Implicit Differentiation, Taylor Expansion and Area

There are only remaining four now, and after that, it's comprehensive test preparation. Now we finish Newton, oh sorry, Euler method. We know what linear approximation is. We also know how to judge whether this is overestimate or underestimate. Do we need more exercises? Can I get a quick confirmation? Do we need more exercises? No, I don't think so. Okay, well, according to the homework you handed in, I don't think so either. So today we're gonna.Go on to the new topic, which is called implicit differentiation. We did this before, but we were under exercised. Under exercised. This is by itself a perennial technique, meaning you are definitely going to have a anywhere between two to four points on simply implicit differentiation. We need to master implicit differentiation. Here, I am even considering maybe we need only one session on it. That's possible. It depends on how much you remember.Okay, let me actually give you the xy tangent of the xy, plus the x to the y's power equals to three x fourth fourth, and I want to find the second derivative of x now because I want to answer the question. It's a function can't give up or can't give down. Before we even go.Let me ask you, what's the domain of this function? By the way, we're permitting whatever inside the tangent to be half of the pi. We permit it. I just said that that's the location of asymptote. Or maybe we we should really say we already know x y cannot equal to half of the pi plus k pi. But in addition to that, is there any other constraint on the domain?```````````` Again, always keep me posted. Tell me what your battle plan, what your line of attack. In fact, have I talked about how do we decide the domain of a function? The spirit is innocent until proven guilty. What does that mean? By default, it's all real, and we begin to exclude the numbers that cannot be put in. The only three principles: what they were seeing, actually regulators, so forth.The three principles is: we can't put the zero on the denominator, obviously. We can't put any non or, well, actually, negative number, negative number. We can't put the zero here in an even root. If we're dealing with real functions, and your AP tests are high school material, I'll always dealing with real functions. Okay, so we can't actually take the square root of a negative number. And if you want to take logarithm, the argument must be strictly greater than zero, otherwise.This comes from the fact that the powers are always strictly greater than zero. By the way, isn't that curious? Because we could have the negative two cubed, right? That just equal the negative eight. Then why can't we take the log base negative two and the argument negative eight? That's going to give us a three. Why can't we do that? In fact, for this particular case, it looks like there's nothing wrong. But if we want to make it into a function.And remember the logarithm is referring to the exponent. Then, well, to make it into a function, the exponent must change. You can take the cubic, the cube of a negative two, but you cannot actually do two point five's power of negative two. Why, Eddie? Could you tell me why immediately? If I swap the three by two point five, and this is illogic within the real domain's why.中文``Ùsing like split it into like two, in like one half. Hmm, very good. We rewrote it as actually five halves. That really just means we're taking.The fifth power, and then take the square root. It's a square root by in the different guise, so we can't do that. It's simply because if you put put something as a base now, which is negative, which is okay for certain exponents, but it's not okay for all the real exponents. Then when we define a function, we're not permitting the base to be negative. So if we want to choose a base, and this is an exponential function, we want the domain for that function to be all reals. Now we have to confine.A to be strictly positive, so you could think of it as actually the fourth criterion of excluding the numbers from the domain. However, that's not quite the domain, okay? This base here right now is just an arbitrary constant. But when we have the x to the y power, both x and y are variables now. It does put a constraint on the domain of x. For this to be legit, meaning if you want to be able to plug in arbitrary y and still have that term well defined. We have to require that x to be greater than zero. Am I being clear? So the graph only occupies half of the plane. There's no appearance of any graph on the left hand side of the y-axis. But now let's actually go about my original question: How do we take the second derivative? The way is that we're doing implicit differentiation, meaning look at the y itself as already a function of x now, and just take the derivative both sides of the equation.However, though, as I said, we're gonna actually look at the second derivative, and there must be some plug-in in the middle. So, well, to get to the second derivative, unavoidably, we do have to take the first derivative first. So we differentiate both sides. Very good. Tell me what we're getting term by term, and I'll put it on the board.D X D over D X Tan, X Y is equal to. Wait. Hmm.