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Problem number six. Let f of x is equal to sine of x squared plus cosine of x. The graph of y is equal to the absolute value of the fifth derivative of f at x is shown above. And I haven't shown it here just so we have some space. I'll show it when we need to show it. I think we have to show it in part D. So first, let's do part A right over here. Write the first non-zero terms for the Taylor series for sine of x about x is equal to 0. And write the first for non-zero terms of the Taylor series for sine of x squared about x equals 0. So let's do this first part. And just a reminder, a Taylor Series is a polynomial approximation of a function. So just to give us a quick reminder. We go into much more detail in this on the videos about Taylor series. But if you have a function that looks like this, and if you wanted to approximate it with a Taylor series around 0, if you only have one term in your Taylor series, you would just literally have it-- it would just be a constant, just like that. If you have two terms in your Taylor series it would be a line that looks like that. If you have three terms in your Taylor series, you'd get to a second degree term. And you'd start approximating it something like that. If you get to the third degree term it might start looking like that. And as you add more and more terms, you get a better, better approximation of your function. And if you add an infinite number, it might actually converge to your function. So let's just remind ourselves, if I have a function-- if I have f of x, this Taylor series-- I can approximate it with this Taylor series. And if we want to center that approximation around 0, it'll be equal to f of 0 plus f prime of 0. The derivative at 0 times x plus the second derivative at 0 times x squared over 2 factorial. You could have divided this term right here by one factorial, which is just 1. Two factorials is just 2. And then plus the third derivative of f at 0 times x to the third over 3 factorial. Plus the fourth derivative-- I think you get the idea here-- at the fourth derivative at 0 times x to the fourth divided by 4 factorial, and so on and so forth. So what they want us to do is find the first four non-zero terms of the Taylor series for sine of x. And some of you might already know this. We actually cover this in the video where we show Euler's identity. But we'll do it again right over here. So if we just take-- I'll call it g of x, because they already defined f of x up here. So let's say that g of x-- let me do this in a new color just to ease the monotony. Let's say that g of x is equal to sine of x. Then we know that g of 0 is going to be equal to 0. And then if we take the derivative, g prime of x is going to be equal to negative-- no, it's going to be positive cosine of x. g prime of 0 is now going to be equal to 1. Cosine of 0 is 1. Then if you take the second derivative, the derivative of cosine of x is negative sine of x. Negative sine of x and the second derivative at 0 once again is going to equal zero, sine of 0, 0. And now let's take the third derivative. Third derivative of our function of g. The derivative of negative sine of x is negative cosine of x. And the third derivative at 0 is now going to be equal to negative 1. And we could keep going. You can already guess where this might lead to, but I'll just keep going just in case. The fourth derivative is once again going to be equal to sine of x. It is now going to be equal to sine of x. And then the fourth derivative at 0, because it's the same thing as the function, is now going to be 0 as well. So this is going to be equal to g to the fourth at 0. And then we could keep going. This is going to be the same thing as g to the fifth, because we start cycling as we take more and more derivatives. So this is going to be the same thing as the fifth derivative at 0. This is going to be the same thing as the sixth derivative at 0. And this is going to be the same. This is an equal sign right here, equals 1. And this is going to be the same thing as the seventh derivative at 0. So if you want the first four non-zero terms, let's just work through it. So first of all, f of 0-- I'll do this in a new color. If we want to approximate g of x, if we want to approximate sine of x. So we'll say sine of x is going to be approximately equal to. So this first term right over here, that is sine of 0. That's g of 0. That's going to be 0, so we don't even have to write it. Then we go to this term right over here. Well, the first derivative, f prime of 0-- or in this case, g prime of 0-- is going to be equal to 1. So it's going to be 1 times x. So you're going to have an x there. Then the next term is going to be 0. We see that right over there. Because the second derivative of our function evaluated at 0 is 0. Our third derivative of our function evaluated at 0 is 1. So that term is going to show up again. Actually, that term is negative 1. I don't want to make a mistake here. That is negative 1, negative 1. It was negative cosine of x. You evaluate negative cosine of x and 0, you get negative 1 right over here. So the third derivative, this thing right over here, is negative 1. So then we have minus x to the third over 3 factorial. The fourth derivative, once again, is 0. The fifth derivative evaluated at 0 is 1. So then you're going to have plus 1 times x to the fifth over 5 factorial. Then the six derivative is 0. So that term is going to disappear. I didn't even write it up here. And then the seventh term, the coefficient is negative 1. Or the seventh derivative evaluated at 0 is negative 1. So you have negative 1 times x to the seventh over 7 factorial. And we have to go all the way to the seven degree term to find the first four non-zero terms for our Taylor Series. And so we're done with at least the first part. We found the first four non-zero terms of sine of x Now, what about sine of x squared? We have to be careful here. Because you might just say, OK, let me just apply this formula. And what you're going to find very quickly is when you start taking the second and third derivatives of this thing right over here, it's going to get really messy. But what you can say is, look, sine of x is approximately this thing right over here. What happens if I just replace the x with the x squared? Then I get sine of x squared is approximately equal to-- instead of an x here, I'm going to put an x squared. Instead of an x to the third here, I'm going to put an x squared to the third over here, over 3 factorial. Instead of an x to the fifth. I can put an x squared to the fifth power over 5 factorial. And instead of an x to the seventh I could write x squared to the seventh power over 7 factorial. So this is a very important thing to realize. Because if you had started directly take the Taylor Series around 0 of this thing right here, you would have taken up all your time trying to take the derivatives. And you probably wouldn't have been able to do it anyway, because it would have gotten really messy. And the key here is just to realize that you substitute x squared for x, you're then going to get the approximation for sine of x squared. We could simplify this a little bit. This is going to be approximately equal to x squared minus x to the sixth over 3 factorial plus x to the 10th over 5 factorial minus x to the 14th over 7 factorial. So that's our second part of the problem.