# Thread: Prove result (continuous derivative)?

1. ## Prove result (continuous derivative)?

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

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

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

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

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

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

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

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

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

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

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

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

$\displaystyle C=f'(0)$

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