[Math] Proof of the product rule. Trick. Add and subtract the same term.

calculusreal-analysis

While I was looking at the proof of the product rule, there was something that I don't quite understand.

Product Rule: $F'(x) = f'(x)g(x) + f(x)g'(x)$

The proof goes like,

$$\begin{align}F'(x)&= \lim_{h\to 0}\frac{F(x+h)-F(x)}{h}\\
&= \lim_{h\to 0}\frac{f(x+h)g(x+h)-f(x)g(x)}{h}\\
&= \lim_{h\to 0}\frac{f(x+h)g(x+h)-f(x+h)g(x)+f(x+h)g(x)-f(x)g(x)}{h}\end{align}$$

Here, we see $f(x+h)g(x)$ is being added and subtracted.

My question is, what does $f(x+h)g(x)$ mean?

And why does it have to be added and subtracted?

I tried to understand it with a little help from google and found it has to do with isolating $f(x+h)$ from $g(x+h)$ but i don't fully understand it.

I also came across what seems to be a useful pic to understand the rule
(pp 2 http://aleph0.clarku.edu/~djoyce/ma120/derivatives2.pdf)
but when it comes to adding and subtract $f(x+h)g(x)$, I don't see the point of doing it.

geometical interpretation

Why does it need to be added and subtracted?

Best Answer

This is the usual proof of this theorem, but I find it works just as well in reverse.

Start with $f'(x)g(x)+f(x)g'(x)$ and see what you get:

\begin{align} f'(x)g(x)+f(x)g'(x) &=\lim_{h\to 0} f'(x)g(x)+f(x+h)g'(x)\\ &= \lim_{h\to 0} \left( g(x)\frac{f(x+h)-f(x)}{h} + f(x+h)\frac{g(x+h)-g(x)}{h}\right)\\ &= \lim_{h\to 0}\frac{f(x+h)g(x+h)-f(x+h)g(x)+f(x+h)g(x)-f(x)g(x)}{h}\\ &= \lim_{h\to 0}\frac{f(x+h)g(x+h)-f(x)g(x)}{h}\\ &= \lim_{h\to 0}\frac{F(x+h)-F(x)}{h}\\ \end{align}

Here, we see that $f(x+h)g(x)$ is something that is cancelled as we routinely simplify the expression. On the other hand, there is a leap of intuition where, rather than adding and subtracting, we rewrite the initial expression as $\displaystyle \lim_{h\to 0} f'(x)g(x)+f(x+h)g'(x)$.

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