Topic 2.2: Defining the Derivative of a Function and Using Derivative Notation

2:49 a.m., one more time

It’s two forty-nine in the morning on March second, 2019, and we’re back on pad thirty-nine A in Florida, where a Falcon 9 is about to launch the very first Crew Dragon. Nine engines roar to life, and up it goes. For a little over two minutes those nine engines push it higher and faster. Then they all shut off at once, the first stage falls away, and the second stage lights up.

Now look at the altitude graph, because somewhere on that blue curve is the exact moment those engines shut off, and you can’t see it at all. By the end of this lesson, you’ll be able to find that moment yourself, using nothing but this one curve.

The trick is to ask how steep the curve is instead of how high. At one minute the slope is about two hundred eighty meters per second, the number you squeezed out last lesson, and we plot it as a height on a brand-new graph. A rocket is a lot to start with, though, so let’s learn the move on a curve where we can check every slope by hand.

Slope becomes height

Here’s the friendliest curve I know, f of x equals x squared, the same U you’ve been graphing since algebra. At x equals three the curve is at a height of nine, and the tangent line there climbs six for every one step to the right, so its slope is six. We keep the same input, three, and on a second graph down here we plot that slope, six, as a height. The slope of the top graph becomes the height of the bottom one.

Let’s move to x equals negative two, where the curve sits at a height of four. Where does the dot go on the slope graph?

At the very bottom of the U the tangent is perfectly flat, so its slope is zero, and that dot sits right on the axis. Those three dots, negative four, zero and six, are already lining up, and I bet you can guess what happens when we fill in every x in between. Let’s let the tangent ride the whole curve. Downhill on the left gives negative slopes, the bottom gives zero, and the right gets steeper and steeper, and the dots make a straight line.

This new graph is a function in its own right, because it takes any x and hands you the slope there. We call it the derivative of f, and we write it f prime. It sure looks like the line y equals two x, but a picture isn’t proof, so let’s find out for sure.

One limit, every slope

In the last lesson you found the slope at one point by squeezing a secant, two points h apart, with h shrinking to zero. The big move today is tiny on paper, because instead of plugging in three, we just leave x as x.

That makes the derivative of f a function, whose value at any x is this limit, as long as the limit exists. We’ll meet a spot where it doesn’t before we’re done. Inside the limit, x is where you’re standing on the curve, and it stays put. The little step h to the second point is the only thing that shrinks.

Let’s do it for x squared, so the top is x plus h, squared, minus x squared, all over h. Multiply out the square to get x squared plus two x h plus h squared, and the x squared at the front cancels the minus x squared at the end. If you plugged in h equals zero right now you’d get zero over zero, which tells you nothing. We don’t have to, because every term on top still has an h we can factor out.

Since the step h gets close to zero but never equals zero, we’re allowed to cancel the h’s, and what’s left is two x, plus the step. Now we shrink the step to nothing, and two x, plus the step, becomes just two x. Notice that x never moved, only the little step did, and f prime of x equals two x, exactly the line our tangent drew.

It matches every dot, and think about what just happened. In the last lesson one limit got you one slope, and this one limit just got you every slope on the curve at once. Let’s try a quick one. If g of x equals x squared plus five, the same U lifted five units up, what’s its derivative?

We just used one limit to find every slope on the curve. We can build the rocket’s climb rate exactly the same way, and in a couple of lessons you’ll learn shortcuts that do that algebra in seconds.

The machine that draws a derivative

Here’s a machine that draws a derivative for real. As the cart rides this tilted track, the paper slides out underneath it at half a meter every second, so the cart’s pen draws its position against time.

Give it one shove, and it climbs, slows down, stops for an instant two meters up, and rolls all the way back. Let’s swing around and look down on the paper, and there’s the graph, s of t, an upside-down U. The second channel is for velocity, and I’ve unhooked that pen.

At two seconds and at six seconds the cart is at exactly the same spot, one and a half meters up. How does its velocity at two seconds compare with its velocity at six?

The position graph knew that all along, in its slope. At two seconds the tangent climbs one meter in two seconds, at four it’s flat, and at six it drops one meter in two seconds. Okay, you’re the velocity pen now, so drag each violet dot up or down until its height matches the slope of the position graph right above it.

Let’s hook the pen back on and run it again. Look at that, the velocity pen goes straight through your three dots, a line from plus one meter per second all the way down to minus one. Everything lines up, too. Where the position graph peaks, the velocity crosses zero, and wherever the position is falling, the velocity graph sits below the axis.

Now let’s flip it around. You drive the cart with the velocity stick, and try to make the blue pen trace this tent-shaped position graph.

Let’s put that first ride on paper. The position is s of t equals t minus t squared over eight. The velocity line starts at one meter per second and drops a quarter every second, down to zero at four, so s prime of t equals one minus t over four. Run the definition on s, the way we did x squared, and you get exactly the same line.

The AP exam loves this kind of question, so let’s answer it the way a grader wants to see it. At t equals six, is the cart moving up the track or down, and how do you know?

