Calculus: Higher-Order Derivatives, Linear Approx

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Practical Application of the Derivative
Linear Approximation
Higher Order Derivatives and Linear Approximations Page [1 of 1]

From that last example of the book falling, we saw that, in fact, we can get velocity by taking the position function and differentiating it, thus producing velocity. And then we saw something new. We saw that if we now take the derivative of velocity that gives us the acceleration function. And if you think about it, the derivative of velocity is just the derivative of position with the derivative of that. So, in particular, we need to take two derivatives of positions to boot us up to acceleration. Position -- one derivative gets us to velocity. And if we take the derivative once again, which would be the second time we take the derivative of this function, we get acceleration.

So, that actually inspires to have a fun idea, mainly the idea of higher order derivatives, which just means taking more than just one derivative. You know, derivatives sometimes, are like potato chips, you just can’t do it once. So, for example, suppose you have f of x equals, oh, I don’t know, let’s say, x to the fourth minus three x squared plus one. Let’s take the derivative, piece of cake, four x cubed minus six x plus zero. But now, this itself is a function and we can just take the derivative of that. If we take the derivative of that what would that be? Well, it’s the derivative of the derivative. So, I have to write sort of, f prime, that’s the function. But now I’m taking the derivative of it. So, I actually put another prime there. So, I call this the second derivative of the original function, and it’s just the derivative of the derivative, which would be twelve x squared minus six. In fact, I could take the derivative of this, which would be the what derivative of this? It would be the one, two, three, it would be the third derivative. And I would mark that, one, two, three, the third derivative, which would be, in this case, twenty-four x. I can take the fourth derivative. Now, of course, when it gets to four, after a while these things start to actually, start taking up some room on your paper. So, what we do now with four and higher derivatives, we actually just denote them by putting a little parenthesis and putting the number of the derivative up there. In this case, it would equal just twenty-four. So, if it’s the fifth derivative, I’d just write a little five in there or whatnot. Anyway, just taking successive derivatives, the derivative of something and then take that something, and take the derivative of it. That gives you the next derivative. So, you can do this all you want.

Let me just show you another little piece of notation. Suppose that, instead of using this notation, I use the y notation. Look how nicely this is going to work. Suppose that I look at y equals x, let’s say cubed, minus two x to the fourth plus three. If I take the derivative we write that as dy/dx equals three x squared, minus eight x cubed plus zero. Now how would I take the next derivative? What would I write? Well, if you think about it for a second and think about what it means when we take a derivative I’ve got to now give you the command differentiate with respect to x. Remember that? That’s the command to take the derivative. And if you look at that what symbol should I use for that derivative of the derivative. Well, here’s what we adopt. We say the second derivative of y with respect to x. In this case, it would just be six x minus twenty-four x squared.

Now, notice how we wrote that. See, there’s two d’s, so we sort of put a little d squared right there, one lone y, so that remains there. But here we have a dx to the dx, so we put d x squared. So the notation is d two y d x two, and it’s read in English, if you’re reading a children’s story, the second derivative of y with respect to x. And you can see where this interesting but perhaps unusual looking -- if you don’t think about it, notation comes from. Why the two is here and then over here, and the answer is readily available, once you realize what we’re doing is differentiating with respect to x. The verb in front of this noun gives me this. And, of course, the third derivative I think you can now guess. It would be the third derivative of y with respect to x would be written this way, and it would be six minus forty-eight x, and you get the idea. So you could take higher and higher derivatives all you want and you can take them -- use the notation this way, notation that way. It’s not a big deal, not a big deal.

Now, here’s the fun thing that I think you’re going to get a kick out of. I have just one sheet of paper left and then we go into, sort of, the, this, wholesome, sort of looking brown. So it will be very exciting to see what’s going to happen. I don’t know. Maybe the staff will actually give me some more paper. We’ll see. Let’s hope so. Keep our fingers crossed.

Okay, now what I want to tell you about right now is the following. We see now how we can use the derivative to find instantaneous velocity. Great. Let’s take a look at sort of the, the other side of the derivative, in particular, the side of the derivative that gives slopes of tangents. How can we actually use that to do interesting kind of things? Turns out that we can actually use the idea of a tangent line to actually compute certain complicated numbers. Let me give you an illustration. What does, let’s say, four point one, the square root of four point one. What does that equal, roughly? Well, of course, roughly speaking, the square root of four point one, that’s actually near the square root of four, and the square root of four we do know is two. So, this is going to be around two. In fact, since four point one is a little bit bigger than four, we expect this to be a little bit bigger than two. But what, how much bigger than two? Well, uh, how would you do this problem?

Well, one solution would be to take out your calculator, turn it on, and then push four point one and push the square root key, which I actually will not do at this moment. And then, somehow, magically something appears and then we believe that’s the answer. Well, how does the computer do that? I mean, you know, let’s look in the back of the computer and look at its mind. How is it actually computing that? Well, that’s an interesting question. How does the computer figure that out? Another interesting question is suppose that you actually wanted to do this and you didn’t want to use a calculator, or suppose that this were just a few years ago before you were born and there weren’t calculators around. How could you actually figure this out? How do calculators figure this kind of thing out? It turns out that one method is actually to use the idea of a derivative, and in particular, remember the basic fact that a derivative represents a slope of tangent line.

