$$\frac{d}{dx}(fg) = \text{?}$$
$$\frac{d}{dx}(fg) \color{red}\neq \frac{df}{dx} \cdot \frac{dg}{dx} \tag{units don't work}$$
$$\frac{d}{dx}(fg) = \quad f'(x)g(x) \quad \text{or} \quad g'(x)f(x) \tag{units consistent}$$
$$\frac{d}{dx}(fg) = c(f'(x)g(x)+g'(x)f(x)) \tag{$f ↔ g$ symmetry} $$
Now I don't know how to procede. I know that $c$ is adimensional and because of that maybe I could prove that it doesn't depend on $f$ and $g$ so it's sufficient to solve for $c$ by choosing arbitrarily $f$ and $g$ (like $g = 1$).