AP Calculus AB — The Derivative: Tangent Lines and Instantaneous Rate of Change
Corrected lesson transcript


## 0:00 — The Derivative: Tangent Lines and Instantaneous Rate

[0:05] Okay, today we're going to learn something called the derivative.
[0:16] This is what calculus is all about.
[0:23] The rest of your calculus career is going to depend on the derivative.
[0:31] I'll define it for you.
[0:32] But let's start with this function here.

## 0:37 — Reviewing Secant-Line Slope

[0:40] A nice little quadratic function.
[0:44] Okay, if we wanted to find the
[0:49] slope of the
[0:52] secant line, let's say between
[0:55] these two points.
[1:01] How do I find the slope of the secant line?
[1:15] Let's say, what is it, between...
[1:32] So between x equals one and three, for example.
[1:35] So this can be f of three minus f of one over three minus one
[1:44] or y two minus y one over x two minus x one.
[1:53] Which gives us, what is that?
[1:56] Three squared minus two times three minus one
[2:00] minus
[2:17] and what is that?
[2:20] 9 minus 6 minus 1 minus 1 minus 2 minus 1 is that 0?
[2:33] Minus...
[2:34] Sorry, minus 1?
[2:38] Minus 2.
[2:39] Minus 2.
[2:46] Anyways, arithmetic and stuff.
[2:54] 2 is that right?
[2:55] Ok, so the slope of that line is 2, and what if we wanted to find the equation of that
[3:01] line, what would we do?
[3:06] Just use point slope formula, doesn't matter if you use this point or this order there.
[3:11] So the equation of the secant line would be y minus, let's say 3, equals 2, the slope times x minus...
[3:30] Was that three?
[3:33] Sorry.
[3:33] Two.
[3:35] This should be a two.
[3:37] Two.
[3:37] This should be a three.
[3:38] There we go.
[3:39] Have you got it?
[3:40] Yeah.
[3:41] Okay.
[3:44] Um, that's easy.
[3:46] Well, we did this in like eighth grade.
[3:50] Yeah.
[3:52] Find equations of secant lines.
[3:56] So, we started calculus with this, from the first day.

## 4:01 — The Tangent-Line Goal

[4:06] We wanted to know, not the slope of the secant line, but what?
[4:11] Slope of the tangent line.
[4:17] So our goal here is to find the slope of the tangent line at x equals 3.
[4:23] We don't have our pair. So first day of school or first day of our lesson we
[4:37] couldn't figure it out so how did we estimate it? We found secant lines like
[4:46] 3 minus f of 2.9 over 3 minus 2.9.
[4:58] OK.
[4:59] And that got an estimate.
[5:01] How do we get a better estimate?
[5:03] 2.9 by 3.
[5:04] And you choose a closer number, closer number.
[5:08] And then you want a better estimate,
[5:10] we choose a closer number, and so on.
[5:26] Let's call this...
[5:40] So this would be the slope of the secant line between two numbers between whatever
[5:51] For any chosen x, this gives the slope of the secant line between x and 3.
[5:58] So wherever X is this gives us the slope of the tangent line between X and 3

## 6:10 — Using a Limit to Find the Slope

[6:11] And now that we learned something called limits
[6:16] We want to know
[6:19] What this is approaching as X approaches 3
[6:25] And we now have this tool called limits
[6:29] as X approaches 3
[6:36] Now all we got to do is if we were able to find this limit
[6:40] this is the slope of the tangent line and that is what we call the derivative
[6:44] which is the slope of the tangent line so in fact let's do this limit that's
[6:52] As x approaches 3, evaluate the limit of the secant slope.
[7:13] one is that it okay if we plug in three what do we get we try direct substitution
[7:43] You get 0 over 0.
[7:47] Right, you plug it in here, you get 2, 2 minus 2 is 0.
[7:51] Okay, so we gotta solve this limit.
[7:54] How are we gonna solve this limit?
[7:58] No.
[8:02] Yeah, let's multiply this out.
[8:05] First off, simplify.
[8:18] plus one. So that's two plus one is three. So negative x squared plus two x plus three.
[8:34] is that right?
[8:41] Probably in 3 you still get 0 over 0. So what are we gonna do next?
[8:51] So very first trick besides direct substitution
[8:57] Factor okay, we got factor this
[9:00] it's usually hard to factor with well not hard but
[9:04] Let me factor out a negative first.
[9:30] And what's this factor 2x minus 3x plus 1?
[9:39] Now these simplify.
[9:53] Now we just use direct substitution.
[9:54] plus 1 is 4. So we took the limit of the slope of the secant line and we ended up
[10:10] with 4 which is the slope of the tangent line. Okay so all we got to do is
[10:21] limits of the slope of the secant line and this gives us the slope of the tangent
[10:28] line so the slope of that line is 4 the equation of the tangent line would be y

