 Welcome back to our lecture series Math 12-10, Calculus I for students at Southern Utah University. As usual, I'll be a professor today, Dr. Andrew Misseldine. This is the second part of a two-part lecture that we began in lecture 22. This is the start of lecture 23, and this two-part lecture has been about the chain rule. If you haven't yet watched lecture 22, are you're otherwise unfamiliar with the chain rule? It would be my recommendation you take a look at that video right now, before we begin proceeding with this video on some more examples of the chain rule. Because what I actually want to do in this video is give you the viewer an opportunity to practice the calculations with the chain rule on your own, and then we can work together with the solution. In this video, we're going to take a couple of different functions and compute their derivatives. Each and every one of these functions will require the chain rule in its calculation. Let's consider this first example, y equals x cubed minus 1 to the 100th power. What I would recommend to you as the viewer is to right now pause your video and try to work this thing out by yourself. No, seriously, pause the video. I know you think I'm going to give you the answer. I am, but the point is for you to try it out yourself first. So be brave, pause the video. We'll do it in just a second. No, seriously, how many times do I have to ask you to pause the video before you actually do it? That's what I'm seriously expecting us to do. Pause the video and we'll do it together in just a second. All right, let's take a look at the derivative of y equals x cubed minus 1 to the 100th power. So the inner function we can see right here to be x cubed minus 1, and this sits inside of the power function x to the 100th power. And so you often see this when you work with these functions here that the inner function is going to be set apart by these parentheses. If you see parentheses in your function notation, it probably means the things inside the parentheses we should visualize as an inner function. And so when we calculate the derivative y prime right here, we're going to take the derivative of the outer function. By the power rule, the derivative of the outer function would be 100 times x cubed minus 1 to the 99th power. So notice what happened here. I brought the exponent out in front. They gave us the coefficient of 100. Then we also lowered the exponent by 1. What goes inside? What am I raising to the 99th power? The function we already had here, which was the inner function for which now if we take the derivative of the inner function x cubed minus 1, its derivative also by the power rule, we're going to get just a 3x squared. Notice that the derivative of a constant is going to be 0 there. So it just disappears. So if you want to put the 3 and the 100 together, maybe 300 times x squared times x cubed minus 1 to the 99th power, that's how I would write the derivative of this function right here. It's to a great advantage to leave derivatives factored. Multiplying that thing out is not going to be worth our time, especially since it's the 99th power. But in general, leave derivatives factored. That's how we want to have them. So don't bother multiplying that out. Leave it in the way we see it in this box here. Let's look at another example here. Let's calculate the derivative of f of x, which equals 1 over the cube root of x squared plus x plus 1. So before I show you the solution, take a few moments. You don't take as long as you need honestly. Pause this video and try to work this out on your own. You are going to use the chain rule in this calculation. So let's try this again together. I think the best thing to observe is, before we start calculating the derivative, is I could write this as a power function, well at least a composed power function. Notice we have a cube root of something in the denominator, have a 1 in the numerator. This tells me that keeping the quadratic x squared plus x plus 1, we could write this to the negative 1 third power. The 1 third power because we have the cube root and then the negative because we're in the denominator. So the reason I like this is that if you use the negative 1 third power, we can avoid having to use the quotient rule, which is a little bit more cumbersome, I would say, than the power rule in this situation. So when you make that recognition, you're still going to see your inner function right here, x squared plus x plus 1. Notice it's the function inside the parentheses. Even if you didn't have the parentheses, if you have something like the cube root of x squared plus x plus 1, it's very nice to identify this inner function as the function inside of the radical. After all, the inner function, we look inside to find it. The outer function is then a power function. Again, we get negative 1 third as that power. So when you compare this to the previous example, we just did a moment ago, the two examples really are not that different. We have a polynomial inside