[Instructor] So as we sort of wrap up this part of the course, what we're doing is we're really sort of gathering these collection of rules, rules for derivatives. And they fall into two categories. There's the, what are our functions that we work with. And so we have a bunch of them. x to the k, e to the x, sine, cosine, tangent, secant. We say, okay, what are the derivatives of those specific functions? And so we have six of them here. By the time we finish with the course as a whole, we're gonna add two more to our list. So don't worry, there's still more functions to come, but those will be in a few weeks from now. But the other thing is to say, look, there's more than six functions that exist out in the world. And so how do we handle other functions? And oftentimes what happens is those other functions are combinations of these. And so we wanna talk about the rules for putting 'em together. That's this bottom block right here. Now, there's a couple rules we've already talked about. So it says if I have, for example, a constant, I can pull the constant out. If I'm taking the derivative of two things being added together, I can take their derivative separately. There's the product rule which says if I have two things multiplying, I can take their derivatives by essentially saying, the way I like to think of it is each one gets their turn. So first, f gets its derivatives, so it's f prime, but g has to wait. So it's f prime times g, but then it's g's turn to shine, g prime, and then f has to take its turn to wait. So f times g prime. And then we have the quotient rule. So we have these rules, but these are sort of good rules, but they're not the best rule, the best rule. That's today. What's today? Well, oh, we can already hear the murmurs. It's the chain rule, you know? So what's the chain rule? Well, chain rule is for when you have a function inside of a function. So there's sort of a different type of rule. Because previously, what we did was we said, sure, we may have two functions, maybe they're being added, or maybe they're being multiplied, or maybe they're being divided. But now it's saying, what if I put one inside of another? It's like a, I guess it's like a turducken kind of deal, you know, a chicken inside of a duck inside of a turkey, and well, now you might have a function inside of a function inside of a function, it can go all the way down. But we just need to know the basics and say, okay, what if I have one function inside of another function? We say, that's f of g of x. And here's the rule. Take the derivative of the f, that's the outside function, but you leave that g of x alone, and you plug it back in, and then you have to multiply it by g prime of x. And that's the derivative. Now, we wanna understand, why does this work? So we're gonna give a sketch of a proof. So I used the word sketch here in quotation marks, 'cause there's a little bit of a technicality, but it's good enough to convince us. So how do we do any of these? Well, if I wanna take the derivative of f of g of x, I go back to the definition, it's a limit definition. Take the limit as h goes to zero of f of g of x plus h, so I'm taking x, and I'm replacing it by x plus h minus f of g of x all divided by h. Now at this point, we're going to do one of our fun tricks of algebra. Now, one of our tricks of algebra is add a good zero. That's what we did when we did the product rule. We added by a really nice zero. This time, we're not gonna add zero, we're gonna multiply by one. So here's what you get. We're gonna multiply by g of x plus h minus g of x over g of x plus h minus g of x. So you'll notice that those we can think of as canceling, and we still have our original two terms. Now if you look, there's some things, they say, "Aha, g of x plus h minus g of x over h, that's g prime." And it is. But the other thing is the first term, we can think of that as being f prime of g of x. Now, why is that? Well, what we can think of is there's a sort of alternative definition of the derivative where we say, instead of having h go to zero, we can talk about things like as b goes to a. So suppose I set a to be g of x, b to be g of x plus h. Well, if I have a nice function, in other words, one that has a derivative, and we're going to assume that we're working with nice functions because I see a derivative of g, so, all right. So that means that g of x plus h is getting close to g of x. So in other words, we have that this expression b is getting close to a. So I say, aha, this is like f of b minus f of a, or b minus a. Well, that's just another way to say the derivative at a, which is f prime of g of x. So there's a sketch of the proof. Now, the reason it's not the proof is there's some sort of small technicality of what would happen if g of x plus h minus g of x were zero. Well, okay, okay, we know how to do that. That's details, details. We'll leave the headaches to the other mathematicians. We just wanna know the intuition. And the intuition is good. So what do we have? Well, a little bit of notational note. Now, why might we call this the chain rule? 