AP Calculus AB — Implicit Differentiation and Related Rates Corrected lesson transcript ## 0:00 — Implicit Differentiation [0:05] The directions are to differentiate. [0:09] What does that mean in calculus, differentiate? [0:15] All that means is to find the derivative. [0:21] Find the derivative. [0:30] And the answer is... [0:38] 8 u to the 7th. [0:49] And what do I put on the left side? [0:56] Y prime. [1:04] However, I want to use Leibniz notation. [1:10] That would be dy over du. [1:16] Why is it dy over du? [1:21] as the representation with respect to u. [1:32] But what I didn't tell you is u is some function of x. [1:44] And I really want dy dx, not dy du. [1:50] Okay, so dy dx, I don't see any x's here. [2:04] Okay, so we assume this variable is some function of x. [2:09] Right now y is a function of u. [2:11] So if I want dy dx, I go 8u to the 7th, and then by the chain rule, we multiply by the [2:22] derivative of u with respect to x. [2:42] This is 8u to the seventh times du over dx. [2:51] The derivative of u with respect to x is cosine x. [3:03] In terms of x, the result is 8 sine to the seventh x times cosine x. [3:11] cosine of x. [3:32] What if the directions asked us to find dy over dt? [3:43] Okay. This implies, right? We only see y's and u's here. But if we're trying to find this, this [3:57] implies that every variable is some function of t, but you just don't even know what it is. And [4:05] we don't even care. So all we do here is [4:09] the derivative of this function is dy with respect to t. We differentiate this 8u to the seventh and then by the chain rule we multiply by the derivative of this function. [4:27] du over dt. [4:30] Okay. You're done. [4:33] Because we don't even know what U is in terms of T. [4:40] We don't... [4:41] Who cares? [4:46] So we can take the derivative with respect to any variable. [4:49] Even if we don't see the variable. [4:52] you take the derivative of whatever this function is and then by the chain rule [4:59] we take the derivative of this function u, since it's just u it's times derivative of u with [5:06] This process is called implicit differentiation. ## 5:23 — A Circle Written Implicitly [5:23] What's this in that equation? [5:31] A circle of radius? [5:39] 3. [5:41] How do I know it's 3? [5:44] It's r squared. [5:45] squared. Okay. And where's the center of the circle? [5:50] Oh, it's, um, zero, zero. [5:53] Zero, zero. Okay. Do you remember this from a couple years ago? [5:59] No. No? Okay. [6:02] The equation of a circle is (x-h) squared plus (y-k) squared equals r squared. [6:05] squared plus y minus k [6:09] squared equals r squared. Is the equation [6:13] have a circle of radius r centered at the e, okay. Okay, anyways, this is a circle [6:23] centered at the origin of radius n. Is this a function? No. Why is this not a function? It [6:38] fails the vertical line test. Okay, but still we call it in calculus called a [6:44] curve. Okay, we don't care if it's not a function. Calculus doesn't really care [6:49] about functions, whether it's a function or not. Okay, this is some curve. It has to be a [6:55] We could solve for y: y squared is 9 minus x squared. [7:05] Then y is plus or minus the square root of 9 minus x squared. [7:12] which is two separate functions now this is a function this is a function okay [7:24] The top semicircle is y equals the square root of 9 minus x squared. [7:34] The bottom semicircle is y equals negative the square root of 9 minus x squared. [7:42] Okay. [7:42] Okay, is y a function of x here? [7:47] It's not. For it to be a function of x, you have to have y equals something. This here, y is a function of x. [7:57] Okay, Y is explicit. Y is explicit. [8:06] A function of X. [8:18] Here, [8:24] We can imply it's a function of x, but we don't know. [8:29] Okay, but here it's explicitly it's written y is explicitly some function of x here it might be might not we don't really care actually but. ## 8:39 — Differentiate Both Sides [8:40] But we can apply it is even though it's not written it want as long. [8:47] So anytime we have y equals some function, this is like explicit, and then here we can [8:54] imply it is or it's not, so on. So we're gonna differentiate this even though y is not explicitly [9:04] a function. Okay, and here's how we do it. The x squared plus y squared equals 9. We're [9:15] to differentiate both sides of the equation not just the right side although we have been doing [9:21] that. We just haven't really been talking about that. Okay. What's it like here? [9:34] We just