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

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 ?

• Hint: The function $g$ defined by $g(t)=f(t)+\frac1t$ satisfies $$g(xy)=g(x)+g(y).$$ Can you solve this? – Did Aug 1 '16 at 17:10

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

• $f'(x)=\frac{1}{x^2}+\frac{f'(1)-1}{x}$ – alans Aug 1 '16 at 22:35
• Is there any general way to do this problem . – Aakash Kumar Aug 2 '16 at 2:54
• You assumed y as constant – Aakash Kumar Aug 2 '16 at 2:58
• @AakashKumar Yes, indeed. For a differentiable function this is a path to try. – egreg Aug 2 '16 at 8:33
• @alans Thanks, fixed – egreg Aug 2 '16 at 8:33

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