'Cause it doesn't look like a chain. Normally, it doesn't look like a chain. Well, the chain rule is about composition. So functions inside of functions. And there's another way that composition can be written, which is this f of g is f circle g. So, well, what if you have a function inside a function inside a function, so forth, and so on. Well, again, you can use the same notation. You just have a series of circles here. So you can think of these as being your links in the chain for putting the functions together. So that could be one way you can think of why is it called the chain rule is that we have this notation which chains these functions together. And again, there's this basic procedure. Take the derivative of the outside, leave the inside alone, plug it back in, multiply on the outside by the derivative of the inside. Now, if you go back over your notes, you've seen we've actually been doing a couple things that really would make the chain rule very easy, or at least the derivatives would be very easy if we had the chain rule. Namely, we've done all of these derivatives before, but we had to work. Now, let's redo them. So suppose I wanted to do the derivative of e to the 2x. Well, what's on the inside? The 2x. So the 2x, that's my inside function. So that would act as my g of x. So what's my outside function? Like the e function, the exponential function. So the derivative of e to the x would be what? e to the x. So the rule says you take the derivative of the outside part, which is the exponential function. So that's e, then you plug in the inside function. So you plug 2x back in. So that's our f of g of x. Then you times that by the derivative of the inside function. Well, the inside function, that's the 2x. Derivative of 2x is two. And that's what we had before. All right, well, similarly, derivative of e to the minus x. Now what's our inside function? Negative x. That's the part on the inside. Okay, so we repeat the process, we say, okay, take the derivative of the outside, so it's sort of the exponential function. That would be e to the x, but what do we plug in? We plug in negative x. So derivative of the outside, plug in the inside, times it by the derivative of the inside, derivative of negative x? - Negative one. - Negative one. And that's what we got before. Ah, two for two. Okay, well, let's see, maybe the pattern continues. The derivative of sine of 2x. Now, what's the inside? - 2x. - 2x, that's the inside. You'll notice one of your clues here is parentheses. Parentheses, all right? That's the inside, the outside function? It's the sine function. So we start by saying, okay, what's the derivative of sine? It's cosine. So we say, okay, derivative of the outside is cosine, plug in the inside, we leave that alone. Times that by the derivative of the inside, derivative of 2x? Two. Ah, three for three, wow. And if you might remember, it was not so easy when we did these before. Okay, and let's try one more. The derivative of tangent 1/2 x. We just barely did this one last time. Okay, the inside, 1/2 x. So going through the process, the outside function is tangent, derivative of tangent? Secant squared. So secant squared of 1/2 x, and then times 1/2. Now, if you go back to the notes from last time, you might remember it took us about a whole page to take the derivative of tangent of 1/2 x and end up with 1/2 secant squared of 1/2 x. So this is a little bit faster, so maybe it's better. Okay, so let's do practice. The more we practice, the better we get. And the chain rule is something that you wanna make it be automatic. You don't even want to have to think about it, because you just want it to be ingrained in how you go. It's a very important rule. It sounds like I'm joking when I say it's the best rule, but it really is probably the most important rule, and it's the foundation of a lot of things that will follow later on. So the derivative of sine of x squared plus e to the cosine x. Now, one thing is, we don't even think about this as a rule anymore, but when we're taking derivative of sine of x squared plus e to the cosine x, what rule do we always use? The sum rule. Yeah, the sum rule. So in other words, we can take these derivatives separately. So I can take the derivative of sine of x squared, and I can take the derivative of e to the cosine of x. All right, so for sine of x squared, all right, well, I think the parentheses give it away. The inside is x squared, the outside is the sine, so taking the derivative of the outside, cosine, to what would we plug in? x squared, times by? 2x, derivative of the inside is 2x, plus? Okay, now here, there's no parentheses, but what's the inside function? Cosine x, the outside function? It's that exponential e. All right, derivative of e to the x? It's e to the x. What do we plug into it? Cosine, and then? Times negative sine on the outside. Now, you'll notice parentheses. Be careful. Sometimes, if you have left off your parentheses, it would look like either cosine x subtracts sine x, and you know, we wouldn't wanna make that mistake. We might lose points, but we wanna lose points in a good way, for good mistakes. Okay, well, now let's try a problem. Find the tangent line to y equals e to the power of x to the fourth minus x at x equals one. So if I'm after a line, what do I need? I need a slope and a point, a y value. Now, we already have the x value. So maybe we can finish off our point. How do we get the y value? Plug in one. So e to the power of one to the fourth