Let $f(xy) = f(x)+ f(y) +\frac{x+y-1}{xy} \forall x,y>0$, $f'(1)=2$, $f$ is differentiable . Find $f(x)$. What I have gotten from $f(xy) = f(x)+ f(y) +\frac{x+y-1}{xy}$ is:

*

*$f(1)=-1$
2.$f(x)+f(\frac{1}{x}) =-(x+\frac{1}{x})$
The information I could get from $f'(1)=2$ is just:
$\lim_{h\to 0} { \frac{f(1+h)+1}{h}}=2$
Since the limit is given, my approach was to find $f'(x)$ and then solve the differential equation. But I'm unable to compute the limit $\lim_{h \to 0} {\frac{f(x+h)-f(x)}{h}}$ with the information given. I have tried stuff like converting $f(x)$ to $f(\frac{x}{h}\cdot h)$ and try to use the original equation and convert it into$ f(\frac{x}{h})+ f(h) +\frac{\frac{x}{h}+h-1}{x}$.But it hasn't been of any use to compute the limit. I need to do it in the way of computing the limit and solving the Differential equation.

The functional equation can be rewritten:
$$f(xy)+\frac1{xy}=f(x)+\frac1x+f(y)+\frac1y.$$
So if $g(x)=f(x)+\frac1x,$ we have $$g(xy)=g(x)+g(y)$$ or $g(x)=C\log |x|$ for some constant $C.$
Then you want $f(x)=C\log |x|-\frac1x.$ Then use the fact that $f'(1)=2$ to determine $C.$

At heart, there is one "obvious" solution, and any other (continuous) solution must differ by a logarithm.
In general, if the functional equation was:
$$f(xy)=f(x)+f(y)+H(x,y)$$ for some function continuous $H,$ and $f_0$ is one continuous solution, then every continuous solution is of the form: $$f(x)=f_0(x)+C\log |x|$$ for some constant $C.$