of an exponent, and so the chain rule will help us with the derivative there. So by the chain rule and the power rule, we're going to first take the outer derivative, which we bring down the coefficient. That is, the coefficient was the old power of negative 1 third. Sitting inside the parentheses was the same interfunction. That part doesn't change. Then we then have to lower the power by 1. If we take negative 1 third minus 1 from it, well, we could actually subtract 3 thirds, because that's the same thing as 1. And so we see that the x component here should actually be a negative 4 thirds. That's the outer derivative. You don't want to forget the inner derivative, though. We still have to take the derivative x squared plus x plus 1 with respect to x right here. And so doing that, I'm just going to copy this thing down again, negative 4 thirds power. The derivative of x squared plus x plus 1, the derivative of x squared will be a 2x, derivative of x is a 1, and the derivative of plus 1, that's since it's a constant, it'll just go to 0. So we get this right here. Since the original function was written as a fraction with radicals, I'm going to write our derivative with that same style in mind here. So we're going to have this 2x plus 1 in the numerator. The denominator, you're going to have this factor of 3, and then you're going to get this x squared plus x plus 1. If you want to raise this to the 4 thirds power, that would be perfectly fine. But like I said, if you want to use more of like a cube root notation, we can write this as the cube root of x squared plus x plus 1, and we're going to raise that to the fourth power. It should be in the denominator, though. So I'm trying to make the derivative resemble the original function in style and theme in that regard. Let's take a look at another example. g of t this time equals t minus 2 over 2t plus 1 raised to the ninth power. Take a moment by yourself, pause this video, and try to work this derivative out on your own using the chain rule. So let's try this one together here. We can see that similar to the previous two examples that we have our inner function, it's marked off by these parentheses right here. Our inner function is going to be this rational function, t minus 2 over 2t plus 1. The outer function is, again, a power function of some kind. So to calculate the derivative, we want to calculate dg over dt in this situation. We have to take the outer derivative first. So we get 9. Then we're going to get 2 minus 2 over 2t plus 1, and we're going to then lower the power to the eighth power. Again, we do that by the power rule. Now we have to take the derivative of the inside function. So we're taking the derivative of t minus 2 over 2t plus 1. So the quotient rule is going to come into play here. So we get, remember in our poem, our low d high minus high d low square the bottom. Here we go. So what happened here? So low, of course, is the bottom function, the denominator, 2t plus 1. One was the derivative of the top. The derivative of t minus 2 with respect to t will be 1. Likewise, we have the top function, t minus 2. And then the derivative of the denominator will be just a 2. Again, taking derivative with respect to t. So combining, let's try to clean this thing up a little bit. We have this coefficient of 9 on top. We have a t minus 2 to the eighth power. Then if we try to put these things together, what do we have there? We have a 2t plus 1. We also have a minus, let's see what we have there, 2t minus 4. I distributed the 2 already. And this sits above. Well, in the denominator, you have a 2t plus 1 to the eighth power, whoops, to the eighth power, which we got from here. But we also have this 2t plus 1 squared coming from over here. We're going to combine those together and get 2t, whoops, 2t plus 1 to the 10th power in the denominator. I'm going to leave that. Again, I want the numerator to be as factored as possible. So 9 times 2t to the eighth. And so what we have here, you'll notice there's a 2t that cancels those 2t. So they're gone. Then you're going to get a 1 plus 4, 1 plus 4. So that's equal to a factor of a 5 right here. I'm going to put that together with the 9. And we're going to then get 9 times 5 times that by t minus 2 to the eighth all over this 2t plus 1 to the 10th. So say that again, our derivative turned out to be 45 times 2, or t minus 2 to the eighth power divided by 2t plus 1 to the 10th power. So in this one, we have to use the chain rule and the quotient rule together. So let's look at another example here. Let's calculate the derivative where y equals 2x plus 1 to the fifth times x cubed minus x plus 1 to the fourth. So I want you to, again, try to calculate the derivative of this on your own. But think about, in addition to the chain rule, what might you need to calculate the derivative of this function that we see on the screen? So let's try it together now. If we want to compute the derivative of y prime, we see that first of all, we have a product of two