minus one. So the x to the fourth minus x is what's in the exponent. So that would be e to the zero, 'cause it's one to the fourth minus one is zero. e to the zero, also known as one. All right, for our slope, what do we need? The derivative. So what is y prime? How do I start it? Yeah, so chain rule, derivative of the outside. e, what goes in the exponent? - x to the fourth. - x to the fourth minus x. Okay, are we done? - No. - No, 'cause remember, it's not just sort of e to the x, it's e, but something's been plugged into it, so you have to do some other stuff. Okay, what's left? Multiply by four x cubed minus one, that being the derivative of x to the fourth minus x, and again, parentheses help us make sure we don't lose anything. Now, is this our slope? Not yet. What do we need? Plug in one. So y prime at one, that's our slope. Well, we're gonna get e to the zero times four minus one, which simplifies to three, 'cause one times three. So our slope is three, and now we're like, woo-hoo, we have our data that we need to make a line, y minus our y coordinate, so y minus one, that's the y coordinate, is equal to our slope, three, times x minus our x coordinate. Well, that also turns out to be one in this case. We could leave our answer like that, or we could say this is y equals 3x, multiply the three through, then you get a minus three over here, you plus one more, there'd be 3x minus two. And there's our line. Well, all right, these have all been pretty straightforward, but you can imagine things can get more interesting. Let's try something a little bit more interesting. So we've talked about things before, which we call these sort of onion problems, problems with layers. And well, what do you do? Well, you work through it one layer at a time. So you just have to think about it. So this is a case that we have functions. So if we think about it, our outermost function is tangent. What's our next layer? Square root. And then we sort of split into two roads. Then we have an exponential with a polynomial. And then on the other side, we have a sine with a polynomial. Ah, exciting. Now, before we take this derivative, let's rewrite it like we would if we were in calculus. When I say like we were in calculus, what's the way that this is currently written that doesn't have a very calculus feel to it? Yeah, the square root symbol. See, calculus really likes the exponent notation, 'cause it's very easy for us to work with the exponent notation. So oftentimes, when we see a square root, we'll say, look, let's rewrite it using exponentiation. So e to the x squared plus five, plus sine of x squared, that's all to the 1/2, inside that tangent, which we're taking the derivative of. Now we're ready. Okay, so when you have a problem like this, the key is don't panic. Don't panic, just work it through a layer at a time. So let's work it through. Start and work on the outside, and go in. So the outside is tangent. What's the derivative of tangent? Secant squared. So we start by saying, okay, it's secant squared of. Now, the next thing is cut and paste. So e to the x squared plus five plus sine of x squared. And this is all to the 1/2 power. Okay, so we said we've written down f prime. So tangent is our f, f prime. Here's our g of x. So f prime of g of x. Now watch our next thing. So the next thing is to multiply by the derivative of this part. So you multiply by the derivative of the inside. And so you say, wait a second, the derivative of the inside, that's another chain rule. And that's okay, you just keep going. It's chain rules all the way down. Okay, so we're gonna multiply this by, now we're gonna take the derivative of this expression. Okay, so how do you take the derivative of something to the 1/2 power? It's a chain rule, what's the outside function right now? To the 1/2 power. You can think about it as a square root function. That's our outside function. How do you take the derivative of something to a 1/2 power? The 1/2 comes down, 'cause think about it as like a derivative of x to the k, derivative of x to the k. In fact, we wrote that down at the very beginning. We said derivative of x to the k, what you do is you bring the k down, and then x to one less power. So here, this is our k, so it comes down, and then we're gonna have the inside stays the same, 'cause that's how the chain rule works, the inside stays the same. e to the x squared plus five plus sine of x squared, and then we subtract one to the negative 1/2. All right, are we done? No, what are we missing? Well, what we're missing is by the chain rule, we've gotta take the derivative of the inside. So we gotta multiply that expression by this derivative. Now this derivative, we're getting closer. Our onion is nearing the core. Okay, derivative of e to the x squared plus five? e to the x squared plus five, and then? Times 2x, and then sum rule. So plus, okay, derivative of sine? - Cosine. - Cosine of? x squared, leave the inside alone, leave it alone. But then we multiply by? - 2x. - By 2x, okay. And then what happens? And then we're done, we're done. So this derivative is this whole expression. I didn't have enough space, otherwise, it would've taken up the width of the screen. And there's some other stuff we're gonna do over here. So this times that times that. But the key