everywhere differentiable function — an interesting problem

Question: Determine all the functions f, which are everywhere differentiable and satisfy 

f(x)+f(y)=f((x+y)/(1-xy)) for all real x and y with xy \neq 1. —- Equation I

Solution:

Let f(x) satisfy Equation I above. Differentiating Equation I partially with respect to each of x and y, we obtain

f^{'}(x)=((1+y^{2})/(1-xy)^{2})f^{'}((x+y)/(1-xy)) Equation II

f^{'}(y)=((1+x^{2})/(1-xy)^{2})f^{'}((x+y)/(1-xy)) Equation III

Eliminating common  terms in Equation II and Equation III, we deduce that

(1+x^{2})f^{'}(x)=(1+y^{2})f^{'}(y)

As the left side of Equation III depends only on x and the  right side only on y, each side of Equation III must be equal to a constant c. Thus, we have

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

and so, f(x)=c\arctan x +d,

for some constant d. However, taking y=0 in Equation III, we obtain f(x)+f(0)=f(x), so that f(0)=0 and

d=0. Clearly, f(x)=c\arctan x satisfies Equation I and so all solutions of Equation I are given by

f(x)=c\arctan x where c is a constant.

Note: In Equation II, the f^{'} is w.r.t. x and in Equation III, the f^{'} is w.r.t. y, yet in equation II 

f^{'}(x+y)/(1-xy) is the same as the corresponding f^{'} in Equation III because the argument (x+y)/(1-xy) is symmetric w.r.t. x and y. 

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