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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

2 Answers 2

up vote 5 down vote accepted

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 $$f' - \frac{f}{x} - xg' = f' - \frac{f}{x} - \left(f' - \frac{f}{x}\right) = 0.$$

What mystifies me is your assertion made in comments that "a lot, but not all" functions satisfy this equation. In fact, all functions that are differentiable satisfy the equation.

Which function did you believe does not satisfy it?

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My bad algebra often mystifies me too. –  ben Oct 31 '11 at 23:33
    
@Robert Israel So, is this relation considered mathematically trivial? –  ben Oct 31 '11 at 23:38
    
@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

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