is just work it through, one layer at a time. What's the rule I need to do right now? What derivative am I taking? And if you do that, it's all a matter of patience, and just applying your rule bit by bit by bit. All right, let's find the derivative of x squared plus 3x squared in two different ways. Now, how are we gonna do it in two different ways? Well, first by expanding, and second, by the chain rule. So what does it mean by expanding? So what that means is we can actually just multiply this out before we take a derivative. So x squared plus 3x quantity squared, that'd be, I think that says a plus b squared. So a squared plus 2ab, that becomes 6x cubed, plus then b squared, 9x squared. So first we expand, and now we say, hey, these are all simple things to take the derivative of. So derivative of x to the fourth? 4x cubed, derivative of 6x cubed? 18x squared, and derivative of 9x squared? 18x. All right, now, you wanna compare that to what if we did it by the chain rule? So by the chain rule, that would be the derivative of x squared plus 3x, squared. Well, the inside is x squared plus 3x, the outside is the square function. So just like over here, where we had the 1/2, what we do is the two comes down, x squared plus 3x to the one. And then what do we do? - [Student] (faintly speaking) - [Instructor] Multiply by the root of the inside, which is? - 2x plus three. - 2x plus three. All right, now, wait a second, I thought, huh? Are these the same? Should they be the same? They should be the same, because if you're taking the derivative of something, there's only one derivative. It's not like, oh, if you take derivative by this method, you're gonna get a really nice slope. But this other method, oh, you're gonna get a crazy slope. But they might have a different expression. Let's actually check. To check if they're the same, let's do some expansion. So x squared times 2x would be 2x cubed, but then there's another two, so that would be 4x cubed. Then we have x squared times three, and then plus 3x times 2x. So that's 3x squared plus 6x squared, which is 9x squared times two. Well, that's 18x squared. 3x times three is 9x, times two is 18x. So the good thing is it's consistent. In fact, this is true in math. If you have two techniques to answer the question, as long as you properly carry them out, you will get an equivalent answer. Like I said, it may take a while, you may have to manipulate it, but you'll get an equivalent answer. Now, what's the point? Both methods right now are about the same level of difficulty. But notice that there's a second part to this question. Find the derivative of x squared plus 3x to the 100. using one of the two preceding methods. So the derivative of x squared plus three x to the 100. Now, our two preceding methods where we could expand this out and take up three or four pages, and then we could take its derivative. And fingers crossed, if we haven't made a mistake, and I will tell you that the calculators that most people have would run out of digits to show the numbers that would show up in the computation. But in theory, it works. But in practice, not very good. But now, let's think about it if we did the chain rule. So expansion, ugh, not so great. Chain rule, can we do it? What would it be? 100 times? x squared plus 3x to the what power? - 99. - 99, and then? - 2x plus three. - 2x plus three. So if you were to actually expand it out, take the derivative term by term and then refactor it, you would get this expression. But this seems a lot simpler. So it's good to have a different method. Now, again, I wanna compare, I'm gonna take a derivative of e to the 31x, but I'm gonna take it in two different ways. So you should check that e to the 31x I claim is the same as e to the x to the 31. And the reason that's true is if you have a power to a power, you'd multiply the two powers together. So let's start by this first one, the derivative of e to the power 31x. Now in this case, what function is the outside function? - e to the x. - The exponential e to the. What's the inside function? - 31x. - The 31x. So the derivative of out function, sorry, the derivative of the outside is e, leave the inside alone, and then you multiply that by 31. So compare that to the other case, where we have e to the x to the 31st power. Now in this case, what's the outside function? Raising to the 31st power. What's the inside? e to the x. So it's a different expression. So what happens now is the functions are, the inside and outside have changed. Hmm. It would be amazing if we got the right answer, or the same answer, but it's math, so we're very suspicious it should work. All right, so derivative of x to the 31st power, what would you do? The 31 comes down, and then you raise it to the 30th power. But what goes on the inside? The e to the x. So this is the generic derivative of x to the k, you know, k, something to the k minus one. The something is our inside function, it's e to the x. Are we done? No, what do we have to do? Multiply by e to the x, which is the derivative of the inside. And if you look at this, lo and behold, you can think of this as e to the x of 30, that's e to the 30x. e to the x is e to the x. And if you have e to the 30x times e to the x, you can simplify it by adding the exponents. And when you