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Moses A.

March 21, 2022

Differentiate (xcosx) in first principles

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0:00

Felicia Sanders

If $f(x)=x+\sqrt{2-x}$ and $g(u)=u+\sqrt{2-u},$ is it true that $f=g ?$

02:07

Fangjun Zhu

00:56

00:38

Amy Jiang

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Okay, so I hope this is not too overwhelming, But let me give you a little preface. So we talked about functions and their derivative functions, and that's going to be a very important thing to be able to do. Given a function find it's two rib. Now we have a definition for the derivative function, but in practice this is very difficult toe work with, especially if you have a function that is very complicated. It consists of combinations of many different, relatively simple functions. And so what we want to develop now, just like we had the limit laws that allowed us to quickly evaluate limits without actually using the definition of the limit. We wanted to develop Cem differentiation rules. So that's exactly what we have stated here. These air differentiation rules, and if you look down this list, most of them are just showing how if I combine a function in a certain way, what happens to the derivative of that combination of functions? The first one, actually the first two are kind of the most basic, So if we have a constant function, so the function whose craft is actually a horizontal line, it's not surprising that the derivative is zero because this slope of a horizontal line is zero for the second one. The identity function. This is just the function whose graph is the line y equals X that has slope one. So it's not surprising that the derivative for each X value is what Now we can get a lot of real estate out of these two basic functions by looking at how functions can be combined so we can multiply a function by a constant. So we see that if we multiply a function by the constant and take the derivative, we just multiply the derivative by the constant. If we add two functions together and take the derivative, we just some the derivative of the two functions. So everything seems to work out the way we would expect. The first thing that gets a little interesting is actually the product rule. Now. The product rule says that if I multiply two functions together, this is what I need to dio. I need to take the derivative of one times the other, plus the one times the derivative of the other. So it's not just a simple is multiplying the derivatives There's actually something a little bit subtle about taking the derivative of a product of two functions. And then if you look at the quotient of two functions, So if I divide two functions and take the derivative of that question now, we're starting to see a really complicated formula. And definitely the question rule is the most intricate. It's also a little bit tricky because you're subtracting it's related to the product rule. But because you're subtracting, you actually have to remember which function to take the derivative of first. And so there is some no Monix to remember this, that I'll kind of go through when we do an example. So you take the bottom times, the derivative of the top, minus the top times, the derivative of the bottom. So I like to think low D hi minus high D Low Lo is the bottom function. De high is the derivative of the top minus top, so high times derivative of the bottom. That's gonna be important just to do several examples toe just memorized, and then you divide by the bottom function square. The power rule is really nice. This one's really quick toe, remember and it's really easy to apply. So if we just have a power function function X to the end, then it's derivative is just in. So you bring down the power and then you take x 21 less than the original power. Okay? And we will actually go through and justify all of these differentiation rules from the definition of the derivative, the only one we're not going to do right now. And I just wanted to make a note. Is that the derivative of this exponential function? Either the eggs is itself, which is really cool. So what that means is, if I want to find the slope of a point on the graph of the X, then it's the same. It's just the evaluation of the function, which is really interesting, and we'll talk a little bit more about the exponential function later. But for now, let's actually go through and see why all of these air true now, I don't think it's important to memorize how to justify all of these now. The first few are going to be really easy. The product rule in question rule. Even the some rule require a little bit of trickery to see why they're true. But I think it's worth while just to see it done one time and say, Okay, the product rule. It's coming from the definition of the derivative. Now I can just use it and it's gonna make finding derivatives and derivative functions a lot easier as we'll see in the examples. Okay, so let's see why the derivative of a constant function is zero. And now this is really straightforward. And in fact you could even just argue like we did before that a constant function is, ah, horizontal line and has zero slope. And that's perfectly valid. But let's see how it goes using the definition of the derivative So f prime of X. Just by definition is the limit is H goes to zero of f of x plus h minus ffx all