$f(x)$ if $f(xy)=f(x) +f(y) +\frac{x+y-1}{xy}$

Question

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 ?

Answer 1

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 $$ f'(x)=\frac{1}{x^2}+\frac{f'(1)-1}{x} $$

Answer 2

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: $$ f(x)=\ln{x}-\frac{1}{x} $$