Let me show you exactly how this would look. Let me just give you the strategy for what this would look like. Let me set up some axes and draw you a picture. So here’s a picture, and suppose I have a function that looks like this, and this is the function, let’s say f of x. So, y equals f of x. And suppose, let’s further suppose that there is a particular value right here, right here, and I want to find out the y value at that particular, at that particular value. Let me call this sort of question mark, question mark right here. Okay. So, so this is an x value and what I want to find out is we want this, we want f of question mark. We want to figure out the height of that function.

Okay, now, how would we go about doing that? Well, if we don’t know what this is, I mean, of course, in, in, some theoretical point of view, it’s real easy, right? You just plug in wherever you see an x you put in question mark. But now I’m actually talking from a computational point of view. You really want to compute it with your hands and calculate it. Well, finding square roots is actually a pretty hard thing to do unless the number was a perfect square to begin with.

So what do you do? Well, let’s just suppose that we knew that nearby this mysterious point was a point where we actually could compute the square root. In this case, the square root of four is actually easy to compute. So what would we do? Well, let me put that in. Let’s suppose that, in fact, there’s a point where we can actually figure things out right here. Oh, look, I think we’re getting more paper, oh, isn’t this great? Let’s say this is actually going to be a known point. Let me call that x not. So x not is a particular value where, in fact, we can, using our own naive method, figure out this value, which I’m going to know is f of x not. So, here’s a particular number where we actually know the value, f of x not. This is the mysterious one where I wish I knew the value, I wish I knew the value there. Well, how could I find the value there?

Well, I can try to estimate it and the idea of estimating it is very simple. It comes down to the tangent line. If we found the equation of the tangent line, right there, at the known point, then what do we notice about the tangent line? A very fundamental fact -- that the tangent line emulates the activity of the function nearby the point of tangency. Notice how close the, the blue curve, sort of, is, is hugging up against the red line, especially near the point x not. They almost sort of look the same. The curve sort of comes down, gently grazes the red line and then slowly drives off. I admit that over here the line and the curve are very, very far apart. But near the point of tangency, they’re almost identical. And that fundamental fact, the fact that near the point of tangency the line, the tangent line and the curve sort of act the same way, allows us to do the following.

We could find the equation of the tangent line and then -- at this point. And then if we know that, we can evaluate the mysterious value, not at the complicated curve, which might be hard to actually compute, but on the line. And computing a line is easy. It’s just a multiplication and an addition. We can always do multiplication and addition. So, in fact, we can then estimate the value by plugging in on the line and seeing, in this case, this estimated answer. The answer we get by plugging in the question mark into the line. It’s not going to be exactly what we want, but it’s certainly going to be approximately, what we want.

So, let me just recap the idea before we go into the details. The idea is if I wanted to actually compute the value of a function at a particular point where I can’t do it computationally, because it’s hard. But nearby there’s a point where I can compute the function because, for some reason, it’s easy. Then I can find the equation of the tangent line on, by the easy point, and then use that line to approximate the harder point.

Let me try to illustrate this with an explicit example before I give you the general formula. So, let’s see if we can actually take a look at how we’d use this. Look what happened, we struck bottom, but not to worry because our staff has provided us with all the paper we want. I don’t want you to think that somehow now we’re not going to do any more math because we ran out of paper. It’s sort of like, you know, when you’re in class and you and you think, “Oh, my god, teacher ran out of chalk, maybe we’ve got to cancel class.” You know, right. That’s going to happen. Okay. So, here we go. Let me just try to get my bearings here. There we are, look at that, brand new. Look, look at all this paper, look at that.

Let’s take a look at the following example. Well, let’s look at the example I just did, in fact, that’s the one I want to look at. So, let’s try to estimate the square root of four point one. Well, the first thing I want to do is figure out what function I should be looking at. What function do I want to evaluate four point one at to get this, this value? Well, it must be the square root function. So, if we let – so the question is what does this equal? And now let’s now define a function, let’s let f of x be the square root function. So, I’m doing this now, inspired by what I want. And now what I do is say, “Okay, now, let’s see, if f of x equals the square root of x,” now I can rephrase, I can rephrase the question that I want to ask. I could say what we want now is to approximate f evaluated at some point, and what number do I want to evaluate f at? Four point one. Because notice that f of four point one is exactly the square root of four point one. So, I’m just rephrasing the question using the function notation. But now what we notice is -- we note that there’s an easy point nearby. In particular, f of four is something we know. That’s the square root of four, which equals two.