## 10:28 — Two Definitions of the Derivative

[10:36] minus 2 is 4 times x minus 3. There's actually two different definitions of
[10:49] the derivative the my favorite one let's say we have any function here okay if we
[11:04] want to know the slope of that tangent line x equals 1 we need two points let
[11:15] Let me call some number right over here x plus some very small number h.
[11:27] The secant slope is f of x plus h minus f of x, divided by x plus h minus x.
[11:41] h minus x. Okay, so that's the slope of the tangent line there. h is the distance
[11:56] The distance between the x-values is h, and we let h approach zero.
[12:03] we're going to do is we want this distance to go to zero. So we take the
[12:08] limit as h approaches 0. And this simplifies, right? h approaches 0. Over h. This is the
[12:36] main definition we're going to use I'll show you the other one a slight
[12:39] difference this is the slope of this tangent line this is the derivative
[12:46] The other definition is the limit for this definition number two.
[12:58] The main notation uses f prime to represent the derivative of f.
[12:59] main notation we're going to use we put this apostrophe up here that stands for
[13:10] the derivative of F we sometimes raise F prime of X or just F prime that means
[13:18] the first derivative or the derivative the slope of the tangent line here this
[13:23] This value is the derivative at x equals a.
[13:32] okay this one is the derivative typically
[13:36] at, it's the derivative, at x equals a. This is the derivative at x. Okay, and I'll show
[14:00] in a minute why it's better.

## 14:04 — Deriving a Polynomial Formula

[14:07] So let's do the same function again.
[14:12] What was it?
[14:13] x squared minus 2x?
[14:16] Minus 2x minus 1.
[14:17] Minus 1.
[14:18] Minus 1.
[14:21] This is the...
[14:29] The definition is the limit as h approaches zero of f of x plus h minus f of x, all over h.
[14:39] The reason this one I like better, well, for two reasons.
[14:44] One is when we try to simplify this, you end up having to factor out a binomial and factoring
[14:54] is hard.
[14:56] Okay here we're only going to have to factor out an H out of here.
[15:01] It's a lot easier when you factor out one term instead of a whole binomial.
[15:07] The other reason I'll show you in a minute.
[15:10] So let's go ahead and do that. Do this.
[15:14] The limit
[15:17] as h approaches zero.
[15:23] Here we plug in x plus
[15:25] h everywhere we see an x.
[15:55] x squared minus 2x minus 1 all over h.
[16:12] This is the slope of the secant line, and we let the distance h between the points approach zero.
[16:19] h, so the distance between those two ordered pairs goes to zero.
[16:24] Okay, now to solve this limit, we just have to multiply out a bunch of stuff and simplify,
[16:29] Eventually every numerator term has a factor of h, which cancels with the denominator.
[16:33] So let's go ahead and multiply that out.
[17:04] okay once you multiply it out a bunch of stuff is going to cancel out but not a bunch some stuff
[17:15] the x squared and the negative x squared is zero
[17:21] negative 2x and 2x is zero
[17:28] negative one positive one is zero
[17:35] Every term in the numerator has a factor of h, so factor out h.
[17:46] Let's do this.
[18:02] And now H divided by H is always 1.
[18:05] as long as we get...
[18:10] as h approaches zero
[18:14] After cancellation, the expression is 2x plus h minus 2.
[18:17] Now we use direct substitution
[18:20] Again, the only thing we plug in zero into
[18:23] is the h
[18:28] And we get 2x plus 0 minus 2 equals 2x minus 2.