over h okay. And here comes the hardest part of this little proof of the derivative of constant function evaluating the constant function. And so when we talked about the pre calculus review, I told you that maybe one of the trickiest things about constant functions is just accepting that no matter what, the input is for a constant function the output is the same. So what is F of X plus H? It's just seen. It doesn't matter what I play in. I always output. See what is ffx? It's just seen all over H. But then this is just the limit is h goes to zero of zero because C minus C is zero age. I must actually assuming it's not zero because I'm taking the limit is h goes to zero. But then the limit of a constant is just that constant. So we get zero and so that justifies why using the definition of the limit, the derivative of a constant function is equal to zero. All right, so let's try to tackle the identity function. F of X is equal to X again geometrically, we can think that the graph of this function is just the line y equals X, which has a slope of one. So it should not be surprising that the derivative is one for all X. But let's see why that's true using the definition of the derivative. So if we take the limit, his h goes to zero. So we're gonna we're trying to find f prime of X. So the limit is H goes to zero of f of X plus age minus F ibex all over age. Well, what's f of X plus age? So the identity function just gives thean put as the output. So since the input is expo sage, the output is just expose age and then f of X is of course, just X over each. But then look what happens. We have X minus X. So that's zero. So we're just taking the limit of H over H, which is just the limit of one, right, because I can cancel H because I'm taking the limit is h goes to zero. So I'm assuming that H is not equal to zero. But then the women of one is just one. So we see the derivative of the identity function is one justified by using the limit definition of the derivative. Okay, so here we're stepping up a little bit in abstraction because we're not dealing with a concrete function and explicit function. We're just saying, given any function that has a derivative, so uh, function, after which that limit exists for each X, and then we have a number C if we multiply f by sea and then take the derivative That should just be C times the derivative of ffx. So let's see why this is the case. So if I look at sea f prime of X, this is going to be the limit is each goes to zero of what? So the function I'm evaluating is CF. So this should be C f of X plus h minus cf of X all over h. And so it might be worth just reminding you kind of what this notation really means. So in parentheses here I have CF. So the function is CF, and what that means is that I take the function, Yeah, evaluated X plus age and then multiplied by C. Okay, so what I get here, lemon, is h goes to zero. This is just going to be see times f of X plus age minus. And then we're going to evaluate Ethic X and then multiplied by seats. So this is going to be C times ffx all over h. But I can factor out a c here and using the limit laws. I can actually factor it all the way out outside of the limit because the limit of C is just see So it's the limit is H goes to zero of f of x plus h minus f of X all over h. But now I'm assuming that f is a differential function. So that means that this limit here right there exists and is actually equal to f prime of X. So then this is equal to see times this quantity or this function, which is f prime of X, and it's a little bit subtle what we're doing here. But the good news is things were behaving exactly like we would expect. Okay, So if I multiply a constant by a function and take the derivative, that's the same thing. Is taking the derivative and then multiplying by the constant You were seeing that here. And so it's a little bit strange because I doubt that you deal a lot with kind of dysfunction, composition notation a lot. Um, and that's really the only difficulty in this constant multiple rule. And in the other examples to follow is just kind of the level of abstraction. You know, in reality, you're going to be given a function, you know, x squared or X plus two or square root of X, and you just want the derivative. You're just going to do it. But these abstract rules kind of just give you a frame worker justification of why you can just do the derivative the way you want to. Okay, so again, this is going to be a little bit abstract to justify. But it's a very useful fact that if I'm just adding to functions together, like if I add together, um, power functions, then when you get a polynomial. So this is kind of give me a way to take the derivative of a polynomial. Then all I do is just take the derivative of each piece that I'm adding together. Okay, so let's see why that's true. So if I look at F plus G Prime and now I'm assuming the F and G are both differential, so this will actually make sense when I get F prime and G prime. So this is supposed to be the limit. By definition, his H goes to zero of F plus gene that function. So the sum of the two functions evaluated at X plus age and then minus dysfunction F plus G evaluated and x all divided by H. Okay, so recall that when you evaluate a some function at a point, you just evaluate each function at that point and Adam together. So this is really f of X plus age plus G of X plus h