AP Calculus BC Topic 2.8: The Product Rule

Topic 2.8 key terms

product rule: For differentiable $f$ and $g$ : $\frac{d}{dx}[f(x)g(x)] = f'(x)g(x) + f(x)g'(x)$ . Each factor takes a turn changing while the other holds still.
quotient rule: For differentiable $u$ and $v$ , wherever $v \ne 0$ : $\left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2}$ , the product rule run backward on $u = \frac{u}{v}\cdot v$ .
derivative of tan x: $\frac{d}{dx}\tan x = \sec^2 x$ , from $\tan x = \frac{\sin x}{\cos x}$ and $\sin^2 x + \cos^2 x = 1$ (radians).
derivative of sec x: $\frac{d}{dx}\sec x = \sec x\tan x$ , from $\sec x = \frac{1}{\cos x}$ .
derivative of cot x: $\frac{d}{dx}\cot x = -\csc^2 x$ , from $\cot x = \frac{\cos x}{\sin x}$ .
derivative of csc x: $\frac{d}{dx}\csc x = -\csc x\cot x$ , from $\csc x = \frac{1}{\sin x}$ .

Transcript

1,294 words · 5 parts

One kilometer offshore

This lighthouse stands on a rock exactly one kilometer off a long, straight beach, and about once every six seconds its beam swings right across the sand. Tonight the whole town is out on the promenade for its festival, and every few seconds the beam paints a bright spot that runs right along the sand in front of them.

I’ve slowed it down ten times so we can follow the spot of light where the beam lands. Right in front of the lighthouse it’s at its slowest, and even then it’s covering a whole kilometer every second, faster than a rifle bullet. Then, the farther down the beach it gets, the faster it runs, and it never stops speeding up, not for a single moment.

On a long enough beach it would leave the space station far behind, and if you swing a beam like this across the Moon, believe it or not, the spot outruns light itself. By the end of this lesson, you’ll be able to work out exactly how fast that spot is moving at any angle, and you’ll see how a spot of light can outrun light without breaking a single law of physics.

Two guesses that fail

Let’s look at the whole thing from above, where the beam has turned an angle theta away from straight ahead. Tan theta is the far side over the near side, x over a thousand, so the spot lands a thousand times tan theta meters down the beach.

This lamp turns exactly one radian every second, so the angle in radians is just the time in seconds. That makes the spot’s speed the derivative of a thousand tan theta, and nobody has given us that derivative yet.

Tan theta is sine theta over cosine theta, and we already know both of their derivatives, because the derivative of sine is cosine and the derivative of cosine is negative sine. The tempting move is to just divide those two derivatives, which gives cosine over negative sine, or negative cotangent theta. Straight ahead, at theta equals zero, that’s negative one over zero, which isn’t even a number. Just past zero it turns huge and negative, as if the spot were racing backward right where we just watched it move slowest. That can’t be right.

Multiplying derivatives fails for products just as badly. Take x times x, which is x squared with derivative two x, while multiplying the two derivatives gives one times one, which is just one. We need real rules for products and quotients, and products come first, because a quotient turns out to be a product in disguise.

Two sides growing

There’s a product hanging right over the bay tonight, with our lighthouse still sweeping away out past it. The town’s drone show has about twelve hundred drones holding a glowing screen in the sky, sixty meters wide and forty meters tall. Now the show starts stretching the screen, so its width grows by two meters every second and its height grows by one meter every second.

The area is the width times the height, two thousand four hundred square meters, and what I want to know is how fast that area is growing right now. Take a guess before we work it out. Right at this moment, how fast is the screen’s area growing?

Let’s freeze the screen and look at five seconds of growth. The width grows ten meters, which adds a tall strip down the right side, ten by forty, four hundred square meters. The height grows five meters, which adds a long strip across the top, sixty by five, three hundred square meters. That leaves one little piece up in the corner, ten by five, where the two growths overlap. Added up, that’s seven hundred fifty square meters in five seconds, a hundred fifty every second.

Now let’s shrink the time step. Over one second the average is a hundred forty-two, over a tenth of a second it’s a hundred forty point two, and over a hundredth it’s a hundred forty point oh two. Look at that corner. Divided by the step, the two strips always give eighty and sixty, but the corner is small in both directions at once. Its share shrinks from ten to two to point two, and melts away to nothing.

What’s left is just the two strips, each rate times the other side, two times forty plus sixty times one. That’s exactly a hundred forty square meters every second. Remember the guess of just multiplying the rates, two times one? Over a one-second step, that two is exactly the corner, two meters by one meter, the one piece that disappears.

The product rule

Let’s make this work for any two differentiable functions. Call the width u and the height v, and let both depend on x. When x changes a little, we’ll call the change in u delta u, and the change in v delta v. The new area comes in the same three pieces as on the screen, delta u times v down the side, u times delta v across the top, and delta u times delta v in the corner.

To turn changes into rates, we divide every piece by delta x. Now let delta x shrink to zero. Delta u over delta x becomes u prime, and delta v over delta x becomes v prime. The corner still has a bare delta v in it, and since a differentiable v is continuous, that delta v goes to zero and takes the whole corner with it.

That’s the product rule, which says the derivative of u times v is u prime times v, plus u times v prime. I remember it as each factor taking a turn, one changing while the other holds still, and on the exam you’ll usually see it written with f and g, like this.

Let’s test it on one where we already know the answer. X squared times x cubed is x to the fifth, so its derivative is five x to the fourth. Which two factors fill the blanks so the product rule agrees?

Here’s one you can’t multiply out first, x squared times sine x. Let u be x squared and v be sine x, so u prime is two x and v prime is cosine x. The product rule says u prime v plus u v prime, so the derivative is two x sine x, plus x squared cosine x.

If you just multiplied the derivatives, you would get two x cosine x, which is the corner all over again, and it’s wrong. On the AP exam you often won’t get formulas at all, just a table of values like this one. Let h of x be f of x times g of x. Use the table to find h prime of two.

Hold it steady

Back over the bay, the show wants to squeeze the screen into a wide panorama, so the width keeps growing two meters every second while the height starts to shrink. The width’s strip still adds two times forty, eighty square meters a second. The shrinking height takes a strip sixty meters long away across the top, so it all depends on how fast the height falls.

You control how fast the height shrinks, so set it to cancel the area we gain from the growing width. At this instant, we want the screen’s area to hold steady.

Notice I said right at this instant. If the height kept shrinking at that same rate, the screen would keep getting wider and shorter, and a moment later the area would start to fall.

So now you drive. The width keeps growing all the way out to eighty meters, and you steer the height’s rate so the area stays at twenty-four hundred the whole way.

Topic 2.8: The Product Rule