## 18:44 — A Function of Tangent Slopes

[18:46] Okay, so
[18:47] But the derivative is the slope of the tangent line.
[18:52] Slopes are numbers.
[18:55] They're not functions.
[19:00] So how is this the slope of the tangent line?
[19:04] If it's not a number.
[19:14] Any thoughts on that?
[19:23] So this is not a slope because it's not a number, but I still call it the slope of the tangent
[19:32] Ok, yeah, this is very important.
[19:34] If we plug in a number for x, we get a slope, we get a number.
[19:41] So, for example, F prime of 3, we get 2 times 3 minus 2, which is 4.
[19:53] That is a slope.
[19:55] That is the slope of the tangent line x equals 3.
[19:58] This is what we did a few minutes ago.
[20:03] If I plug in 1, f prime of 1, we get 2 times 1 minus 2 is 0.
[20:12] And we get the slope of the tangent line.
[20:16] We get the derivative at x equals 1.
[20:19] If we plug in, let's say, zero, we get two times zero minus two is negative two.
[20:31] We get the slope of the tangent line right here.
[20:37] The definition is the limit as h approaches zero of f of x plus h minus f of x, all over h.
[20:46] 0 f of x plus h minus fx over x plus h, this formula.
[20:53] Okay, we end up getting another function of x,
[20:58] which simply tells us all we do is plug in a number
[21:01] for x to give us the slope of the tangent line.
[21:05] Okay, so this gives us the slope of the tangent line,
[21:09] depending on the value of x, which is nice.
[21:12] Now we can find the slope anywhere.
[21:14] Just plug in a different x value,
[21:15] we get a different slope of the tangent line. If we use if we were to use the
[21:23] other formula the limit is X approaches we usually put a number in here like 3
[21:31] f of X minus f of 3 over X minus 3 this will give us a specific number this
[21:41] gives us 4 it does not tell us the slope of the tangent line anywhere okay so
[21:47] that's the main reason I like the other definition better okay let's talk about

## 21:55 — Meaning and Notation

[21:55] It gives us the instantaneous rate of change.
[22:22] One notation is f prime of x.
[22:38] Another one is dy dx.
[22:49] If we're given like a function f of x equals whatever,
[22:56] x squared plus 3 or something, we usually use this notation.
[23:01] If we were just given like y equals, we usually use this notation.
[23:09] Okay, the other notations would be
[23:17] D dx of f
[23:22] That also means the derivative of our function f with respect to our variable x
[23:33] That the only other one
[23:40] The expression dy over dx is another derivative notation, not an ordinary fraction.
[23:47] our function y and tells us with wit with respect to what variable we're
[23:53] eventually going to have multi very multiple variables so it's gonna be
[23:57] important which variable we're taking the derivative with respect to but we
[24:03] all three of these notations questions so far at Leibniz which he used this
[24:22] notation and I believe Renee Descartes used this notation but either one and
[24:35] we're gonna use them both all year round so let's do another one

## 24:40 — Radical Function and the Conjugate

[24:40] It's called g of x. Let's just keep it simple. Square root of x plus 1.
[24:46] Let's find the equation of the tangent line.
[25:01] Half tax equals 8.
[25:19] To find the equation of the tangent line, what do we have to find first?
[25:29] Uh oh.
[25:32] To find any equation, what do we need to know about the equation?
[25:42] We need to find the slope of the tangent line before we can find the equation of the tangent
[25:46] line, that's what I was asking for.
[25:48] So we need to find the slope of the tangent line.
[25:50] The slope of the tangent line is the derivative.
[26:02] We use g prime of x.
[26:09] Use g prime of x, the limit as h approaches zero of g of x plus h minus g of x, all over h.
[26:12] h approaches zero.
[26:17] g of x plus h minus g of x all over h. That is simply the definition of the derivative.
[26:39] Now we plug in this right here.
[26:43] Substitution gives the square root of x plus h plus 1, minus the square root of x plus 1, all over h.
[26:55] minus the square root of X plus H X plus one all over H.
[27:11] Every time we use the definition, you try direct substitution, you always get 0 over
[27:17] 0.
[27:18] So we need to do some trick.
[27:20] What's going to be the first, what trick we're going to use here?
[27:23] Define this limit.
[27:32] Multiply numerator and denominator by the conjugate.
[27:36] conjugate which is what rad x plus h plus 1 minus plus rad x plus 1 over
[27:55] x plus x plus one.
[27:59] get this and this minus
[28:31] The middle terms cancel when multiplying conjugates.
[28:35] milli-turns. I always can't tell when you multiply
[28:38] by the derivative. That's zero.
[28:58] minus minus okay the x's cancel is zero the one and the negative one is zero
[29:12] And we're only left with H, right?
[29:33] Now the H is simplified.
[29:38] Direct substitution now gives 1 over 2 times the square root of x plus 1.
[29:43] Yeah.
[29:58] OK, so this is the slope of the secant line,
[30:02] but depending on the value of x.
[30:06] So what x value did we want to find today?
[30:11] At x equals 8, g prime of 8 is 1 over 2 times the square root of 9, or 1 sixth.
[30:33] This is the slope of the tangent line at x equals 8.
[30:47] and to find the equation now that we got the slope it's simply we always do point
[30:52] slope formula