okay. And then minus. Let's put this in, brackets because we're really going to be subtracting everything over here ffx plus g of X because that's what's happens when I evaluate F plus G. It exits just f of x plus g of X, but I'm subtracting both of those quantities. And then, of course, we're all divided by H. And now So I'm trying toe show that this is equal to f prime of X plus g prime of X. But really, all the pieces are in place. I really just need to reorganize this this question here and then to supply the limit laws into the two different pieces. So what I'm gonna do is I'm going to say this is the limit is h goes to zero. I'm just going to kind of re associate things, So I'm gonna put f of X plus h and then I have minus ffx I'm gonna put the minus ffx over here. And then that's, of course, divided by age. And then I have this g of X plus age, and I'm going to take this minus G of X and associate it there and then that's divided by H. Okay, so notice that I split up the fraction, but everything is still okay, right? Because these fractions have a common denominator. I could just re add them together and kind of move things back around, and you see that this really is the same thing. But now I can take the limit of this difference question right here. The limit of that, plus the limit of that. But the limit of this is f prime of X. In the limit of this is H coast to zero is G prime of X. That's where I'm just using the limit loss. So this really is breath prime of X plus g private banks. So nothing crazy going on here. Just a little bit of reorganization to see that Yes, if I add two functions and then take the derivative that's the same is taking the derivative of each function and adding those derivatives together All right, here's where things start to get really interesting. So if you find yourself lost, just hold on. That is gonna be okay. Um, yeah, but here, Thio show that the derivative of a product is given by this kind of strange. You take the derivative of each function and then multiplied by the other. I'm together. We're gonna have to use some really mathematical trickery. And we're also going to have to use something that we mentioned in the last lecture. And you probably just brush right over the fact that if a function is differential, it is continuous. Okay, so that's going to be important. But first of all, you know, with any problem like this, if you're trying to show something, you just start by writing down what you know. Okay, So I'm trying to show that the derivative of this product is this. So let's just write down. By definition, what is the derivative of this product of two functions? So it's the limit is H goes to zero of f times G evaluated at X plus age minus this function F times G evaluated X all over H. Okay. So just using how we know how to combine functions. Let's just evaluate what this is saying. So f times g evaluated at X plus h is the same thing as F of X plus h times G of X plus age, and then f times g of X is the same thing is ffx times g of X and then all over h. Okay, so now with the some rule, all we had to do was just move things erect. Now, here it is not so easy, because all of the pieces air not here to give me this. So if if I'm looking for something that looks like F prime of X, I'm looking for something like f of X plus H minus f of x. And then I also need something that looks like G of X plus H minus G of X. So this is where the trickery comes in. So I'm going to do something that we've kind of done before, and I'm just going to strategically add zero. Okay, So adding zero to anything well, doesn't change it. So what I'm gonna dio is I'm going to write that this is f of X plus age times G of X plus age. And then I'm going to subtract ffx times g of X plus age. Okay, so that's coming from nowhere. So I'm subtracting something. Well, then I better add it back to make sure that I'm not actually changing the value of this expression. And then finally, I have minus f of x times GLX and then all of this over H. Okay, so step back. Make sure you believe that this is actually true, that this is actually equal. And just to kind of help you notice that this middle part is actually just zero I'm adding and subtracting the same thing. Okay, so why did I do that? Well, now I'm gonna factor. So notice that the first two terms here have a factor of G of X plus age in the second two terms. Have a factor. Ffx so I can factor. Okay, so I'll take The limit is h goes to zero Now have g of X plus itch. Okay. And if I factor that out, I'm left with what f of X plus age minus ffx that comes from just factoring out g of X plus h here. Just looking at the numerator and then plus well, let me factor out F F x and then one of my left with G of explicit minus G of X. And now I need to divide everything by age. But let me just multiple are just divide each of these terms by H okay. And so because both of these terms have the same denominator, I could just write it as a single fraction like this. So you should be the same thing. But I'm just reorganizing in in a way that it's starting to look a lot like this. What we wanted to be equal to. In fact, there's Onley. One problem. Everything looks beautiful. So if I take the limit, I'm just gonna apply the limit laws to take the limit of everything here. The limit of this guy is f prime of X, okay? And then the limit of this guy is g prime of X. The limit