# The derivative of $f(x)$ if $f(x)+f(y)=f\left(\frac{x+y}{1-xy}\right)$

derivativesfunctional-equations

Question. What is the derivative of $$f(x)$$ if $$f(x)+f(y)=f\left(\frac{x+y}{1-xy}\right)$$?

So my solution is the following:

Differentiating with respect to $$x$$ gives

$$f'(x) = f '\left(\frac{x+y}{1-xy}\right)$$

Let $$x=0$$ then $$f'(0)=f '(y)$$ for all $$y$$ in $$\mathbb{R}$$. So the derivative is constant.

So my problem is that I can't solve this but also I don't know what I've done wrong. This seems perfectly fine for me.

#### Best Answer

Ok so I have almost solved it.

$$f'(x)=f'\left(\frac{x+y}{1-xy}\right)\left(\frac{(1-xy)-(x+y)(-y)}{(1-xy)^2}\right)$$

Let $$x=0$$, then $$f'(0)=f'(y)(1+y^2)$$, thus $$f'(y)=\frac{c}{1+y^2}$$ for all $$y$$ in $$\mathbb{R}$$.

So we have $$f(y)=c*\arctan y+C$$ where $$C=0$$ since $$f(0)=0$$.

I know the function that I've gotten satisfies the functional equation if $$xy<1$$.

So if I want to show that the functional equation is satisfied, must I prove that $$\arctan x+\arctan y=\arctan\left(\frac{x+y}{1-xy}\right)$$ for $$xy<1$$ or is it enough that I got $$f(y)$$ based on $$x=0$$ and $$y$$ is any real number so $$xy<1$$?