said this equals this before we went to you. I told you that. We actually are [9:42] The derivative of y with respect to x is dy over dx. [9:50] The derivative of 8u to the eighth is 8u to the seventh times du over dx. [9:57] this kind of makes sense so we're gonna do the same thing except now we have [10:03] variables on both sides so we just use all the same rules on the left hand side [10:10] we have a sum. So how do I take the derivative of a sum? It's just the sum of [10:18] each derivative. Okay so we start with this one. Derivative of x squared is 2x, [10:27] except x is, could be a function of any variable we want. So we do need, I should [10:35] We are differentiating with respect to x. [10:47] with respect to x so it's 2x times the derivative of x with respect to x dx dx [11:00] The derivative of y squared is 2y times dy over dx. [11:14] dy dx [11:17] we're done with the left hand side [11:22] and the derivative of 9 is [11:25] 0 we don't need any chain rule because there's no variables [11:32] okay and what is dx dx [11:37] 1 the rate of change of x [11:39] with respect to the change of x is one. The rate of change is the same. So that's one. [11:45] You can take it as a fraction as well. We get this. [11:57] Okay, and if I want to find dy dx, we now just solve for this. [12:04] Two y times dy over dx equals negative 2x, so divide by 2y. ## 12:23 — Tangent Slopes on the Circle [12:23] So what does this tell us? [12:30] Okay. [12:38] That's not a very good circle, but if we want to find the slope of the tangent line, which [12:45] is our dy dx, it is our x and y. [12:48] Okay. [12:49] Okay. [12:57] At x equals 2. [13:00] So find the slope. [13:23] So I need to plug in x equals 2 into here. [13:28] At x equals 2, the derivative is negative 2 over y; we still need the y-coordinate. [13:43] What's that? [13:49] So this is a little this is a little weird because we normally don't do this very [13:52] often. We need a y coordinate just like we plugged in an x coordinate. So how do I find [13:58] this y coordinate right here? The original function right? That's this. Okay, so x squared [14:12] At x equals 2, y squared equals 5. [14:32] Thus y equals plus or minus the square root of 5. [14:46] Yeah, when x is 2, we got one value here and we got one value here. So this will [14:53] The two points are (2, square root of 5) and (2, negative square root of 5). [15:08] The two slopes are negative 2 over square root of 5 and positive 2 over square root of 5. [15:13] is going to be dy dx of 2, square root of 5 is negative 2 over square root of 5 and the other one [15:45] So the slope of this line is exactly negative 2 over square root of 5 and the slope of this line is [15:57] positive 2 over square root of 5. ## 16:01 — The Explicit Method Compared [16:04] The harder method is to solve for y first and then differentiate. [16:14] So dy dx would be what? [16:17] That's 9 minus X squared. [16:21] The 1 hat. [16:26] And then 1 half times the derivative of the inside, which is negative 2x. [16:37] And then if we wanted at x equals 2, we get 1 half 9 minus 2 squared 1 half times negative [16:54] two times two [16:57] And if you do a bunch of work comes out to be [17:01] negative two over ## 17:02 — Differentiate with Respect to Any Variable [17:03] Five or something like that [17:05] And we would only gotten one of the tangent lines slope to the tangent lines instead of two of them [17:10] Then we have to do the same thing with the other one [17:13] Okay, a lot of functions. There's no way to [17:18] Change it to make it explicitly [17:21] Okay [17:24] Now take the same curve and find dy over dz. [17:35] Now let's find d y d z I [17:48] Don't even see these [17:51] And we don't care. This implies that every variable is a function and it's a function of z. [18:04] Okay. So we just start on the left hand side, differentiate the whole left hand side, which again is a sum. [18:11] So we take the derivative of each piece separately. [18:14] We get 2x times dx over dz. [18:22] chain rule we multiply by the derivative of that function with respect to whatever variable. [18:34] Plus 2y times dy over dz. [18:41] to the variable z. [18:46] And the derivative of constant is 0. [18:53] And then we just solve for [18:55] dy dz, which is this one right here. [18:59] So we have two while. [19:27] And then what those simplify and that's it. [19:31] You will still have a, you have this rate of change in our answer and that's okay. [19:38] We can differentiate with respect to any variable we want. [19:44] If we