中文```We can use the chain rule on tan x y. So it becomes secant squared x y times d over.dx, xy, very good indeed, which gives you a xy derivative plus y, and what is the x derivative? Oh, x derivative. Oh, yes, and however, but fundamentally, you're just taking the.Total a derivative regarding x. Now, absolutely, this is a product rule. Knowing y itself is a function of x. And you did perfectly right. So we're getting actually fixing the y, take the derivative of x, that's the y. Fixing x, take the derivative of y, that's the x y derivative. Perfect. This is actually the derivative regarding x of the first term. And Elaine, can you pick up the second one? This is a harder one, so much harder. Remember the two.Variables and their appearing on two different locations. We covered the principle before. Whenever there are multiple appearances of things, now you always fix the others and just take one at a time. Ah, we could try using the chain rule. Yes. ```中文嗯。嗯，x y，specific，e y log in，natural log effect. Yeah, and first.In fact, Eddie is considering this as an exponential function. He's thinking of the x as a constant now. Consider x just a number, y is a variable. If we do that, it's sort of like the e to the y, except for it's a different base. We covered it. We rewrite this into the e to the natural log of the x to the y, right? And so the y jumps to the front. Natural log that's serving as just a coefficient. So derivative applying the exponential rule, it's just itself. But on top of it, there's another.So, log of the x, and I did finish it for me. Maybe you set it, but is there anything else to the derivative of the first part considering the y as a variable? Then is there something else? I'm still working that part out. Good.Yeah, I got that far. I'm just trying to find the by part. Oh, very good indeed.Y X to the Y minus one. Eight. Wait a second. Um, right. Very good. You haven't quite finished this one, but you.Started by looking at the x as a variable. At this moment, you then you did well. The y is just a number, so power rule applies, and we just lower the y and the x to one. Very good. But yet we have to finish the other part. It's not quite done yet. When we look at this experimental function as a y on the exponent. ``````中文嗯。中文嗯。中文嗯。Tell me what you where you guys are. Eddie, you started off, you get it almost perfect now.It's lacking. You got it right earlier. Whenever you're actually differentiating regarding why, oh no, not regarding why, but look at the y variable. You remembered to do the y prime because remembered. Finally, there's inner functional y itself. It's a function of x, isn't it? So shouldn't we multiply this by the y prime? Oh yeah, Prisma.And Elaine, do you agree? Yeah. And okay, don't forget that. Well, in fact, there's a marker. If you're dealing with implicit differentiation, which really just means you don't have an isolated y, x and y are coupled together, intertwined in a function. Now, whenever you're addressing the term y, now the necessary is going to be the y prime. It showed up earlier. Now it's showing up here. Alright, now final. So now we're getting to the right hand side of the equation. Wait a second, a why do I do equal? I mean, add it on to. They're on the same side, so we're adding on to that. Would equal to the right hand side of the equation now, which is relatively simple. We're just getting actually the twelve of a x cube, right? I said it because I trust you guys definitely know how to do it. Well, what did we accomplish? Did we solve the y prime? We did almost.At this moment, you can isolate a y prime, but actually, you may not have to. And because even if you isolate it, you're going to end up with a huge fraction. And when you take the second derivative, you have to plug in again. So, but that well, nevertheless, though, we have solved what is a y prime. The y prime here. Go ahead and do your own algebra, and after you do it totally your own, and then you could match the answers with mine.```Do I get a confirmation?中文``````呃，我 got the same thing. 好的，and Elaine you. 呃，I'm almost done. Okay. 好的，and you could.We go forward and take the second derivative. I'll tell you it's a nightmare. Eventually, we don't have to really simplify it, but I want to cover the entire procedure so that you know what's necessary. So at this moment, how would you take the second derivative? I got the same thing. So let's get it together. How do we finish the second derivative? We can do what we did. There's no qualitative difference, except for it's just more complex. And when we take the second derivative, we apply several rules. Okay, there's a quotient rule, there's a product rule, and there's also the chain rule. Quotient rule first. We're going to just square.The denominator, and then we're going to take the numerator's derivative, and there will be several appearances of the y prime. Well, in fact, for that purpose here, and look at how I do it because I just want to organize the terms to single out the presence of the y prime. Well, first we just want to look at the x that's the variable. I'm dealing with the numerator now, and if that's the case, I'm getting thirty six x squared, the minus the y y minus one, and x of the.Y minus two at the moment. I'm passing the y as if it's just a number, and then we're getting actually the y squared, then the second, the second of the xy, then the squared tangent of the xy, and that's