Why the Product Rule
Many students learning calculus will struggle with the product rule. It violates an instinct that many other math topics have trained the student to expect. Let’s start with an example. Take the function x and the function sin(x). We already know (at this point in a calculus class) that the derivative of x is 1, and that the derivative of sin(x) is cos(x).
The assumption that most students make is that therefore, if you’re going to find the derivative of x sin(x) then it must be \(1\cdot \cos(x)\). Put as a general formula, suppose you have functions f and g. The belief that new students have is that \((fg)’ = f’g’\).
Here is an easy proof that this is wrong. Consider the example of the function \(x(x-1)\) and look at its graph. The main thing to notice from its graph is that it goes down and then up. Since the derivative is the change in the function, then anywhere that the function is going down, the derivative should be negative. Anywhere that the function is going up, the derivative should be positive.
Now let’s compute \(f’g’\) and see if it shows us a function that goes from negative (where the original function is decreasing) to positive (where the original is increasing). We know that \(x’=1\) and also \((x-1)’=1\). Therefore if we followed this calculation we would think the derivative of \(x(x-1)\) would be 1. But that does not go from negative to positive, it just stays positive—so it can’t be the derivative.
Now let’s compute the correct product rule. The correct product rule tells us that
$$ (fg)’=f’g + fg’ $$
If we apply this to the example \(x(x-1)\) we would find that the derivative is
$$ 1(x-1)+x(1) = 2x-1 $$
Notice that this does go from negative to positive. In fact it is negative where \(x<1/2\), and this is exactly where the original function \(x(x-1)\) is decreasing. This is also positive where \(x>1/2\) which is exactly where the original function is increasing. Therefore the rule \((fg)’=f’g+fg’\) seems to get the right answer.
Note that this is not a proof that the product rule is \((fg)’=f’g+fg’\). I’ll give the proof below, even though that proof is not really the point of this post. The point of this post was just to demonstrate to a very beginning student that the product rule works, whereas the naive calculation \(f’g’\) does not.
Theorem: For differentiable real valued functions of a real variable, f and g
$$ (fg)’ = f’g+fg’ $$
Proof: The definition of the derivative is, recall,
$$ f’(x) = \lim_{h\to 0} \frac{f(x+h)-f(x)}{h} $$
That is of course stated for a single function, so we need to apply this instead to the function fg. Therefore we are trying to investigate
$$ (fg)’(x) = \lim_{h\to 0}\frac{(fg)(x+h) - (fg)(x)}{h} $$
Notice that \((fg)(x) = f(x)g(x)\). This is just boring notation and is not an interesting statement. But it means that we can now do a famous real analysis technique (in fact it comes up in non-analysis settings a lot too, so that it’s more like a general technique) of what I call “wiggling a term in”. By that I mean: Add it in and subtract it back out. This does not change the quantity, which is a good thing, because it means that doing this is valid.
You might wonder, if it doesn’t change anything, how could it possibly be useful? But you move too fast! I didn’t say it doesn’t change anything! I said it doesn’t change the quantity. It does change how the quantity is represented, and THAT is what we use! In the steps below I “wiggle in” the quantity \(f(x)g(x+h)\).
$$ \begin{aligned}(fg)’(x) &= \lim_{h\to 0}\frac{f(x+h)g(x+h)-f(x)g(x+h)+f(x)g(x+h)-f(x)g(x)}{h} \\ &= \lim_{h\to 0}\frac{f(x+h)g(x+h)-f(x)g(x+h)}{h} + \lim_{h\to 0}\frac{f(x)g(x+h)-f(x)g(x)}{h} \\ &= \left(\lim_{h\to 0}g(x+h)\right)\left(\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}\right)+\left(\lim_{h\to 0}f(x)\right)\left(\lim_{h\to 0}\frac{g(x+h)-g(x)}{h} \right) \\ &= f’(x)g(x)+f(x)g’(x) \end{aligned}$$
where the final equality is due to continuity of any differentiable function, as well as the definition of the derivative.