are asked to find the slope of a tangent line, that's always dy ds. [19:51] But if they just say find dy dz, that's all we do. ## 20:00 — Combine Product and Chain Rules [20:00] Yeah, these are actually. [20:06] Okay, this is definitely not a function. [20:12] Who cares? Some curve. [20:16] Let's see what it looks like. [20:17] In fact, just for fun. [20:32] The left side is sine x times e to the y. [20:44] What did I say? Y cubed? That's kind of cool. Definitely not a function, but who cares? [21:02] It's occurred. Okay, if we wanted to find the two slopes of the tangent lines at x equals 2, [21:09] we're gonna have to differentiate. We'd have to plug it, figure out those y-values. [21:16] It's not the point. [21:23] Okay, let's find [21:27] dy dx. [21:32] Okay, we just differentiate the left hand side separately from the right hand side. [21:36] So on the left hand side, what do we have here? [21:42] A product. [21:43] A product. [21:44] So what rule do we got each? [21:45] A product rule. [21:47] Okay. [21:47] Okay. [21:58] Use the product rule: the derivative of sine x times e to the y, plus sine x times the derivative of e to the y. [22:03] t to the y, [22:06] sine t to the y. [22:17] The derivative of sine x is cosine x, and the derivative of e to the y is e to the y times dy over dx. [22:26] derivative of x with respect x times e to the y. The derivative of e to the y is e to the y [22:42] times the derivative of y, which is dy dx. [22:51] The derivative of y cubed is 3y squared times dy over dx. [22:59] 3y squared times the derivative of y with respect x. [23:20] And now if we want to find dy, dx, we have to solve for these. [23:25] How are we going to do that? [23:29] We got to bring both to one side. [23:32] Okay. And dx, dx is? [23:36] One. [23:37] 2 months, bud. [24:09] Okay, try them both to one side. [24:12] There is the implicit derivative. It is not pretty, but it gives a tangent slope at any point on the curve. [24:16] But if we note any ordered pair that's on the curve, [24:21] we plug in the x-coordinates into the x's, the y-coordinates into the y's, [24:26] and we're going to get some slope of the tangent line. [24:29] Okay, up that crazy looking curve. [24:34] Okay, what questions we got on this? [24:38] Differentiate as usual, and every time a variable depends on the differentiation variable, include the chain-rule derivative factor. [24:45] by the chain will multiply by the derivative of that inside function with respect to any [24:51] variable we want. [24:53] Okay, if at the very beginning... [25:02] If I said find [25:06] dy, dt, [25:10] instead of dy, dx, [25:12] it would be the only thing we do different. [25:18] The only thing different is [25:20] And this would be DX DT. [25:25] This would be DY DT. [25:28] This would be DY DT. [25:33] And then everything, this would not simplify to one. [25:40] It just stay in there part of that product. [25:43] Makes sense? [25:44] Kind of? [25:46] OK. ## 25:46 — A Cylinder and Changing Variables [25:46] Let us do another problem. [26:04] What is this formula for? [26:08] The volume of the classroom here? [26:12] The volume of a cylinder is the area of the circular base times its height. [26:16] This is the area of the base times our hives. [26:23] Here volume is a function of what? [26:32] Volume is a function of both radius and height. [26:37] Do we see any X's? [26:38] What is the variable? Radius and height? Yeah, here the volume is a function of both the radius and the height. [26:53] We could find dV over dr, dV over dh, or dV over dt. [27:08] We will find dV over dt, so volume, radius, and height are all functions of time. [27:23] and A's are functions of T's. Okay, so let's start the left-hand side. We differentiate this with [27:36] respect to time and it is or with respect to T. That was the left side easy. Now the right-hand [27:49] hand side. What do we have? [27:53] A product. We do have two products, but it's [27:57] just a constant. So, let's keep it with this guy right here. [28:04] Okay, [28:05] that's that product right there. [28:09] So, we need what role? [28:16] Okay, so. ## 28:19 — A Preview of Related Rates [28:19] The derivative is dV over dt. [28:26] The derivative of pi r squared is 2 pi r times dr over dt. [28:34] Then apply the product rule and include dh over dt on the second term. [29:01] And that's it. Okay. Why would we ever want to do this? [29:13] So soon we're gonna have cylinders pretend this is a cylinder. [29:29] It does not. [29:33] Water is not coming out of it. [29:35] Water is going