do that, what do you end up with? e to the 30x plus x, which is 31x. And you'll see they both agree. So again, you have options. Things work, work really well. All right, good. Now, let's get a little bit more abstract. So given that y equals 4x minus six is tangent to y equals f of x at x equals two, find the tangent lines to the following functions at x equals two: y equals f of x squared minus x, y equals, in parentheses, f of x quantity squared, and y equals f of f of x. All right, so we wanna go through and look at all of these, hmm. Now, this is a good time to sort of remind ourselves of a general philosophy, which I don't know if we've explicitly stated yet, but just in case we haven't, let's explicitly state it right now, so that we're like, okay, we know for sure that this is true. So here's our sort of overarching philosophy. If we have a tangent line, let's say to y equals h of x, at a point x equals a. So that's sort of my data. I have this tangent line. This is a tangent line to a function at a particular point. What's happening is that that tangent line is actually storing two pieces of information about the function h. Two pieces, no more, no less. So if I have a tangent line to a function at a particular point, I claim there are exactly two things I can say about that function and nothing else. Now, when I say something about the function, I mean values about the function. So if I know the tangent line to h of x at x equals a, what do I know? - (faintly speaking) - Right, but think about it in terms of h, express it in terms of h, what do I know? Okay, so h prime of a corresponds to that rate of change. That gives us the slope. So I'm gonna add that onto our list. That's one of two things that we know. What's the other thing that we know? - h of a. - h of a. And the reason we know h of a is it comes from the word tangent. Tangent means I'm touching. So I have to match what the value of the function is. And here's the great thing. The tangent line stores that information, but if we have that information, we can recover the tangent line. So you should think of the tangent line as saying, oh, I've coded up information about the function and the derivative at the point of tangency. So we can apply it in our case. So in our case, we have our function is f of x, and our point is at two. So let's write down what we know about our function f so we can have it for reference. So according to this, our function is f, our point is two, we should know what is f of two and what is f prime of two. Let's start with f of two, what is f of two? What? It's two. Now, why is it two? Plug in x equals two into your tangent line, 'cause it has to match. Two times four is eight. Eight minus six is two, what is f prime of two? Four, and how do we get that? You can take the slope. If you want, you could even say, oh, I just take the derivative of 4x minus six and I get four. Now, that actually turns out to be the same. So they could have phrased the problem, instead of saying given that y equals 4x minus six is tangent to y equals f of x at x equals two, they could have said given that f of two equals two, and f prime of two equals four, then find these other things. So once you understand this philosophy, you can jump back and forth between tangent lines and these two pieces of data. Then a lot of the problems that you'll see in the future at some point, maybe, like tomorrow during the review as an example, will be much simpler. All right, so let's do this. We have three functions. So let's start with this function here. So I'm gonna call my function h of x, f of x squared minus x. And if I'm after the tangent line, what do I need? Well, I need to know what is h of two and what is h prime of two. So probably the simplest thing is starting with what is h of two? That's f of two squared minus two. Well, two squared minus two is two. So that's f of two, which is? Which is two. Okay, not the most exciting thing, but it's true. h prime of two. Well, if I wanna find h prime of two, that says I should plug two into h prime. Do I have h prime? Not yet, but can we find it? Yes, we can. We have the technology. So h prime of x would be, what rule do we need? - Chain rule. - The chain rule. Cha cha cha chain rule. Yes, all right. Derivative of the outside? - f prime. - f prime of? x squared minus x with the inside alone, times by? - 2x. - 2x minus one, the derivative of the inside. Now what do we need? We need h prime of two, so we should plug in two. So that would be f prime two squared minus two, and then we're gonna times it by two times two minus one. Well, this'll become f prime of two. f prime of two is four. Two times two minus one? Is three, four times three gives 12. So h prime of two is 12. And now, we have our two pieces of information. And so we can reconstruct the tangent line. So we say, okay, y minus the y coordinate two is equal to the slope, 12, times x minus the x coordinate, which is two. So there's the first one. So for f of x squared minus x, that's what we get. All right, now let's do a different one. f of x squared. So I'm gonna call h of x just for convenience f of x squared. And remember, we are still playing the same game. We want h of two, we want h prime of two. So how do we begin? Plug in two. So h of two would be f of two squared. Well, f of two is two, so two squared, or four. So h of two is four. Good, progress. Now, next step? Find the derivative. 