AP Calculus BC Topic 2.9: The Quotient Rule

Topic 2.9 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

774 words · 2 parts

Thinner or denser?

There’s a problem show designers really worry about. As the screen grows, the drones have to spread out to cover it and the picture gets grainier, so the barge keeps launching extra drones, forty-six every second.

We can measure how tightly packed the picture is by dividing its number of drones by its area, and we’ll call that density D. Right now, twelve hundred over twenty-four hundred gives half a drone for every square meter.

More drones arrive every second, but the area keeps growing too. Right now, which way do you think the picture’s density is heading?

To stay just as dense, every new square meter needs half a drone. The screen gains a hundred forty square meters every second, so it needs seventy new drones a second just to keep up. Only forty-six arrive, so the show falls twenty-four drones short every second. Spread that shortfall over twenty-four hundred square meters, and the density drops by a hundredth of a drone per square meter, every second.

That little argument is the quotient rule in disguise, and it’s the tool our lighthouse has been waiting for. If D is N over A, then N is D times A, which is a product, and we know how to handle those now. The product rule says N prime equals D prime A, plus D A prime.

Solve for D prime by subtracting D A prime and dividing by A. That D A prime is the seventy drones a second the screen needs just to keep up, and it’s exactly where the minus sign comes from.

Put N over A back in for D, so the top is N prime minus N over A times A prime. Then multiply the top and the bottom by A to clear that little fraction, which gives N prime A minus N A prime, all over A squared. With any two differentiable functions, that’s the quotient rule, u prime v minus u v prime, over v squared, wherever v isn’t zero.

A lot of teachers sing it as low d-high minus high d-low, over the square of what’s below. High means the top of our fraction, low means the bottom, and d-high means we differentiate the top.

If you ever forget which way the subtraction goes, test it on one over x, which you know goes down as x grows. The rule gives negative one over x squared, which is negative, so that order is right.

With the show’s numbers, that’s forty-six times twenty-four hundred minus twelve hundred times a hundred forty, over twenty-four hundred squared. That works out to negative one hundredth of a drone per square meter per second, the same answer we reasoned out.

An AP question might ask whether the density is increasing or decreasing at this instant, and ask you to justify your answer. A full-credit answer connects the derivative’s value and sign to what the density is doing, in one complete sentence. Because D prime is negative one hundredth, which is less than zero, the density is decreasing at this instant.

Now you take the barge’s controls. Set the launch rate so that, right at this instant, the picture is getting neither denser nor thinner.

Quotients on the exam

Let’s do a symbolic one, e to the x over x. The top is e to the x, which is its own derivative, and the bottom is x, whose derivative is one. Low d-high is x times e to the x, minus high d-low, e to the x times one, all over x squared. Pulling out e to the x leaves e to the x times the quantity x minus one, over x squared.

We can sanity-check it on the graph, because that derivative is zero when x is one, and that’s exactly where the graph of e to the x over x bottoms out and turns around. Now one in exam style, and this time pay close attention to which function is on top. Let k of x be g of x over f of x. Use the table to find k prime of one.

We’re squaring negative two, so the bottom is positive four. Keep those parentheses around the negative two, because leaving them out costs people points on the exam. Not every quotient needs the quotient rule, though. For x cubed plus five x, all over x, the bottom divides evenly into the top, so simplify first, which leaves x squared plus five, and its derivative is two x, as long as x isn’t zero. The quotient rule would get you there too, just the long way around, so it pays to look before you differentiate.

Topic 2.9: The Quotient Rule

Topic 2.9 key terms

Transcript

Thinner or denser?

Quotients on the exam

Unit 2: Differentiation: Definition and Fundamental Properties