of just ffx is just ffx that's constant with respect to age. But then I also have this which I'm gonna leave for a second. So I also need to take the limit of G of X plus age and so everything is great as long as the limit is h goes to zero of g of X plus age equals G of X. So how do I know this is true? And what is this even saying? Think about it for a second. Let me just rewrite it for you. Let's let X plus age equal say why then his h goes to zero. Why is approaching X So this is the same thing is saying the limit is why approaches x of g of why is equal to G of X. But this is starting to look familiar. This is just saying that G is continuous at X Now remember, we need this to be true. So if she is continuous X, we're done. So why is g continuous it x? Well, here's the key. She is differential below decks and we've shown that if a function is differential, wherever it's differential, it has to be continuous there. So because JI is differential Jesus continuous and we can really conclude that this guy is g of X, which is exactly what we want. Alright, so let's talk about the cushion rule. So the first thing to notice about the question rule is that it's not so different from the product rule if we define a new function. Okay, So what I can do is I can say f over Jean is really the same thing. Is f times the function one over G so I can actually apply the product rule to this function F times one over Jean. But what? I really need to know how what to do is to take the derivative of one over g. Okay, So, actually, what we're gonna do is we're just gonna show that the derivative of one over G is given by a particular formula, and that is compatible if we just supply the product rule to this situation. Okay, so let's see what the derivative of one over G is. So, for one of her, Jean, the derivative by definition is given by this limit as a church goes to zero of one over G. That function evaluated at X plus h and then minus the function one over G evaluated X and then all over h. Okay, so let's just fill in what this actually means. You're one over g evaluated exploitation. That's kind of symbolic. Just to mean Okay, this is one over G of X plus age and then minus one over G of X all over each and the trick here. So again, we have this complicated mess in the numerator. We want to get a common denominator and right, this is a single fraction. So this is the limit is h goes to zero. And now I want to multiply this on the top and bottom by g of X and this in the top and bottom by G f x plus h. So we're going to be left with G of X minus G of X plus age over G of X plus h times G of X and then that all over H. But of course, this h is, like over one, and so we can do a little bit of manipulation here. And actually we can just put age up in the denominator like this. Okay, so, you know, dividing a fraction by a number, that number just goes into the denominator. That's a really useful trick to remember algebraic Lee when you're simplifying things using algebra. So let's write this in a very suggestive form. So I'm gonna factor out this one over G of X plus h and the one over G of x toe have won over G of X plus age times, you fax And then times I want to write this as well. I'm going to factor out a minus one. So I'm gonna make this minus here because I'm going to switch the order of these two. I'm gonna write G of X plus age minus G of X all over H. Okay, so, you know, take a minute. Make sure you believe what I'm doing here. So if I multiply this negative one back in notice, I'll get back to G of X minus G of X plus H. That's why I have to factor that negative one out. Everything else looks okay. And so what we have is the following. No justice before G of X is continuous. And why do I know that? Well, it's right here because G of X is differential. I'm assuming jeez, differential. And we have this nice result that, uh, differential bility is stronger than continuity. So function is differential. It has to be continuous wherever it's different. Trouble so g of X is going to be continuous. So when I take this limit is h goes to zero g of x plus h really just goes to G of X. So this is giving me minus one over g of x times g of X so g of X squared and then times. This limit here is simply g prime of X so I can write. This is just G prime affects divided by G of X square. And if you just apply the product rule F times one over G, then you'll see you'll get exactly the formula that we have in the question rule. Okay, so the power rule is probably everybody's favorite rule because it's so algorithmic and so straightforward to apply. But actually, I would say that the proof or the justification for why the powerful works is probably the hardest. And I'm not saying it's the hardest because there's anything deep. There's just not necessarily like a very elementary way to attack. Okay, now, actually, in my own personal notes, I have four different proofs of the powerful, so one uses mathematical induction. One uses the binomial ethereum, and I always wrestle with kind of which way to present the students in the way I think is the simplest use is something that we've actually talked about before. And that's the idea of factoring differences of powers. Okay, but in order to do that, we sort of have to manipulate our definition of the derivative just a little bit, But I think you'll follow. So here's our function. Ffx and now in is a