Derivatives of products are tricky

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Two young mathematicians discuss derivatives of products and products of derivatives.

Check out this dialogue between two calculus students (based on a true story):

Devyn: Hey Riley, remember the sum rule for derivatives?
Riley: You know I do.
Devyn: What do you think that the “product rule” will be?
Riley: Let’s give this a spin: $\frac {d}{dx} \left (f(x)\cdot g(x)\right ) = f'(x) \cdot g'(x)?$
Devyn: Hmmm, let’s give this theory an acid test. Let’s try $f(x) = x^2+1\qquad \text {and}\qquad g(x) = x^3-3x$ Now $\begin{align*} f'(x)g'(x) &= (2x)(3x^2-3)\\ &= 6x^3-6x. \end{align*}$
Riley: On the other hand, $\begin{align*} f(x)g(x) &= (x^2+1)(x^3-3x)\\ &=x^5-3x^3+x^3-3x\\ &=x^5-2x^3-3x. \end{align*}$
Devyn: And so, $\frac {d}{dx} \left (f(x) \cdot g(x)\right ) = 5x^4-6x^2-3.$
Riley: Wow. Hmmm. It looks like our guess was incorrect.
Devyn: I’ve got a feeling that the so-called “product rule” might be a bit tricky.

Above, our intrepid young mathematicians guess that the “product rule” might be: $\frac {d}{dx} \left (f(x)\cdot g(x)\right ) = f'(x) \cdot g'(x)?$ Does this ever hold true?

Answers will vary. A partial answer is that this will hold when either $f(x)$ or $g(x)$ are zero, or when both are constants.