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This question already has an answer here:

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 ?

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marked as duplicate by naslundx, Claude Leibovici, user91500, Dragonemperor42, Johannes Kloos Aug 12 '16 at 11:39

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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    $\begingroup$ Hint: The function $g$ defined by $g(t)=f(t)+\frac1t$ satisfies $$g(xy)=g(x)+g(y).$$ Can you solve this? $\endgroup$ – Did Aug 1 '16 at 17:10
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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 $$ f'(x)=\frac{1}{x^2}+\frac{f'(1)-1}{x} $$

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

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