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

Question

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.

Why does it need to be added and subtracted?

Answer 1

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 $\displaystyle \lim_{h\to 0} f'(x)g(x)+f(x+h)g'(x)$.

Answer 2

The question "what does $f(x+h)g(x)$ mean" is easy to answer: It means the product of the numbers $f(x+h)$ and $g(x)$.

Now why are they added and subtracted? We can split that into two parts: Why is it added? And why is it then subtracted again?

The second part is easy to explain: We want an equality, so if we add something (other than $0$), we obviously have to subtract it again, or we'd not arrive at the same value as before.

But then, why are we adding it in the first place?

Well, we want to express the derivative of $f(x)g(x)$ by the derivatives of $f(x)$ and $g(x)$. Now the derivative of $f(x)$ is the limit of $(f(x+h)-f(x))/h$ and likewise for $g$. But we don't see those terms in the expression. However, we already see parts of those expressions. Let's transform the expression slightly top more clearly see that: $$\frac{F(x+h)-F(x)}{h}=\frac{f(x+h)g(x+h)-f(x)g(x)}{h} = \frac{f(x+h)}{h}g(x+h) - f(x)\frac{g(x)}{h}$$ As you can see, there's the expression $f(x+h)/h$ which looks almost like the expression we have to take the limit of to get $f'(x)$, but there's something missing: The $-f(x)$ part. Well, if it is missing, let's just add it:

$$\frac{f(x+h)\color{red}{-f(x)}}{h}g(x+h)-f(x)\frac{g(x)}{h}$$

Now we have a nice difference quotient on the left side. But we changed the expression! Our expression is no longer the same we started with, and certainly doesn't equal $(F(x+h)-F(x))/h$. How to fix it? Well, simply add again what we just subtracted:

$$\frac{F(x+h)-F(x)}{h}=\frac{f(x+h)-f(x)}{h}g(x+h) \color{red}{+ \frac{f(x)}{h}g(x+h)} -f(x)\frac{g(x)}{h}$$ Now we are back at the original value, but still have the nice quotient. Moreover, we see that on the last two terms, we can move $f(x)$ out, to get $$\frac{F(x+h)-F(x)}{h}=\frac{f(x+h)-f(x)}{h}g(x+h) + f(x)\frac{g(x+h)-g(x)}{h}$$ Now, that's even better! Now we also have the quotient in the right form for $g$! All we now have to do is to take the limit for $h\to 0$, and find $$F'(x)=f'(x)g(x)+f(x)g'(x)$$

Answer 3

If you want a proof that follows the picture more closely, you could start with \begin{equation} f(x+h)g(x+h) = f(x)g(x) + (f(x+h)-f(x))g(x) + (g(x+h)-g(x))f(x) + (f(x+h)-f(x))(g(x+h)-g(x)). \end{equation}

Now bring $f(x)g(x)$ to the other side, divide by $h$, and take the limit as $h\to 0$.

Answer 4

The point of adding and subtracting $f(x+h)g(x)$ is to get two terms that we can use the distributive law with. We have $$f(x+h)g(x+h)=f(x+h)g(x+h)-f(x+h)g(x)+f(x+h)g(x)-f(x)g(x)\\ =f(x+h)[g(x+h)-g(x)]+g(x)[f(x+h)-f(x)]$$ The two terms in square brackets are ones in the definition of $g'(x)$ and $f'(x)$. In your figure, the first term in the last line is Area B and the second term in the last line is Area A. We then argue that Area C is small enough we can ignore it as it has a factor of $h$ in each dimension.

Answer 5

I'm assuming you have the rest of the proof in front of you, but the point of adding and subtracting is so that you only have one $+h$ difference to deal with, rather than two. The reason why we only want one $+h$ is because it becomes simply the definition of the derivative for one function, rather than two.

For example, the first two terms of the numerator only has the $+h$ difference in $g$: $$f(x+h)g(x+h)-f(x+h)g(x)=f(x+h)[g(x+h)-g(x)]$$ and the $g(x+h)-g(x)$ on the right side looks very familiar: it's part of the definition of the derivative of $g$.

In terms of the picture, the goal is to find another way to express the sum of the areas of $A$, $B$, and $C$. You could express it as the area of the whole rectangle minus the largest rectangle (i.e., $f(x+h)g(x+h)-f(x)g(x)$),

or you could express it as the area of $B$ and $C$ (i.e., $f(x+h)g(x+h) - f(x+h)g(x)$) added with the area of $A$ (i.e., $f(x+h)g(x)-f(x)g(x)$). The reason for doing this, though, is what I stated above: to make the proof work and to get the desired result.