it. So this is only getting done by regarding the x as a variable, and after that, I'm going to actually only regard the y as a variable. But we do know because of the.Finally, all of these would be multiplied onto the y prime, and that actually includes. There's still product rule out. Still pretty complicated. X of the y minus one, plus the y, and the natural log of the x, and times the x of the y. Come on, x minus y minus one, and the minus the second squared of xy, then minus.The y twice of the second xy squared tangent, xy, and finally x. I know it's just a bundle, and that's only part. We have done only the derivative numerator. I'm going to put that into a curly bracket. This is going to be multiplied onto the denominator. I'll call that k to d now. Just copying, and then we have to subtract by the numerator multiplied by the derivative denominator in the same.But I want you to, and then we don't bother, okay? We just discuss what to do to do the final algebraic simplification. At this moment here, I do require you to do it term by term, to the point that you could match every every single term with my denominator or with my numerator. Okay, we do it only once, and we put it behind us. So please do take the trouble, and the minute I'm following a certain order as I was spoken, as I was speaking of it.But in case there's a one single place that there's a mismatch, stop me right away. You may not understand the logic behind it. Oh, as a reminder, and Eddie, do you remember what's the derivative of a second? If you don't remember, rewrite it as actually the cosine negative first.那倒不 mean inverse， it means ninety first power， and use your chain rule. So what is the derivative of the second x Find the EnglishSin over cosine squared. Yes, that's correct. Or you can rewrite it. Usually, we write it as the tangent x times the secant x. We repeat the secant itself and we multiply the tangent x. The way that you mentioned, you talked, you said it. It's also correct. So please go ahead. We don't usually do such complicated derivatives.But it's a necessary skill. And give me a hint where you are. Don't make me wait for too long, before I can hear something from your progress.对，你有女朋友。``````中文嗯。``Ìn EnglishIn it, where are you? How many terms have you gotten? I just started taking the zero derivative, considering why is the variable. Oh, so you match the all the x terms already. Yeah, I got the same thing for the x terms. Oh, good. You know that's not small feat. Very good. And we do understand once you switch to y, definitely every single term would be multiplied down to the y prime, right? Yeah, good, very good. You can go ahead.````````````嗯。`````` 中文```中文嗯。``Ànd Eddie, where are you?I can't meet you out. I'm still solving. Are you on Ice Park or White Park? Uh. What what is.So, can you hear me? No, I haven't heard the answer. Ah, um. What do you mean by a x and y part? Ah, because in fact, bundle the map into, for example, this whole bundle here. Hey, come on, my pen turned itself off. So for.For this whole bundle, come on. They are all concerning the y as a variable. That's why they will multiply down to a y prime. And I gathered all the appearance of the axes here. So that's this part. So I recommend you actually handle them in parts as well. Checking like I'm done taking the derivative, but I'm checking a few of my signs. Good.All the terms except for like a few ones with like the like the signs match, so I'm checking those. Okay. Oh, good. Ah, that that bodes really well. ```中文```中文嗯。``Ì got the same thing. Excellent, and that's really great. And Eddie, you, but Eddie's where you are. I'm getting the same.Everything. Yes. Okay, great. Or if that's the case, we are not going to do it anymore because just that I randomly wrote something which is really really complicated. My point is made. Finally, this is not legit answer yet because we haven't totally written the derivative in terms of variables alone. We have that very quirky y prime that's not that's not quirky but it's it's a illegitimate.So what do we do? We plug in, because we did work out the first derivative. That's why this kind of problem is very tricky. Meaning, if you miss something really tiny in the first derivative, it's going to wreck everything else. Eventually, when you plug it in, and what the correctness of the final answer depending depends on getting the first derivative right. We're good. Are we? Hello. Yeah. Yes. Yeah. So that's the procedure. The same.这个技术也适用于斜坡场。所以，让我实际给你演示。现在，我 gonna make it into an application problem in a way it's sort of a word problem now. So, just combine this with what we learned lately, which is, we were really just looking at a linear approximation to figure out a nearby point. So, if I give you a slope field where the y derivative equals to a yaya, let's do the cosine of the x y. What. Do I like x y so much? Well, there's nothing wrong. Let's do x squared away and then plus the i, let's say, y cubed over x. Okay, this is a derivative, and we want to do. And a easy initial condition will be, if I plug in, for example, x equal to one. And y, I'm giving you that initial condition.Why could it be? And I want to evaluate a point nearby. We want to find the x equal to one point two. Oh, you guys know how to do it because it's zero point two. It's really decently small. You don't need to stepwise. You just need to do linear approximation. That's not hard. But more importantly, is it under or overestimate? Can you tell me?