things, right? Notice we have a 2x plus 1 to the fifth power times that by an x cubed minus x plus 1 to the fourth power. So the first thing we want to do is apply the product rule. So we're going to take the derivative of 2x plus 1 to the fifth power times that by x cubed minus x plus 1 to the fourth power. And we're going to add that to it, add to it the other possibility. Oh, I forgot to write the derivative. We're going to take the derivative of this friend right here, the first one. Next, we're going to take the 2x plus 1 to the fifth power. And then we're going to take the derivative of x cubed minus x plus 1 raised to the fourth power. So let me emphasize that we were taking the derivative of this one right here. Now, with those here, the functions we have to now take the derivative, that's where the chain rule comes into play here, right? We have our inner function 2x plus 1. We have our outer function, which is the fifth power. For the other one, we have the inner function of x cubed minus x plus 1. That's our inner function. Our outer function is the fourth power right there. So let's compute these derivatives individually. So taking the derivative of 2x plus 1 to the fifth power, we're going to get 5 times 2x plus 1 raised to the fourth. So we lower the power by 1. Then this is the part you don't want to forget. You have to take the derivative of the inner function, the derivative of 2x plus 1. That's going to be a 2. And then we're going to times that by the x cubed minus x plus 1 to the fourth that we already had. Next part, we're going to have a 2x plus 1 to the fifth power. Now taking the derivative of the next part, we're going to get a 4 times x cubed minus x plus 1 raised to the third. So we lowered the power by 1. The coefficient came out in front like so. Now we have to take the derivative of x cubed minus x plus 1, the inner derivative, which we see is going to be 3x squared minus 1, like so. So this does give us the correct derivative. What I would like to do next is to attempt to factor this thing, because again, we'd like our derivatives to be factored. So what can we offer from these two products here? Notice they both have a 2x plus 1. The first product has it to the fourth power. The second one has it to the fifth power. We can't take more than what the lesser one has. So we're going to take out 2x plus 1 to the fourth. We have a x cubed minus x plus 1 to the fourth in the first product. We have a x cubed minus x plus 1 to the third power there. Again, taking the lower power, we're going to take 3. And then also notice we have a factor of 2, and 4 itself can factor as 2 times 2. So we can take out a 2 from both of these things. That's going to be our greatest common divisor there. So let's take out the 2, the 2x plus 1 to the fourth, times x cubed minus x plus 1 to the third. So what's then left behind? Well, we left behind. So if we're looking at the first product right here, we didn't take away the 5, so we got that. We took away all the 2x plus 1s, we took away the 2, and we took away all but one of the x cubed minus x plus 1s. That's all that's left behind there. Looking at the second product now, we took away 4 of the 2x plus 1, so there's still a 2x plus 1. We took away one of the 2s, we left another 2, so there's still a 2. We took away all of the x cubed minus x plus 1s, but we also still have this 3x squared minus 1, like so. So again, in this process of cleaning up, I would suggest we try to expand this piece right here a little bit. Distribute the 5. We're going to get 5x cubed minus 5x plus 5. Here, if we distribute the 2, we're going to get here 2x plus 1, times that by 6x squared minus 2. We got to foil that thing out there. I'm just going to copy this part down. Foil that out. We're going to get a 12x cubed minus 4x plus 6x squared minus 2, and so what like terms can we find? There are x cubes together, and so at this stage, I'm going to actually write down back in Y again where we're at 2 times 2x plus 1 to the fourth, times x cubed minus x plus 1 to the third, and so now when we combine the x cubes together, we get 5 plus 12, that's going to be 17x cubes. Let's see. In terms of x squares, do we have any? There's one right here, so then there's not other ones to count, right? We already took out the x cubes. We're going to take away this x squared, so we get a 6x squared. How about the linear terms? What do we have there? There is a negative 5x and a negative 4x. That'll combine, of course, to give us a negative 9x. That doesn't look like a 9. There we go. We did it that time. And then lastly, we have a 5 and we have a negative 2. That combines to give us a plus 3. So a little bit of arithmetic there, algebra and play there, but this is what the derivative would look like in its factored form. We're not going to worry about factoring that cubic polynomial. It's irreducible. All right, let's take a look at one last example in this video where, again, I want you to give it a good try before you see me do it. We want