in. Water is going in. [29:36] Oh! [29:38] But it does have a leak. [29:40] So it's going out as well. [29:42] Guys, why are you freaking out? [29:45] This is it right here. [29:49] This is how... [29:51] This is the rate of change of volume with respect to time. [29:56] These factors describe how radius and height change with time. [30:05] And they're all related. Why are they all related? Or how are they all related? [30:11] They're related by exactly this equation. [30:17] Okay, this is the relation right here. [30:20] So if I know how fast I'm pouring water in, that would be the rate of change of the volume. [30:27] Let's say I'm putting water in at three gallons an hour. [30:34] It's pretty slow, but who cares? [30:38] This would be our dv dt. [30:44] Okay. [30:46] And so for example if the radius is not changing right, fix cop the radius doesn't change. [30:56] So would be the rate of change of the radius would be DR. That'd be zero. Okay the height [31:05] would be changing as we add water and it's changing exactly this whatever that is. So [31:13] If we plug in three actually right here, [31:17] and we knew the radius of the top, whatever, [31:20] and we could find exactly how fast the height is changing. [31:26] This topic is called related rates. [31:31] It is super, it was easy. [31:33] We just did it. ## 31:33 — Differentiate with Respect to Height [31:38] This is exactly how the different rates are related. [31:41] the rate of change of the volume the rate of change the radius and the rate of change of the height. [31:45] It's not complicated. It's that easy. Okay. [31:54] What if, instead of dV over dt, we find dV over dh? [32:04] What would be the only difference? [32:09] We have an 8 here, an 8 here, and an 8 here. [32:15] And then this would simplify to 1. [32:17] Well, it's not really simplifying to d1. [32:19] That's easy peasy. [32:25] And then we're done. [32:30] This is why I like the fractions. [32:33] Okay, what does R prime even mean? What does H prime mean? The ray of change, but with respect to what variables? [32:47] But how do we know that? How do I know that H prime is equal to 1? [33:00] Why is x prime equal to 1? Only when we're differentiating with respect to x. Only when we're differentiating with respect to h. [33:09] Leibniz notation records the variable with respect to which each derivative is taken. [33:26] Implicit differentiation uses the same derivative rules, with a chain-rule factor for each dependent variable. [33:35] derivative of the function you multiply by the derivative of the variable with respect to whatever [33:40] favorite. [33:42] Can you try one more? [33:44] Where should we start at home? [33:50] One more. [33:51] Let's do one more. [33:52] I like it. ## 33:54 — Combine Quotient, Product, and Chain Rules [33:54] We will find dx over dn. [34:00] Whatever that means. [34:04] It's the rate of change of X with respect to the change in n. [34:11] So we started the left hand side. [34:13] The left-hand side requires the quotient rule. [34:16] Quotient rule. [34:17] We've got to use the quotient rule. [34:30] So, the [34:31] The derivative of x to the fourth is 4x cubed times dx over dn. [34:36] 4x cubed. [34:37] Times the derivative of our inside, [34:40] which is [34:43] dx with respect to n. [34:47] We don't care what goes. [34:52] And then the second function [34:57] minus the first function. [35:00] Differentiate square root of y as one over 2 square root of y times dy over dn. [35:08] y to the negative one half times the derivative of y which is dy dn divided by the square root of y squared. [35:30] The right-hand side is a product. [35:39] So we got to use the product pool. [35:50] The derivative of x cubed is 3x squared times dx over dn. [35:59] dx dn. [36:02] Then keep x cubed and differentiate y as dy over dn. [36:08] So, dy over dn. [36:14] You could think of it as the derivative of y as 1, but then we still have to multiply [36:18] by the derivative of that variable, which is dy over dn. [36:26] Okay. [36:28] If I said find [36:30] dy dx, what would be the only change? [36:37] All the dn's would be [36:39] dx's and then the dx over dx [36:42] simplifies to one. [36:46] What was it? I asked for dx dn. [36:54] We could collect all terms containing dx over dn and solve, but we will stop after setting up the derivative correctly. [37:03] We'd have to clear the denominator here, bring all everything with the DX on one side, everything without DX, XDN on the other side. [37:13] But we're going to stop there. That wasn't bad, right?