$f(x)$ if $f(xy)=f(x) +f(y) +\frac{x+y-1}{xy}$ Let $f$ be a differentiable function satisfying the functional time $f(xy)=f(x) +f(y) +\frac{x+y-1}{xy} \forall x,y \gt 0 $ and $f'(1)=2$
My work 
Putting $y=1$
$$f(1)=-1$$
$$f'(x)=\lim_{h\to 0}\frac{f(x+h) - f(x)}{h}$$
But I don't know anything about $f(x+h)$  so what to do in this problem ?
 A: Differentiate both sides with respect to $x$:
$$
yf'(xy)=f'(x)-\frac{1}{x^2}+\frac{1}{x^2y}
$$
For $y=1/x$, we get
$$
\frac{f'(1)}{x}=f'(x)-\frac{1}{x^2}+\frac{1}{x}
$$
so
$$
f'(x)=\frac{1}{x^2}+\frac{f'(1)-1}{x}
$$
A: Let $u=xy$. Differentiate the given function with respect to $x$:
$$
    y\frac{df(u)}{du}=\frac{df(x)}{dx}+\frac{1}{xy}-\frac{x+y-1}{x^2y}
$$
Then differentiate w.r.t. $y$:
$$
    xy\frac{d^2f(u)}{du^2}+\frac{df(u)}{du}=-\frac{1}{xy^2}-\frac{1}{x^2y}+\frac{x+y-1}{x^2y^2}
$$
The right-hand side simplifies:
$$
    xy\frac{d^2f(u)}{du^2}+\frac{df(u)}{du}=-\frac{1}{x^2y^2}
$$
Substituting $u$ for $xy$:
$$
    u\frac{d^2f(u)}{du^2}+\frac{df(u)}{du}=-\frac{1}{u^2}
$$
Re-writing the left-hand side:
$$
    \frac{d}{du}\left[u\frac{df(u)}{du}\right]=-\frac{1}{u^2}
$$
Integrating and applying the given boundary condition gives:
$$
    u\frac{df(u)}{du}=1+\frac{1}{u}
$$
Integrating again gives:
$$
    f(u)=a+\ln{u}-\frac{1}{u}
$$
This is a solution when $a=0$, so:
$$
    f(x)=\ln{x}-\frac{1}{x}
$$