to calculate the derivative of y. We have e to the x squared plus 1. That's the exponent of the e times it by the square root of 5x plus 2. We can see there's some products and compositions of functions going on there. Pause the video for a moment and give it a try yourself. All right, trying this together. Notice we have a product of two things. You have the exponential function times the square root function. So the first thing we're going to do is use the product rule. y prime is going to look like e to the x squared plus 1 prime times the square root of 5x plus 2. And then the next one, we're going to get e to the x squared plus 1 times that by the square root of 5x plus 2 prime. So we have to take the derivative of these things. So with the first one, what's the composition? The chain rule is going to come into play now. Our inner function, when you have an exponential function, is the x one itself, x squared plus 1. The outer function is the natural exponential, which if you take the derivative of e to any power, you're going to get back itself. So when you take the outer derivative, you're going to get e to the x squared plus 1. When you take the inner derivative, that is the derivative of x squared plus 1, you're going to get a 2x. Then you're going to multiply that by the square root of 5x plus 2 that was already there. Then the next one, you don't do anything with the e this time because we're not taking its derivative. But for that square root of 5x plus 2, notice the thing inside of the square root is the inner function, 5x plus 2. Then the outer function is the square root function, or if you prefer, it's the one-half power. So by the power rule, we're going to get one-half times 5x plus 2 to the negative one-half power. And then we're also going to multiply that by the inner function, which is 5. That is the derivative of the inner function, the inner derivative. The derivative of 5x plus 2 will be a 5. And so try to rewrite this thing a little bit. What are some things we can see here? We're going to have a 2x e to the x squared plus 1 times the square root of 5x plus 2. Really not much you could do there. But the next one, we're going to have a 5e to the x squared plus 1. That's in the numerator and denominator. We're going to get 2 times the square root of 5x plus 2, like so. So now it's going to be my goal to try to factor this thing. Again, we like things to be factored when it comes to a derivative. So what can I afford to take out? Both of the terms, we have this e to the x squared plus 1, e to the x squared plus 1. We're going to factor that thing out, definitely. So we're going to take out e to the x squared plus 1. But the other thing I want to take out is I want to take out, well, let's say it this way. I want to take away this 2, right? Basically, I want to factor out the whole denominator. So how can I afford to get a 2? Well, here we have a 2, but a 2 is the same thing as 4 halves. Notice, I just wrote that as a fraction. And so that way, if we take out the 2, right, that'll then leave behind with these other terms. The first group, you'll actually get a 4x now. And we'll come back to the square root in a second, what's going on there. And then for the second one, since we already had a 2 in the denominator, you're going to end up with a 5 right there. But so what are we doing in the denominator? Keep on going here. We want to also focus on this square root of 5x plus 2. I want to factor that out. So again, I'm factoring everything away here. So how do I accomplish that? Well, what I have to do is I have to think of the first one. If I have a square root of 5x plus 2 on the bottom, that means I must have a square root of 5x plus 2 on the top. Which this is actually fortuitous for us. Because when you take the square root of 5x plus 2 and times by itself, you end up with just a 5x plus 2 with no square roots anymore. And you'll still have, of course, the one in the denominator. Like so. So this does give us the factored form of these type of expressions here, for which then the thing in the parentheses could be expanded and simplified. We have e to the x squared plus 1. Distribute the 4x. We're going to get a 20x squared. We're going to get a 4x times 2, which is an 8x. And then there's that constant 5. There's nothing for it to combine to. So there we go. We're already good to go. And then the denominator will look like 2 times the square root of 5x plus 2. So this derivative is now calculated. And we can see that it's factored. The numerator is factored. The denominator is factored. And when I say it's factored, if there's fractions and all, that means we just want a single fraction. We should add all of those things together so that we could then identify very quickly from this function. What are the horizontal, where are the horizontal tangents of our function? Where are the vertical tangents of our function? Which is something we've talked about here or there. And we'll talk about this, of course, more in future videos.