'Cause we need h prime, so we better figure out what h prime is. All right, what rule do we need to take this derivative? Yeah, cha cha cha chain rule. All right, so derivative of f of x squared will be? Two f of x to the one. Oftentimes we don't write it down, but we'll go ahead and just get into the habit. Anything else? Times f prime of x. Next? - Plug in two? - Plug in two. So that'd be two times f of two times f prime of two. Well, two, f of two? - Two. - Two, f prime of two? - Four. - Four, which leaves us with a grand total of? 16. So again, y minus our y coordinate, which is four, is equal to our slope, 16, times x minus our x coordinate, which is the two. All right, one to go. Okay, for our last one, we have f of f of x. All right, so again, I'll just call h of x, f of f of x. And what is h of two? That's f of f of two, but f of two is two. So that simplifies the f of two, which is two. Okay, so all right, h of two is two. That's fine, that's fine. h prime of two? I don't know that yet. What is h prime of x, just in general? Okay, f of f of x. f prime of f of x, good. Anything else? Times f prime of x. Plug in two, h prime of two, that's f prime of f of two times f prime of two. And so you just, there's no rush, you just start evaluating. f of two is two, and then you have another f prime of two, and f prime of two is four, and four, so this would make 16. So what do we get? y minus the y coordinate, which is two, is equal to our slope, 16, times x minus the x coordinate, which is two. And there's our line. Okay, really close to the end, but I think we all, we squeezed it in there. All right, well, do a little bit more interesting things. So given the table below, determine h prime of x, and I think this is a kappa prime, but it looks kinda like a k, so maybe I'll say k. k prime of x for x equals one, two, three, and four. Given that h of x is f of g of x, and k of x is g of f of x. All right. Well, this is a case where there's several things that we can do. When we're giving information about the function, one of the things we can do is give an explicit function. In other words, say, here's this function, take its derivative. Something else we can do is you can say here's a tangent line, and the reason we can do that, it's like the previous problem. It says, aha, tangent line is encoding information about the function, and then the derivative. Alternatively, you can just say, oh, here's a bunch of information about the function and the derivative. The other thing we can do, and we won't have time to do it today, we'll do it next time, is we can do pictures. We can say, look, here's the picture of the function, and let's extract information. So you should be comfortable working in all these formats, but in each case, what it all boils down to is saying if I really care about my tangent lines, I'm really just saying, what are these two pieces of information? That's what I care about in all cases when I'm after tangent lines. All right, well, hmm, determine h prime of x and k prime of x, so h of x. So the nice thing is you don't even have to find a tangent line, we just have to find a bunch of derivatives. So h of x, that's f of g of x. So what is h prime of x? So derivative of the outside, f prime g of x, and then? g prime of x. And so we get to find the derivative for a bunch of points. So this is gonna be a little bit of practice here of learning to read a table. Isn't that exciting? Ah, so if I wanna find out one, I plug one in for everywhere I see an x, and now I start the table read. g of one. So here's g, g of one? It's one. So this is f prime of one times g prime of one. Well, f prime of one? Two, g prime of one? Four, two times four is eight. All right, h prime of two, that's f prime of g of two times g prime of two. g of two? Three, so this is f prime of three times g prime of two. f prime of three? One, over there. g prime of two? Two, one times two? Two. h prime of three? Well, that would be f prime of g of three times g prime of three. So I'm just plugging three into my function for h prime of x. g of three? Two. So this is f prime of two times g prime of three. f prime of two? Four, g prime of three? Three, giving us 12. And h prime of four? Well, that's f prime of g of four times g prime of four, and g of four? Four, that's f prime of four times g prime of four. Well, f prime of four is three, g prime of four is one. Three times one gives three. Now, we're actually only halfway done, right? We just did h prime. What else do you have to do? - k prime. - k prime. Well, if you're gonna do k prime, and we won't have time to finish it, but you notice it's g prime of f of x times f prime of x, let's just do one value. k prime of one would be g prime of f of one times f prime of one. f of one is three. And so we have g prime of three, f prime of one. g prime of three is three. f prime of one? Two, which gives us six. The point here is, look, they're not the same. Six doesn't show up anywhere. So you gotta really pay attention to order. Order matters. f of g of x, g of f of x, they are not interchangeable. All right, well, this is a good place to stop, and we'll do some more next time.