``````中文嗯。```Can you talk me through what you guys are doing. I want to find if it's an overestimate or an underestimate. I'm taking the second derivative and seeing if it's positive or negative. Very good. Very good. In fact, then once you're on the track, and I would say I'm happy. We don't need to do it again because we have just done that before. And then definitely when you take the second derivative, there will be surely appearance of the y derivative, so we have to actually plug that in. Other than that.We're fine. That's exactly how you judge. And once you know, for example, if the second derivative turns out to be positive, and then what does that indicate? Is it an over or underestimate? Overestimate. Good. And I oh, whoa, whoa, whoa, whoa, whoa, whoa, I said the second derivative is positive. Then it's underestimate if it's positive, right? Because in fact, the tangent line will be under the curve itself. And when we do a linear approximation, we're picking a point on the tangent line, not on the curve itself, so that's under. Well done. Okay, we're done with this technique now. Let's actually move on to the next one. And the next one though, it's an integrating well, it is integrating technique, but we actually we have covered more integrating technique than ever necessary.For the AP, let's just say that the integrating techniques required by the AP—it's a tiny subset of everything we covered. However, I want to highlight something which is not a proper technique, but it's very useful, which is symmetry. Okay, that's going to save you a lot of trouble. We have just a little bit of time—that's enough to cover this technique. What do we mean? It works if you're dealing with a definite integral, meaning we're aiming at producing a final answer instead of a.Having a find a antiderivative. We actually don't care what antiderivative is. That means when we use substitution, we don't need to substitute back. We can change our variable. Or let me actually talk concretely. If we want integrate from negative two to positive two, and let's go for sine x cubed i times the let's say.哎呀呀，natural log， okay， natural log， the absolute value of the x minus, hmm, e to the x, cosine x, dx, hey come on, why doesn't it write cosine x dx now? Then how do we work this out? I'm not going to say that I let you think about it.I'll tell you, it's not meant to be hard. But if you just look at the first part, don't even try to find the antiderivative. We won't be able to. It's not because we're not smart. In fact, it just doesn't have an analytic function. As its antiderivative, so what do we do?中文嗯。`````` ``Ànd then may I know what you guys are thinking?I don't really have any ideas. And the I D U, I said it's not necessary. This is what we mean by symmetry. No matter what, uh, we understand that the definite integral is definitely going to be the area covered under the curve, right? And there's something probably we.Can take advantage of the integrating boundaries from negative two to positive two. That's a symmetrical region. And can we find some pattern to this integrant? For example, is that even function or odd function?``Ìs it odd? Absolutely. So no matter how weird the graph looks like, it will be odd function. In fact, when you plug in, it's gonna actually look something probably like that. Yeah, accurately.But at least they were getting the zero points right. And then does that help? Can we immediately draw some conclusion about that part of the integral? We only need to find for like one part of it, and then we can just multiply by two. Oh, so we're saying that integral means the area, not the signed area. Wait, wait, oh.It's zero. Yeah, whenever you integrate all the function over a symmetrical region. Of course, later I can get fancy. I can change it, do a shape, and make it somewhere else, so that maybe you have to compute the cube to recognize the symmetry. But nevertheless, IP problems are easy. Okay, they're for symmetries. So for this one, and oh, I don't mean they're for symmetries like you. You guys are so young. There are still challenges. Nevertheless, though, they're not supposed to be. Too hard. So for our function immediately, your total area would be zero. So in the future, if you're doing a definite integral, the goal is just find a number. Then the minute you're seeing something that can be cancelled out, just cancel them. No doubt about it. Are we good? There are actually a lot of ingenious applications to really really interesting problems here. I'm not going to go through them yet because we have a long time in front of us. Okay, in the future we.Can play with the calculus right now to get you adequately prepared for it. I'm only showing you the fundamental techniques, so watch out for even functions and odd functions, and that's my point. Makes sense. Yeah, okay, and Eddie, you've been quiet. Any question from you? And are you totally comfortable with it? Yes, beautiful. Now we're gonna actually move on to a technique that we covered repeatedly before, except for.I just didn't. I should bring it down to the to the AP. Well, I mean the test level, and not to mention the ways that we we did