# Need rule for which functions satisfy this equation

Computationally, it has dawned on me recently that a lot of polynomial type functions (not trig functions, etc.) satisfy the following:

$f'-g-xg'=0$

where

$f=f(x)$

$g=\frac{f}{x}$

I'd appreciate it if someone could help me come up with a rule for which functions will satisfy this and which ones won't.

Thanks

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It's not clear what you are asking for. What does "rule for which functions will satisfy this" mean? Are you looking for sufficient/necessary conditions? Or something else? –  Srivatsan Oct 30 '11 at 17:57
If $f(x)=x g(x)$ then $f'(x)=g(x)+x g'(x)$. It's called the product rule. –  Christian Blatter Oct 31 '11 at 9:57

If $g(x) = f(x)/x$ then $g'(x) = f'(x)/x - f(x)/x^2 = (f'(x) - g(x))/x$.

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If $g(x) = \frac{f}{x}$, then $$g'(x) = \frac{xf' - f}{x^2},$$ so $$xg' = x\left(\frac{xf'-f}{x^2}\right) = \frac{xf'-f}{x} = f' - \frac{f}{x},$$ so naturally, if $f$ is differentiable, you have $$f' - \frac{f}{x} - xg' = f' - \frac{f}{x} - \left(f' - \frac{f}{x}\right) = 0.$$
@ben: I don't think Robert will get a ping from a comment here, since this question was not written by him and he has not commented here. The @ functionality is somewhat limited. You should post your question as a comment to his answer. Speaking for myself, I would say that it is as "mathematically trivial" as the relation that says that $2\times(3+4) = (2\times 3)+(2\times 4)$; it is a simple consequence of the properties of the derivative. –  Arturo Magidin Nov 1 '11 at 3:05