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If $F(x)$ is a differentiable function that satisfies the above functional equation for all real x and y except -1. If $F(0) \ne 0$ and $F'(0) = 1$ then $F(x) = ? $

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  • $\begingroup$ I assume "!=" means "not equals"? $\endgroup$ – Ethan Dlugie Apr 23 at 16:52
  • $\begingroup$ Yes. Sorry, I did not realise that it could be misinterpreted as factorial. Will edit. $\endgroup$ – Mamta Kumari Apr 23 at 17:06
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Hint

Let $x=\tan x_1$ and $y=\tan y_1$ with $|x_1|<{\pi\over 2}$ and $|y_1|<{\pi\over 2}$. Then by defining $$g(x)=e^{f(\tan x)}$$ we have $$g(x_1+y_1)=g(x_1)+g(y_1)$$

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  • $\begingroup$ I don't think that's working! But I did find the answer by exploiting the fact that the function is said to differentiable. F(x) turned out to be ((1+x)/(1-x))^(0.5) $\endgroup$ – Mamta Kumari Apr 23 at 17:34

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