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Let X be the set of differentiable functions in real line. Find an

infinity set $H$ in X such that

$$ f,g\in H ~~\rightarrow (fg)'=f+g $$

1 Answers 1

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We can find these functions by noting that the condition has to be fulfilled for $f=g=h$, which leads to a simple differential equation:

$$ \begin{align} (hh)'&=h+h\;, \\ 2hh'&=2h\;. \end{align} $$

Unless $h=0$, this implies $h'=1$, with solutions $h=x+c$, and indeed

$$ ((x+a)(x+b))'=(x+a)+(x+b) $$

for all $a,b\in\mathbb R$.

With hindsight, another way to find this solution (but without showing that it's unique) would have been to look at the product rule, $(fg)'=f'g+fg'$, and note that it yields $f+g$ if $f'=g'=1$.

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    is there distinct functions f and g such that $(fg)'=f+g$ and f or g is not of the form $x+a$ ? http://math.stackexchange.com/users/6622/joriki2012-12-16
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    @joriki is there distinct functions f and g such that $(fg)^′=f+g$ and $f$ or $g$ is not of the form $x+a$ ?2012-12-16