Let’s put six into the velocity function. One minus six fourths gives negative one half meter per second. Then we say what it means. Because s prime of six is negative, the cart’s position is decreasing at six seconds, so it’s moving back down the track. It has the derivative’s value, what its sign means, and a conclusion about the actual cart, and that’s what earns the point.

Four ways to write it, and the tangent line

The AP exam writes the derivative in four different ways, so let’s line them all up on one function, y equals x squared. We already found f prime of x equals two x, and you’ll also see the same derivative written as y prime equals two x or d y d x equals two x.

Read this last one as d d x of x squared equals two x. It’s an instruction to take the derivative of x squared with respect to x, and the answer is two x. The d y d x form carries its units with it, change in y per change in x, so for our cart, d s d t comes out in meters per second.

Plug in a number and you get a single number, so f prime of three is six. In d y d x form you add a little bar at the bottom that says at x equals three.

Watch the order, though, because plugging in three first is a classic mistake. Since f of three is nine, you’d be taking the derivative of the number nine, and a constant like nine is a flat line with a slope of zero. That zero belongs to the flat line y equals nine, not to our parabola. For the parabola’s slope at three, take the derivative first, two x, and then plug in three to get six.

Let’s make sure that sticks. If y equals x squared, what is d y d x at x equals negative five? Now we can write the tangent line, because a line needs just two things, a point and a slope, and we know how to get both.

The point comes from f, since f of three is nine, and the slope comes from f prime, since f prime of three is six. Point-slope form puts them together, y minus nine equals six times the quantity x minus three. The trap here is writing two x as the slope. Two x is a different slope at every x, and a line needs one number, the slope right at the point, which is six.

Last lesson’s tangent didn’t cross its curve, and that’s common, but it isn’t a rule. This flat tangent on x cubed crosses right through the origin, and it’s still the tangent, because it matches the curve’s direction at that point. Okay, you write one. What’s the tangent line to y equals x squared at x equals negative one?

We do 1/x, you do √x

Let’s try a trickier curve, r of x equals one over x. Before any algebra, look at the part where x is positive. Where do you think its slope graph will sit, compared with the axis?

Let’s check that prediction with the definition. We take one over the quantity x plus h, subtract one over x, and divide the whole difference by h. Two fractions on top means a common denominator, x times the quantity x plus h, and the top becomes x minus the quantity x plus h. Before I simplify, what do you think x minus the quantity x plus h comes out to?

The minus sign hits both terms, so it’s x minus x minus h, which is negative h, and that negative is exactly the downhill we predicted. Dividing by h puts an h in the bottom, and the h’s cancel because h isn’t zero. As h shrinks to zero, the bottom, x times the quantity x plus h, becomes x squared, so we get negative one over x squared, for every x except zero.

The picture agrees, because the slope graph sits below the axis on both sides, even on the right, where the curve itself is up above. At x equals two the curve is at one half, with a slope of negative one over two squared, a negative quarter, and over at negative two, on the other branch, the slope is that same negative quarter.

Now it’s all yours. I’ve set up q of x equals the square root of x in the definition for you. The trick for square roots is to multiply the top and bottom by the conjugate, the same two roots joined by a plus, which makes a difference of squares, so the roots disappear from the top.

Work out q prime of x, and then tell me the slope of the square root curve at x equals four. The root curve agrees, because its tangent at four rises a quarter for every step. Down at x equals zero is the spot I promised where the limit fails.

This tangent near zero is steep, and the slopes grow without bound as x approaches zero from the right, so the square root has no finite derivative at zero. The next lesson is all about places like that. The AP exam likes to hide a derivative inside a limit like this one. What’s its value?

Find the moment

Let’s go back to that rocket, because now you have everything you need. The slopes of a curve make a graph of their own, and the highest point of that graph is where the curve is steepest.

Before you measure anything, sketch on the empty graph how you think the climb rate changes over the whole flight. Drag the tangent along the flight, watch its slopes pile up on the graph underneath, and stop at the moment the rocket is climbing fastest.

Now compare it with what you expected. If yours kept climbing all the way to the end, you’re in good company, because the altitude never stops rising. Its slopes do turn around, though, at a hundred fifty-eight seconds, and that’s no accident.

Now watch what the rocket does at exactly that moment. The engines shut off. That’s main engine cutoff, MECO for short, two minutes and thirty-eight seconds into the flight, and you found it with nothing but the altitude curve.

The rocket doesn’t stop climbing, because H prime, its climb rate, is still positive. From here on, though, it climbs a little slower every second, because without the first stage pushing it up, gravity wins the up-and-down part of the trip. You built a whole graph out of slopes, and its peak gave you one number, about fourteen hundred meters per second. On the board you found formulas like two x, and the fourth way to show a derivative is in words.

In words, you’d say that a hundred fifty-eight seconds after liftoff, the rocket’s altitude is increasing at about fourteen hundred meters per second. That sentence, with its units and its moment, is exactly what the AP exam wants. The engines shut off right where the slope graph peaks, and you found that from one curve. Next time, we’ll go hunting for places where a curve has no slope at all.