# Thread: Prove result (continuous derivative)?

1. ## Prove result (continuous derivative)?

Suppose that 'f' has a continuous derivative on $\mathbb{R}$ and that $f(x) + f(y) = f\left(\frac{x + y}{1 - xy}\right)$ for each x and y such that xy < 1. Prove that, for some constant C, $f(x) = C\tan^{-1} x$.

2. Put $x=y=0$ to obtain $f(0)+f(0)=f(0)$ and thus $f(0)=0$

Put $y=-x$ to obtain $f(x)+f(-x)=f(0)=0$ and thus $f(-x)=-f(x)$

$f'(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}$

$f'(x)=\lim_{h\to 0}\frac{f(x+h)+f(-x)}{h}$

$f'(x)=\lim_{h\to 0}\frac{f(\frac{h}{1-hx+x^2})}{h}$

$f'(x)=\lim_{h\to 0}\frac{f(\frac{h}{1-hx+x^2})}{\frac{h}{1+hx+x^2}}$. $\frac{1}{1+hx+x^2}$

$
f'(x)=\frac{f'(0)}{1+x^2}
$

since $\lim_{h\to 0}\frac{f(\frac{h}{1-hx+x^2})}{\frac{h}{1+hx+x^2}}$ $=\frac{f(0+\frac{h}{1-hx+x^2})-f(0)}{\frac{h}{1+hx+x^2}}=f'(0)$

$f(x)=\int\frac{f'(0)}{1+x^2}dx$ $=f'(0)tan^{-1}x+k$

$k=0$ since $f(0)=0$

$f(x)=f'(0)tan^{-1}x$

$
C=f'(0)
$

$f(x)=Ctan^{-1}x$