that. Usually, it's a little unconventional. Remember how to find the limit of the four indeterminate forms. Do you guys still remember the four indeterminate forms? We're finding the limit now. Usually, if you're finding the limit of a function, you just plug in the number. I don't remember. Okay, let me actually give you the big picture now.Type out the caption for this sentenceType out the caption for this sentence. Doesn't at all. No, I believe. I sort of remember it, but I don't remember like solving it. We positively solve them as the application of a power expansion. I said we want to leave the lowest order infinitesimal for both numerator and denominator so that you could combine them. Currently, what is making it non-trivial is that if you plug in x equal to zero, you're getting zero over zero, right? So we can't compare. We don't.No, eventually, which zero? Not all zeros are equal. You know, some zeros are infinite bigger than others. It could possibly be approaching infinity, zero, or anywhere in between. So go ahead, use our power expansion technique. We'll use power expansion. Yes, please. Although, um, AP, I'll teach you a new technique. One minus cosine squared x is just sine squared x.Oh, what a, oh indeed, indeed, that's correct. And I, I, I actually didn't think of it. Let me actually do it differently. What mine's called science cube. Sorry, I made a trivial. I didn't mean to.中文嗯。 We could try, like using the expression of one over.One minus x, and just plug in cosine cubed x for the x. We can do that. We can do that, but we can be smarter. Remember, our goal is to isolate the infinitesimals. There are plenty of this algebra out here. That's not even infinitesimal. They're innocent, meaning you can just plug in your x is equal to zero. For that purpose, for the purpose of narrowing down what's really causing the zero, we factor the denominator. This is a totally general technique.Okay, okay, it's universalizable. So what we want to do is to rewrite it as a sine x denominator would be one minus the cosine x, then multiply by the part that's not the infinitesimal. I hope you remember, and this is a difference of the cube formula. When we do the one minus cosine x cubed, and what would be the other component.One plus cosine x plus cosine squared x. Yes, very good. One plus cosine x and then plus cosine squared x. Now, does it help? Yes, because remember x is approaching zero. What is its approach? Its approach is something quite simple. Its approach is three, right? Then can I simply scribble.Replace it by a simple three. Our job is really finer. This component, it's zero over zero. May I do that? Yeah. Alright, Ah Ela, do you agree?Yeah. Okay. Well, let's go ahead. At this moment, you really want to do plug in your power expansion now. Can I get water? Yes, you can.中文```中文嗯。中文嗯。Then you're back off.So, can we finish it?嗯，嗯。And Elaine, where are you? I'm using the power expansions. Right.And remember, they are the imaginary and real parts of the power expansion of the exponential function, meaning e to the x power. So, do you remember what is the power expansion of sine? Oh, we need only keep the lowest order infinitesimal, right? Ah, the power expansion of sine is x minus ah x cubed over three factorial plus x to the fifth power.Um, over five factorial, and so on. You're absolutely right. So in fact, the numerator simply becomes x. That's a vision. Without one minus equals x. Negative x squared over two factorial. Yeah, that's correct. I just me actually. Well, then once we cancel out this one minus now, in fact, and a well, you're saying after we cancel, it just x squared over two. So do we know what?这是 final answer。 ```Hello, what is the final answer?中文嗯。对呢？What is your final answer? When I'm plugging in zero, it's still zero over zero. I know, but then they have different orders. Though you couldn't know which one to cancel, what is left over.I'm getting a two word react. Wait. Well, we're getting actually after cancellation. It's first order on the top, but second order. So don't we just have a, oh yes, but this one is no longer indeterminate. It's very determinate, isn't it? Yeah. Yeah, we're done. In fact, the final answer is the perfect infinity.Are we totally crystal clear about this now? It's approaching infinity because two over three x is approaching infinity, right? Yes, because we have a left infinitesimal on the denominator.That's why the overall is a protein affinity. Okay. Alright, we're done now. Thank you for your very good effort. Well, next session, we're going to continue. There are two more techniques to cover. Now have fun, and if there's a, it is time that you start systematically doing the review now and start looking at the College Board online archive and just gradually go over the entire structure of the AP.Really, it's not ready to do so. Keep me posted of your progress. Would you have no homework today? The homework gets just start doing AP prep and by doing the Arkhaven College Board. No, the material today reached the closure. I was confident that you all got it, so no homework to finish from there. Thank you for asking. The, of course, take care. Bye.

Implicit Differentiation, Taylor Expansion and Area

Implicit Differentiation, Taylor Expansion and Area

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