lready here, today, used this way of rewriting du/ dx. I think when I was talking about d/dt(uv) and so on, I pulled that d / dt outside and put whatever function you're differentiating over to the right. So that's just a notational switch. It looks good. It looks like good algebra doesn't it? But what it's doing is regarding this notation as an operator. It's something you apply to a function to get a new function. I apply it to the sine function, and I get the cosine function. I apply it to x^2, and I get 2x. This thing here, that symbol, represents an operator, which you apply to a function. And the operator says, take the function and differentiate it. So further notation that people often use, is they give a different name to that operator. And they'll write capital D for it. So this is just using capital D for the symbol d/dx.
So in terms of that notation, let's see. Let's write down what higher derivatives look like. So let's see. That's what u' is. How about u''? Let's write that in terms of the d/dx notation. Well I'm supposed to differentiate u' right? So that's d/dx applied to the function du/ dx. Differentiate the derivative. That's what I've done. Or I could write that as d/dx applied to d/dx applied to u. Just pulling that u outside. So I'm doing d/dx twice. I'm doing that operator twice. I could write that as (d/dx)^2 applied to u. Differentiate twice, and do it to the function u. Or, I can write it as, now this is a strange one. I could also write as like that. It's getting stranger and stranger, isn't it? This is definitely just a kind of abuse of notation. But people will go even further and write (d^2)u/dx^2. So this is the strangest one. This identity quality is the strangest one, because you may think that you're taking d of the quantity x^2. But that's not what's intended. This is not d(x^2). What's intended is the quantity dx^2. In this notation, which is very common, what's intended by the denominator is the quantity dx^2. It's part of this second differentiation operator. So I've written a bunch of equalities down here, and the only content to them is that these are all different notations for the same thing. You'll see this notation very commonly. So for instance the third derivative is (d^3)u/dx^3, and so on. Sorry?
Professor: Yes, absolutely. Or an equally good notation is to write the operator (D^3)u. Absolutely. So I guess I should also write over here when I was talking about d^2, the second derivative, another notation is do the operator capital D twice. Let's see an example of how this can be applied. I'll answer this question.
Professor: Yeah, so the question is whether the fourth derivative always gives you the original function back, like what happened here. No. That's very, very special to sines and cosines. All right? And, in fact, let's see an example of that. I'll do a calculation. Let's calculate the nth derivative of x^n. Okay, n is a number, like 1, 2, 3, 4. Here we go. Let's do this. So, let's do this bit by bit. What's the first derivative of x^n? So everybody knows this. I'm just using a new notation, this capital D notation. So it's n x ^ (n -1). Now you know know, by the way, n could be a negative number for that, but for now, for this application, I wanna take n to be 1, 2, 3, and so on; one of those numbers.
Ok, we did one derivative. Let's compute the second derivative of x ^ n. Well there's this n constant that comes out, and then the exponent comes down, and it gets reduced by 1. All right? Should I do one more? D^3 (x^n) = n(n-1). That's the constant from here. Times that exponent, (n - 2), times 1 less, (n - 3) is the new exponent. Well, I keep on going until I come to a new blackboard. Now, I think I'm going to stop when I get to the n minus first derivative, so we can see what's likely to happen. So when I took the third derivative, I had the n minus third power of x. And when I took the second derivative, I had the second power of x. So, I think what'll happen when I have the n minus first derivative is I'll have the first power of x left over. The powers of x kee