It transforms products into sums

By making $x=y=1$ into the given functional equation we get that $f(1)=0$ . Then you can see that if $x$ is any positive real number and $y=\frac{1}{x}$ , we obtain that $f(x)+f(\frac{1}{x})=f(1)=0$ and therefore $f(\frac{1}{x})=-f(x).\:\: \: \: \: \: \: \: \: \: \: \: \:\: \: (1)$ Using (1) and the functional equation we get that for any real numbers $x$ and $a$ the following equality holds: $f(x)-f(a)=$ $f(x)+f(\frac{1}{a})=f(\frac{x}{a})=f(\frac{x}{a})-f(1).$ Therefore the continuity of $f$ at 1 implies the continuity at any positive number $a.$

Now let us prove by Mathematical Induction that for any positive number $a$ and any natural number $n$ the following is true $f(a^{n})=nf(a).\:\: \: \: \: \: \: \: \: \: \: \: \:\: \: (2)$ Indeed, making $n=1$ we obtain that $f(a)=f(a)$ . Assuming that $f(a^{k})=kf(a)$ , and using the functional equation given in the problem, we get that $f(a^{k+1})=f(a^{k}*a)=f(a^{k})+f(a)=kf(a)+f(a)=$ $(k+1)f(a).$ By replacing $a$ by $a^{\frac{1}{n}}$ in (2) we obtain that $f((a^{\frac{1}{n}})^{n})=nf(a^{\frac{1}{n}})$ , or equivalently, $f(a)=nf(a^{\frac{1}{n}})$ for any positive number $a$ and any natural number $n.$ Dividing both sides by $n$ we obtain $f(a^{\frac{1}{n}})=\frac{1}{n}f(a)\:\: \: \: \: \: \: \: \: \: \: \: \:\: \: (3)$

Now, we can prove that for any real number $r,$ $f(a^{r}) =rf(a). \:\: \: \: \: \: \: \: \: \: \: \: \:\: \: (4)$ First we are going to assume that $r$ is a positive rational number, so it can be written like $\frac{n}{m},$ where $m$ and $n$ are natural numbers and then using (2) and (3) $f(a^{\frac{n}{m}})$ $=f((a^{\frac{1}{m}})^{n})$ $=nf(a^{\frac{1}{m}})=\frac{n}{m}f(a).$ The equality (4) is also going to be true in the case the $r$ is a negative rational number. Actually, if $r$ is a negative rational number then using (1) and assuming that (4) is true for positive rational numbers, we obtain that $f(a^{r})= -f(a^{-r})=-(-r)f(a)=rf(a).$ The case when $r=0$ it is easily derived from the fact that $f(1)=0.$ To prove that (4) is true for any real number $r$ it is enough to consider a sequence of rational numbers $(r_{n})_{n}$ , such that $\lim_{n\to \infty}r_{n}=r.$ Then taking limits in both sides of $f(a^{r_{n}})=r_{n}f(a)$ and using the continuity of $f(x)$ we obtain that $f(a^{r})=rf(a).$ This completes the proof of (4).

The rest is very simple. If $x$ is any real positive number, from (4) we obtain that $f(x)=f(3^{\log_{3}x})=(\log_{3}x)f(3)=7\log_{3}x.$ Therefore, $f(x)$ is differentiable at any positive number and $f'(1)=\frac{7}{\ln 3}\approx 6.37.$

My solution is in no way half as good as the other solutions, but I just want to share it.

Putting $x=y=\sqrt3$ into the equation would give us $f(3)=2f(\sqrt3)$ , which gives us $f(\sqrt3)=7/2$ .

Then, induction can be used to prove the following statement: $f\left(3^{1/2^n}\right)=\frac7{2^n}$

Since $3^{1/2^n}$ approaches $1$ and $\dfrac7{2^n}$ approaches $0$ and it is continuous, this can be used to compute $f'(1)$ .

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The limit sign will be omitted.

$\quad f'(1)$

$=\displaystyle\frac{7/2^n-7/2^{n+1}}{3^{1/2^n}-3^{1/2^{n+1}}}$

$=\displaystyle\frac{7/2^{n+1}}{3^{1/2^n}-3^{1/2^{n+1}}}$

$=\displaystyle\frac72\frac{1/2^n}{3^{1/2^n}-3^{1/2^{n+1}}}$

$=\displaystyle\frac72\frac{1/2^n}{3^{1/2^n}(1-3^{-1/2^{n+1}})}$

$=\displaystyle\frac72\frac{1/2^n}{1-3^{-1/2^{n+1}}}$

$=\displaystyle\frac72\frac{\ln(1/2)\times(1/2^n)}{-3^{-1/2^{n+1}}\ln3(1/(2^{n+1})^2)(\ln2)(2^{n+1})}$

$=\displaystyle\frac72\frac{1/2^n}{3^{-1/2^{n+1}}\ln3(1/(2^{n+1})^2)(2^{n+1})}$

$=\displaystyle\frac72\frac{1/2^n}{3^{-1/2^{n+1}}\ln3(1/2^{n+1})}$

$=\displaystyle7\frac{1}{3^{-1/2^{n+1}}\ln3}$

$=\displaystyle\frac7{\ln3}$

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The limit from the negative side can be checked by a similar method, based on $f(1/3)=f(1)-f(3)=-7$ .

Let $f(x) = g(\log{x})$ , so that $f$ and $g$ are defined over positive real numbers. If $f$ is continuous, then $g$ is also continuous.

We get $f(xy) = g(\log{xy}) = g(\log{x}+\log{y})$ [1]

and $f(x)+f(y) = g(\log{x})+g(\log{y})$ [2]

[1] = [2]; $g(\log{x}+\log{y}) = g(\log{x})+g(\log{y})$

Substituting $x$ as $e^{x}$ and $y$ as $e^{y}$ we get

$g(x+y) = g(x)+g(y)$ .

Given that $g$ is continuous, $g$ satisfies Cauchy's Functional Equation.

We get $g(x) = cx$ for some constant $c$ .

Which gives $f(x) = g(\log{x}) = c\log{x}$ .

Solve for $c$ we get $c = \displaystyle \frac{7}{\log{3}}$ .

Therefore, $f'(x) = \displaystyle \frac{c}{x}$

At $x = 1$ , $f'(1) = c = \boxed{\displaystyle \frac{7}{\log{3}}}$ . ~~~

If one differentiates the above functional equation with respect to x and y, one will obtain:

y f'(xy) = f'(x) (i),
x
f'(xy) = f'(y) (ii).

and equating (i) with (ii) gives:

f'(x)/y = f'(y)/x, or x f'(x) = y f'(y) = A (real constant) (iii)

and integrating (iii) with respect to x yields:

f(x) = A*ln(x), f(3) =7.

which finally results in f(x) = (3/ln(7))*ln(x).

Certainly f(x) is differentiable at x = 1, and f'(1) = (3/ln(7)) * (1/1) = (3/ln(7)) = 6.372.

PS: Great proof, Arturo Presa!

I came up with a very simple solution, although I don't think it is a rigorous proof:

When I saw that the function had the property of $f(xy)=f(x)+f(y)$ , I thought that a logarithm function would have that property.

So I defined $f(x)=a\ln{x}$

$f(3)=7$ , so $a=\frac{7}{\ln{3}}$

$f(x)=\frac{7\ln{x}}{\ln{3}}$ , and so $f'(x)=\frac{7}{x\ln{3}}$

$f'(1)\approx6.37167$