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In differential calculus, the product rule is both simple in form and high in utility.  As such, it is typically presented early on in calculus courses — soon after the linearity of the derivative, in fact.  Moreover, the product rule is easy to derive from first principles:

Theorem (Product Rule): Let $f$ and $g$ be differentiable on the open set $U$. Then $fg$ is differentiable on $U$, and we have

$(fg)'(x)=f(x)g'(x)+g(x)f'(x)$  for all  $x \in U$.

Proof: For $x \in U$, we have (by definition of the derivative)

$\displaystyle(fg)'(x) = \lim_{\epsilon \to 0} \frac{f(x+\epsilon)g(x+\epsilon)-f(x)g(x)}{\epsilon}$

$\displaystyle = \lim_{\epsilon \to 0} \frac{\left(f(x+\epsilon)-f(x)\right)g(x+\epsilon)+f(x)\left(g(x+\epsilon)-g(x)\right)}{\epsilon}$

$\displaystyle = \lim_{\epsilon \to 0} \frac{g(x\!+\!\epsilon)\left(f(x\!+\!\epsilon)-f(x)\right)}{\epsilon}\!+\!\lim_{\epsilon \to 0} \frac{f(x)(g(x\!+\!\epsilon)-g(x))}{\epsilon},$

under the assumption that each of these last two limits exists.  This of course holds, as these limits are $g(x)f'(x)$ and $f(x)g'(x)$, respectively.   $\square$

All in all, then, the product rule is easy to prove and easy to use.  But — and this is of utmost pedagogical importance — is the product rule intuitive? By this proof alone, I would argue not; the manipulation of the numerator is weakly-motivated and our result falls out without reference to more general phenomena.

In this post, we’ll explore the merits of a second proof of the product rule, one that I hope presents a motivated and compelling argument as to why the product rule should look the way it does.