## 30:52 — Equation of the Tangent Line

[31:00] There it is.
[31:17] Let's see if that worked out.
[31:55] Okay, and there it is.
[31:57] That is the slope of the tangent line.
[32:00] That is the equation of the tangent line through the point 8 comma 3.
[32:07] Pretty cool.
[32:08] We need no calculators.
[32:10] We didn't have to keep plugging in values closer and closer and closer to 8 to the hard
[32:19] secant line formula.
[32:20] We were able to do it all with no calculator.

## 32:22 — When a Derivative Does Not Exist

[32:23] We wanted any other point.
[32:34] What x value is this not defined at?
[32:41] Definitely not negative one, and then everything to the left of negative one.
[32:49] okay is our function defined at negative one it is okay but the slope of the
[32:59] tangent line is not defined the endpoints well this one's a little
[33:06] different but it will not have a derivative at endpoints because the
[33:13] limit as H approaches 0 from the left the function doesn't even exist okay so
[33:21] derivatives don't exist at endpoints the other problem is this is a vertical
[33:30] tangent line which the slope won't exist anyways okay let's do one more what time

## 33:37 — Instantaneous Rate of Flow

[33:38] was V of T negative minus 1 squared plus 5. This gives us the volume, the amount of
[33:53] water in the tank, the volume of the tank in gallons. So this is in gallons. Yeah.
[34:05] So the question was, what's the instantaneous rate of flow at three hours?
[34:15] And t was in hours, right?
[34:19] We need to find the slope of the tangent line at t equals three.
[34:32] So, we need to find V prime of t, or also known as, well, dV dt, same thing.
[34:49] the limit as h approaches 0
[34:57] V of
[35:00] T plus
[35:02] H minus V of T all over H.
[35:07] Here.
[35:10] So it's going to be negative t plus h minus h.
[35:16] 1 squared plus 5.
[35:28] Negative t minus 1 squared plus 5 all over h.
[35:40] Actually, before we finish this letter,
[35:44] There's going to be units for this answer. What are the units going to be?
[35:50] Gallons per hour. Why is that?
[35:59] This is the change in gallons. All these values are in gallons.
[36:06] And this is the change in x, this is going to be in hours.
[36:13] Okay.
[36:14] You can always look at this, it's always the units of this function divided by the units
[36:19] of this variable.
[36:21] So gallons per hour.
[36:25] Now we just have to multiply this out.
[36:28] This is going to be a little ugly.
[36:32] Expand t plus h minus 1, squared.
[36:39] What is that equal to?
[36:41] T squared.
[36:44] S t h minus t.
[36:52] plus TH plus H squared minus H, is that right?
[37:05] Minus T minus H, that's this guy right here.
[37:18] So we have negative t squared.
[37:29] THTH plus 2TH minus 2T minus 2H minus 1, is that right?
[37:51] Distribute the negatives and combine like terms.
[38:07] 3t plus 1 minus 5 hopefully I did that right let me distribute this negative
[38:21] negative negative positive positive positive and we should have a bunch of
[38:32] here negative 2t squared t squared 2t minus 2t plus 1 that's not right 5 and
[38:54] Is that it?
[38:56] Where did I mess up?
[38:58] Did I?
[39:04] Uh, this should have been plus one right here, yeah?
[39:12] And then this becomes minus one.
[39:15] There we go.
[39:16] Now these cancel out.
[39:30] Is that all we got left?
[39:34] More rh.
[39:35] And we factor out an h.
[39:51] We get negative 2t plus 2.
[40:10] And then we wanted that V equals, or T equals 3.
[40:19] Negative 2 times 3 plus 2 times 6, negative 4.
[40:25] And again we said it's gallons per hour.
[40:32] The negative rate means the volume is decreasing at 4 gallons per hour.
[40:41] per hour.
[40:43] Okay, so all the derivatives for homework tonight.
[40:49] Alright, you gotta use either of the definitions.
[40:56] Can you use the power rule?
[40:58] I have no idea what the power rule is.
[41:01] We gotta use the definition.
[41:05] We will learn some shortcuts later on.
[41:08] With the homework we gotta use the definition.
[41:10] Okay, the uh...