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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.

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