Determine all continuous $f : \mathbb R \rightarrow \mathbb R$ that satisfies $$f(xy) = xf(y) + yf(x)$$
I tried rewrite the equation as $f(xy) + f(x)f(y) + xy = (f(x) + x)(f(y) + y)$ and I know that $f(0) = f(1) = 0$. Thanks in advance!
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3$\begingroup$ The equation looks a lot like the product rule.......$f(x)$ could be something related to the derivative operator.......Maybe : $f(x)=\frac{dx}{dt}$ ($t$ be some dummy variable) $\endgroup$– Saket GurjarMay 30, 2020 at 10:25
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2$\begingroup$ See artofproblemsolving.com/community/c6h1667141p10595936. The idea is $g(x)=f(x)/x$ for $x\neq 0$ gives $g(xy)=g(x)+g(y)$. Then $h(x)=g(e^x)$ gives $h(x+y)=h(x)+h(y)$ (Cauchy's functional equation). Also $f(x) \equiv 0$ is an obvious solution. $\endgroup$– SilMay 30, 2020 at 10:30
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1 Answer
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Hint $\frac{f(xy)}{xy}=\frac{f(y)}{y}+\frac{f(x)}{x}$.
Set $h(y)=\frac{f(y)}{y}$.
Can you finish from here?
here is complete solution.
answered May 30, 2020 at 10:27