)$ such that for all $x, y \in (-1, 1)$,
$$f(x+y) = \frac{f(x)+f(y)}{1-f(x)f(y)}$$
Suppose that $f$ is differentiable at $x = 0$
(i) Show that $f$ is differentiable on $(-1, 1)$.
(ii) If $f'(0) = \pi/2$, find the explicit expression of $f(x)$.

To establish $f(0)$:
$$f(x+y) = \frac{f(x)+f(y)}{1-f(x)f(y)}\implies f(x+0) = \frac{f(x)+f(0)}{1-f(x)f(0)}$$
$$\implies f(x)\left\{1-f(x)f(0)\right\}=f(x)+f(0)\implies f(0)=0$$
To find the derivative:
$$\frac{f(x+h) -f(x)}{h}= \frac{1}{h}\left(\frac{f(x)+f(h)}{1-f(x)f(h)}-f(x)\right)=\frac{(1+f(x)^2)}{1-f(x)f(h)}\frac{f(h)}{h}$$
$$=\frac{(1+f(x)^2)}{1-f(x)f(h)}\frac{f(h)-f(0)}{h}\to \frac{(1+f(x)^2)}{1-f(x)f(0)}f'(0)\text{ as }h\to0$$
So:
$$f'(x)=f'(0)\cdot\left(1+f(x)^2\right)$$
If $f'(0)=\pi/2$, then:
$$f'(x)=\frac{\pi}{2}\left(1+f(x)^2\right)\implies \frac{f'}{1+f^2}=\frac{\pi}{2}$$
$$\frac{d}{dx}\tan^{-1}f(x)=\frac{\pi}{2}\implies \tan^{-1}f(x)-\tan^{-1}f(0)=\frac{\pi}{2}x\implies f(x)=\tan\left(\frac{\pi}{2}x\right)$$
