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It's given that $$f(xy)=\frac {f (x)}{y}+\frac {f (y)}{x}$$ Also $x,y>0$ and $f(x)$ is differentiable for $x>0$ such that $f(e)=\frac{1}{e}$. By the look of the functional equation I am sure it does involve log at some point . By common substitutions I have been able to deduce that f(1)=0 and f(1/e)=-e but I am not sure how to proceed . Any hint is appreciated.

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  • $\begingroup$ Substitute $$y=1$$ $\endgroup$ Mar 20, 2019 at 14:13
  • $\begingroup$ I got f(1)=0 from that $\endgroup$ Mar 20, 2019 at 14:36

2 Answers 2

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Multiplying with $xy$ yields $$(xy)f(xy) = xf(x) + yf(y).$$ Define $g(x) := xf(x)$ to get $g(xy) = g(x) + g(y)$. Dou you know a/the function that satisfies this?

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  • $\begingroup$ Great answer! Thank you for sharing it :). $\endgroup$ Mar 20, 2019 at 14:17
  • $\begingroup$ got it . There is a log there. and $f(x)=\frac {logx}{x} $ $\endgroup$ Mar 20, 2019 at 14:48
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Let $g(x)=xf(x)$ so we have $$g(xy) = g(x)+g(y)$$

Let $h(x)=g(e^{x})$, then $$h(x+y) = g(e^{x+y})=g(e^x\cdot e^y)= g(e^x)+g(e^y)=h(x)+h(y)$$

So $h$ is Cauchy function and since it is differentiable it is linear, so $h(x)=ax$ for some real $a$.

Since $h(1)= g(e)= ef(e)= e{1\over e} = 1$ so $a=1$ and thus $$x = g(e^x)\implies g(x) = \log x$$

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  • $\begingroup$ My I ask on what bases you decide some answer is better than other? $\endgroup$
    – nonuser
    Mar 20, 2019 at 14:51
  • $\begingroup$ So you are not going to answer me? $\endgroup$
    – nonuser
    Mar 20, 2019 at 15:51

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