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We know from calculus the product rule for differentiation $$(f(x)g(x))' = f'(x)g(x) + f(x)g'(x)$$

A beginner who has not yet learned this may try doing something like this:

$$(f(x)g(x))' = f'(x)g'(x)$$

Now to my question, can we determine for which class of functions this will "accidentally" work? In other words what can we demand of $f,g$ for this to hold?

mathreadler
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    See this https://math.stackexchange.com/questions/10927/when-do-the-freshmans-dream-product-and-quotient-rules-for-differentiation-hold?rq=1 – Zain Patel Jun 10 '17 at 10:42
  • @ZainPatel great, thanks, but that question demands to hold for the quotient rule too, I don't think they are equivalent. – mathreadler Jun 10 '17 at 10:44
  • yes - just posting it here because it's related. The first answer deals with it holding for only the product case at the end. – Zain Patel Jun 10 '17 at 10:49
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    When $f,g \ne 0$, the condition is equivalent to $(f'/f - 1)(g'/g - 1) = 1$. Since $g'/g = [\log |g|]'$, this shows that once a function $f$ has been arbitrarily chosen, there will generally be a corresponding function $g$, determined up to a constant factor, at least on some interval. – user49640 Jun 10 '17 at 10:54
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    Cool, would you mind making an answer of it? – mathreadler Jun 10 '17 at 11:05

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