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If $f$ is a differentiable function on $\mathbb{R}$ and $f'(0)=2$ satisfying $$f(x+y) = \frac{f(x)+f(y)}{1-f(x)f(y)},$$ then to prove that $f(x)=\tan 2x$.

I know that we must prove using the first definition(principle) of differentiation but I am not able to proceed. I got $f(0)=0$ and I also proved function is odd.

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    $\begingroup$ $f(x)=2\tan x$ is not defined on the whole of $\mathbb{R}$ (it is undefined at $x=(n+\frac12)\pi$, $n\in\mathbb{Z}$), let alone differentiable. Also, it does not satisfy the function equation. $\endgroup$ – user10354138 Jun 23 at 5:37
  • $\begingroup$ Ok but how to prove the function is 2tanx (x,y is not of form (2n+1)π/2) $\endgroup$ – Tarun Elango Jun 23 at 5:40
  • $\begingroup$ “the first definition(principle) of differentiation”? $\endgroup$ – let's have a breakdown Jun 23 at 6:47
  • $\begingroup$ Lim h-->0 [f(x+h)-f(x)]/h = f'(x) $\endgroup$ – Tarun Elango Jun 23 at 6:49
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As suggested by user marty cohen, I'll expand his hint into a full answer. We have:

$$\begin{array} \nbsp f'(x) &=& \displaystyle\lim_{y\to 0} \dfrac{f(x+y)-f(x)}{y}\\ &=&\displaystyle\lim_{y\to 0} \dfrac{f(x)+f(y)-f(x)[1-f(x)f(y)]}{y[1-f(x)f(y)]}\\ &=&\displaystyle\lim_{y\to 0} \dfrac{f(y)+f(x)^2 f(y)}{y[1-f(x)f(y)]}\\ &=&\displaystyle\lim_{y\to 0} \left(\dfrac{f(y)}{y} \;\cdot \dfrac{1+f(x)^2}{1-f(x)f(y)}\right)\\ &=&f'(0) \cdot(1+f(x)^2)\\ &=&2(1+f(x)^2) \end{array}$$

where in the next-to-last line we use the definition of $f'(0)$, as well as continuity of $f$ and $f(0)=0$.

Now the differential equation

$$y' = 2(1+y^2)$$

has the general solution $f(x)= \tan(2x+C)$; the condition $f(0) = 0$ forces $C=0$, so

$$f(x) = \tan(2x)$$

is the only possible solution for such an $f$. However, this $f$ is not defined at $x \in \lbrace \dfrac{k \pi}{2}: k \in \mathbb Z\rbrace$, so if one is formal and interprets "differentiable on $\mathbb R$" as "defined and differentiable on all of $\mathbb{R}$", the question as stated has no solution.

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    $\begingroup$ I like this technique of functional equation leading to the function via getting an equation for the derivative. Works for log, power, arctan, probably others. $\endgroup$ – marty cohen Jun 24 at 4:33
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If a solution exists it has to be $f(x)=\tan(2x)$, (not $2\tan\, x$). Let $g(x)=\tan^{-1}f(x)$. Then we get $g(x+y)=g(x)+g(y)$ which implies (by continuity) that $g(x)=cx$ for some $c$. Hence $f(x)=\tan (cx)$ and we can find $c$ from the fact that $f'(0)=2$.

As already pointed out $tan(2x)$ is not defined at certain points. This proves that there is no differentiable function on $\mathbb R$ satisfying the given equation

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  • $\begingroup$ Sorry for the mistake but how to prove the function $\endgroup$ – Tarun Elango Jun 23 at 6:00
  • $\begingroup$ Using first principle? $\endgroup$ – Tarun Elango Jun 23 at 6:01
  • $\begingroup$ @TarunElango You cannot do that because there is no solution. $\endgroup$ – Kavi Rama Murthy Jun 23 at 6:19
  • $\begingroup$ Ok thank you sir $\endgroup$ – Tarun Elango Jun 23 at 6:22
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Find $(f(x+y)-f(x))/(y)$ and let $y \to 0$. Use $f(0)=0$.

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  • $\begingroup$ I am unable to proceed after this particular step $\endgroup$ – Tarun Elango Jun 23 at 5:44
  • $\begingroup$ I tried using f(x+y)=f(x)+f(y)/1-f(x)f(y) then I also tried partial differentiation on the original equation $\endgroup$ – Tarun Elango Jun 23 at 5:45
  • $\begingroup$ Can you provide some more steps from here? $f'(x)=\lim_{y\to 0} \frac{2f(x)+f(y)-(f(x))^2f(y))}{(1-f(x)f(y))y}$ $\endgroup$ – Archis Welankar Jun 23 at 5:48
  • $\begingroup$ @ArchisWelankar: Check your computation. Then, using $f'(0)=2$, you should get $f'(x)=2(1+f(x)^2)$. $\endgroup$ – Torsten Schoeneberg Jun 23 at 6:37
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    $\begingroup$ @ArchisWelankar: $f'(x) = \lim_{y\to 0}\frac{f(x+y)-f(x)}{y} = \lim_{y\to 0}\frac{f(x)+f(y)-f(x)(1-f(x)f(y))}{y(1-f(x)f(y))}=\lim_{y\to 0}\frac{f(y)+f(x)^2f(y)}{y(1+f(x)f(y))}=\lim_{y\to 0}\frac{f(y)}{y}\frac{1+f(x)^2}{1+f(x)f(y)}\stackrel{f(0)=0}=f'(0)(1+f(x)^2)=2(1+f(x)^2)$ $\endgroup$ – Torsten Schoeneberg Jun 23 at 17:08

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