Is it possible to turn this 'proof' of the product rule into a rigorous argument?

Question

I have often found linear approximation to be useful in understanding the main theorems of calculus. I tried using it to 'prove' the product rule, as I find the typical proof for it to be unintuitive. However, I'm not sure that the substitution I made can be properly justified: $$ (f \cdot g)'(a) = \lim_{h \to 0} \frac{f(a+h)g(a+h)-f(a)g(a)}{h} $$ Here is where I use my questionable substitution: replace $f(a+h)$ with $f(a)+f'(a)h$; make a similiar substitution for $g(a+h)$. As $h$ approaches $0$, the linear approximation becomes better and better. $(f \cdot g)'(a)$ becomes \begin{align} \lim_{h \to 0} \frac{\bigl(f(a)+f'(a)h\bigr)\bigl(g(a)+g'(a)h\bigr)-f(a)g(a)}{h} &= \lim_{h \to 0} \frac{f(a)g'(a)h+g(a)f'(a)h+f'(a)g'(x)h^2}{h} \\ &= \lim_{h \to 0} f(a)g'(a)+g(a)f'(a)+f'(a)g'(x)h \\ &= f(a)g'(a)+g(a)f'(a) \end{align}

There were many things about my 'proof' that pleased me. For instance, it lines up very closely with the common visual explanation of the product rule:

(This image is taken from 3Blue1Brown's video on visualising the chain and product rule. Check it out.)

However, I'm still unsure about my substitution. I've heard people use similar arguments to this, e.g. $$ \lim_{x \to 0}\frac{\sin x + \tan x}{\sin x}=\lim_{x \to 0}\frac{x+x}{x}=2 $$ because $\sin$ and $\tan$ are 'locally linear', but I am yet to see a formal justification for this kind of substitution.

Answer 1

Short answer: big-& little-O notation.

Your strategy can work by stating that, for small nonzero $h$, $f(a+h)\in f(a)+hf^\prime(a)+o(h)$ etc. What's more, the $h$ coefficient is unique; this can be taken as a definition of the derivative, equivalent to the usual one. Since the product of two $O(h)$ terms is $O(h^2)$, it's $o(h)$. So$$\begin{align}f(a+h)g(a+h)&\in(f(a)+hf^\prime(a)+o(h))(g(a)+hg^\prime(a)+o(h))\\&\subseteq f(a)g(a)+h[f(a)g^\prime(a)+f^\prime(a)g(a)]+o(h).\end{align}$$Then we just read off the $h$ coefficient.

Answer 2

Assuming that you can replace $f(a + h)$ by $f(a) + f'(a)h$, and $g(a + h)$ by $g(a) + g'(a)h$, is equivalent to assuming that the limit $$\lim_{h \to 0} \frac{f(a+h)g(a+h) - [f(a) + f'(a)h][g(a) + g'(a)h]}{h} = 0,$$ which ends up requiring some messy algebra to rigorously justify--messy algebra which, when all is said and FOILed, is substantially equivalent to the traditional verification of the Product Rule (see, for instance, Paul's Online Notes).

That said, I could see this argument working really well as an informal justification, a "heuristic" of what we should expect the Product Rule to be. Especially if your audience doesn't need 100% rigor, or is already familiar with the Product Rule, this is definitely an attractive and intuition-building way of presenting it.

Answer 3

I think I have found an alternative to my approach. Instead of using the fact that $f(x+h)-f(a)\sim f(a)+f'(a)h$, I will instead write $\Delta f$ for $f(a+h)-f(a)$ and $\Delta g$ for $g(a+h)-g(a)$. It follows that \begin{align} (f \cdot g)'(a) &= \lim_{h \to 0} \frac{f(a+h)g(a+h)-f(a)g(a)}{h} \tag{1} \\ &= \lim_{h \to 0} \frac{\bigl(f(a)+\Delta f\bigr)\bigl(g(a)+\Delta g\bigr)-f(a)g(a)}{h} \tag{2} \\ &= \lim_{h \to 0} \frac{f(a)\Delta g+ g(a)\Delta f +\Delta f \Delta g}{h} \tag{3} \\ &= f(a)\lim_{h \to 0} \frac{\Delta g}{h} + g(a)\lim_{h \to 0} \frac{\Delta f}{h}+\lim_{h\to 0}\frac{\Delta f \Delta g}{h} \tag{4} \\ &=f'(a)g(a)+f(a)g'(a) \tag{5} \end{align} This proof has the advantage of being more intuitive without the disadvantage of worrying about whether or not a certain substitution is rigorously justified.

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$(1)$ uses the definition of the derivative.
$(2)$ rewrites this in terms of $\Delta f$ and $\Delta g$: since $\Delta f=f(a+h)-f(a)$, we know that $f(a+h)=\Delta f+f(a)$.
$(3)$ expands the brackets. Note that the $f(a)(g(a)$ term is cancelled out.
$(4)$ uses the addition limit law to tidy up.
$(5)$ uses the fact that $$\lim_{h\to 0}\frac{\Delta f}{h}=\lim_{h\to 0}\frac{f(a+h)-f(a)}{h}=f'(a)$$ and that \begin{align} \lim_{h\to 0}\frac{\Delta f \Delta g}{h}&=\lim_{h\to 0}\frac{\bigl(f(a+h)-f(a)\bigr)\bigl(g(a+h)-g(a)\bigr)}{h} \\ &=\lim_{h\to 0}\bigl(f(a+h)-f(a)\bigr)\lim_{h\to 0}\frac{g(a+h)-g(a)}{h} \\ &=0 \cdot g'(a) \\ &=0 \end{align} Note that since $f$ is continuous at $a$, we know that $$ \lim_{x \to a}f(x)=f(a) $$ Letting $x=a+h$, we see that $$ \lim_{h \to 0}f(a+h)=f(a) \implies \lim_{h \to 0}f(a+h)-f(a)=0 $$ This provides a rigorous justification for why $$ \lim_{h\to 0}\frac{\Delta f \Delta g}{h}=0 $$