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In general, $\frac{d}{dx}(f(x) \cdot g(x)) \neq \frac{d}{dx}f(x) \cdot \frac{d}{dx}g(x)$

When does this result hold true? My first try is to use product rule on left side and compare the two sides, but this hasn't helped at all. Any suggestions?

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If $f(x)=g(x)=e^{2x}$, this is true. – Your Ad Here Feb 20 '14 at 20:57
is that an exhaustive list? – jj77646_piyy Feb 20 '14 at 20:58
See the answers here for other examples. – David Mitra Feb 20 '14 at 21:00

well assuming one of the two functions, e.g. $g$ is given, then finding $f$ is just solving a homogeneous linear equation.

$$f'(x)g(x) + f(x)g'(x) = f'(x)g'(x)$$ which is the same as $$f'(x)(g(x)-g'(x)) + f(x)g'(x) = 0$$ given $g$ then you can solve it by $$f(x) = f(x_0)\exp \left\{ -\int_{x_0}^x \frac{g'(y)}{g(y)-g'(y)}dy\right\}.$$

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If you accept your own answer, you obtain the 'self-learner' badge. – Cloudscape Sep 20 '15 at 16:29
It's not my own answer. Apparently this person didn't accept any answer... which is a problem, people ask and don't read the answer :) – Martingalo Sep 20 '15 at 16:39

One can try the exponential of the form $f(x) = e^{ax}$ and $g(x)= e^{bx}$, then

$(fg)' = f'g+fg' = e^{ax}e^{bx}(a+b)=^! f'g'= ae^{ax}be^{bx}\Leftrightarrow a+b = ab\Leftrightarrow a,b = 0 \lor a,b = 2. $

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