non negative integer. Okay, so it could be 01234 etcetera. And so, by definition, what's the derivative of dysfunction? It's just going to be The limit is h goes to zero of X plus h to the end minus X to the end. All over H Okay, so you see the problem that we're really trying to show this for a general. Now, if it was X squared, we could just foil this out and simplify and be done, or even if it was execute. If it was X to the fourth, it might take a little bit longer, but we could do it. But the problem here is that this is an arbitrary in. And so what we would have to do if we wanted to continue is probably invoke the binomial theorem to expand this, Okay, But I wanna kind of avoid doing that. And what I want to do is notice that I can actually rephrase this in the following way. I can let X plus age equal. Why? Okay, Some things. We've seen a similar idea. When we talked about functions that are differential being continuous and taking the limits of a differential function we kind of want to do this idea. Well, it's the same thing here. And so what I can dio is note that is h goes to zero. Why is going to be approaching X so I can write? This is the limit is why approaches x of why to the end minus X to the end over what is H where h is just gonna be why And then I can subtract X. So why minus X? Okay, Yeah. And why did I do this? Because we actually know how to factor this in general. So this was a difference of squares. This will be like why, minus x times y plus X. If it was a difference of cubes, it would be why minus x times why squared plus y x plus X squared? But the important thing to note Is that this? Why did the n minus X to the N for whatever value of n will always have a factor of why minus X and what's gonna be left is something relatively nice to deal with. So what we can do is factor the numerator as fathers. So we're gonna have why to the n minus one plus Why, to the end minus two times X and then plus etcetera. And then so you see the powers of why are going down one while the powers of X we're going to be going up one so eventually we're gonna have why Times x to the N minus two plus x to the N minus one. Okay. And then all of this is divided by why minus x. And that's really convenient because these old cancer okay, and so all we're left with is the limit is why approaches x of why to the n minus one plus y to the n minus two times x plus. Why to the n minus three times X square. Plus, however many terms there are between then, why times x to the n minus two plus x to the N minus one. So, how many terms are there here? Well, just look at the powers of X. So here in the first term, this is X to the zero. This is X to the first, this is X to the second, and then we're just gonna count up. This is X to the N minus two. And then this is X to the N minus one. So, how many terms are there? Well, if we didn't have the X to the zero term, this would literally just be counting. 123456 All the way up in minus two in minus one. So there's in minus one terms, not counting the X to the zero term. That means that there's in terms in total. And when I let why, good X notice that this is gonna be X to the N minus one. This is going to be X to the n minus one x to the n minus one. Because I'm adding so in minus three plus two is in minus one in minus to plus one is in minus one, of course X to the n minus one. So I really do get in terms each of those terms being exactly x to the n minus one upon applying the limit loss. And so that's the idea behind justifying the power rule. And this is a little tricky, I think. E think it's kind of it does stretch your mind in a lot of ways, but I do think it's the easiest way to see why it's true. I think if you try to use, you could see you can use the power rule multiple times and use kind of an inductive argument. But that sort of uses an idea that's beyond the scope of this class, just like the binomial theorem. You know, you may be familiar with it and you can use it. But again, this is probably the most elementary in terms of the algebra required to see why the power rule is true. So we've seen and actually justified power role for non negative integers. But I want to make a note that there's actually a more general statement that you should know and it's called the General Power rule and unfortunately we don't have the tools to justify this quite yet, but it's so useful we might as well, go ahead and start establishing it. So the general powerful states that if Alfa is any riel number and ffx is equal to X raised to the Alfa Power, So here Alfa is not just a manager or a positive image er or non negative integer, it could be square root of two. It could be a rational number one half, 3/4 etcetera. Then the derivative follows Theis exact same formula that we had for the case. For a non negative integer. It's just Alfa Times X to the Alfa minus one. So this is something to keep in the back of your mind because it's very useful because a lot of times you'll have a power function that's not just X to some non negative integer like X to the one half square root of X, for instance, falls under this general power rule. So just so you know it's there. And when we talk about long, arrhythmic differentiation later on in the course will actually go back and justify this general power rule

Applications of the Derivative

Integrals

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