So, let me show you a picture of this. Now, I happen to know what the graph of the square root function looks like, and maybe you don’t, and if you don’t, it’s okay. You’ll know it later, but if you don’t know it now don’t worry about it. But let me draw it for you so you can see it. It looks like this. It looks like, actually looks like a parabola, but something on its side. You see that’s the side ways parabola basically. That’s what we’re looking at. And what I’m interested in is looking at the point four point one. So, suppose this is one, two, three, four, and this is five. This is four here, and this is five. The point I’m interested in, actually, is the point four point one, which maybe, is right here, very close by, four point one. And I’m interested in actually journeying up to this square root function and seeing what that equals. That’s the square root of four point one, which is the thing I’m trying to figure out. But I can’t. I mean, of course, you know, philosophically I can, but in practice, I don’t know how to do that. So, what I’m going to do instead is, notice that actually at four, which is very, very close to four point one, I can figure out the graph. I can figure out the function. I know exactly what it is. Let me put that in. If you go up here, what you see is this equals two because the square root of four is two.

So, what I’m going to do right now is find the equation of the line that’s tangent to the this curve at this known easy point, namely this point right here. So, I’m going to draw in the tangent line for you right now live. Pretty good, but not great, I have to admit, but that’s okay. There’s the tangent line. This is the line that’s tangent, tangent at the point x equals four, the easy point, the one where we actually know all the information about. And if you look at this picture, you can see that the tangent line and the curve agree nearby. And, so, if I now evaluate, if I now evaluate four point one at the line. Go all the way up to the line and go over, I don’t get the exact value I want, but I will get a good estimate. This is a good estimate. This is a good approximation to four point o, to four point one, square root of four point one. So the strategy here is when you have a very subtle curve that you can’t evaluate, if you can evaluate and find the tangent line nearby, then evaluate a line. A line is easier to evaluate than a curve.

Okay. Let’s try to execute this right now. So, let’s find the equation of the, of the tangent line to this curve at the point four. Well, this is actually an old problem, to find the equation of the tangent line, what do I need to do? I need two pieces of information. I need slope and I need a point on the line. Well, a point on the line I already have. We know the line contains four comma two. So, four two is on the line, because it’s a tangent line at that point. All that’s missing is the slope.

Well, let’s find the slope of the tangent line. How do I do that? I take the derivative. I take the derivative and then what do I do with the derivative? Oops, I take the derivative and I evaluate the derivative at, at the point four. Let me just put this here for now. And so let’s now take the derivative, and you can see the function is over there. The function, remember, is the square root of x. The derivative is what? Well, the square root of x falls right there, it’s equal to x to the one half power. And now you remember what to do. You bring down the one half. So, this is one half x to the – subtract one from the exponent minus one half. And if you simplify that, as you can see, that just gives us one over two square root of x. So, there, in fact, is the derivative.

And what do I want to do? I want to find the slope at the point four. So, I plug in four there. So, that’s two over the square root of four, which equals one over two times the square root of four, which is two. So, one over two times two, which is four. So, the slope of that line is a fourth. The line passes through the point four comma two. And so if I use y minus y one equals M times x minus x one, I see that y minus two equals one fourth times x minus four. And if you now solve this for y by bringing the negative two over, what would I see? I would see that y equals one-fourth x, and when I bring this –distribute this, I see a minus four over four, which is a minus one. But then I bring over this, becomes a plus two. So, plus two minus one gives me a plus one. So, that’s the equation of this line. So, this is y equals one fourth x plus one.

Well, let’s now evaluate this at the point one point four. So what happens if you now evaluate that at one point four? Well, let’s see what happens. So, if we plug in one point four, so at x equals – I’m sorry, four point one, not one point four, what would that equal? By the way, that’s just the number forty one over ten. And if I look at that, I’ll see y equals one fourth times forty-one over ten, plus one, which equals forty-one over forty plus one. One is actually forty over forty, and so what I see here is eighty-one over forty. So the value of the line at the point four point one, the point four point one turns out to be eighty-one over forty. So, the good approximation is eighty-one over forty. And so we’re thinking that eighty-one over forty should be a pretty good approximation to this.

How did I do that? Well, I took the tangent line at the point I know and evaluated the tangent line at that nearby mysterious point, and said that approximately will equal the actual value. So, let’s see how close we really are.

So, remember, we’re trying to find the square root of four point one, and our guess is eighty-one over forty. Let’s see how good we did. When I take this calculator, which somehow, you know what’s sort of interesting. We sort of believe that whatever the calculator says must be true. How do we know the calculator is right? Sort of interesting point. We just believe that on complete faith. Now, of course, if it’s just addition, subtraction, or multiplication of integers, regular numbers, it’s not a big deal because we can just check it. But this kind of stuff, you can’t check that. We just take the calculator’s word for it. Anyway, let’s see what the calculator says, uh, this value is. So, I put in four point one and I finally push the square root key, and you can see what it says there. It says two point zero, can you see that, two, four, eight and so forth. So this is the calculator guess is two point zero two four and it keeps going.

Let’s see what our guess is. What’s eighty-one over forty? Eighty-one divided by forty. That equals two point zero two five. Look how amazingly accurately we’ve actually estimated this number. Pretty good. Almost to three decimal places just using calculus, just using calculus.

Okay, well, up next what I’ll do is give you the general formulation for how you actually find approximations. Basically, all we’re going to do is do this previous example, but I’m not going to tell you what the function is. I’ll just call it f of x. And we’ll work through and find the actual formula that you can always use to help you approximate these difficult numbers where a known by number is standing by. So, I’ll pick that up next. Okay.