Topics covered: Chain rule - Higher derivatives
Instructor: Prof. David Jerison
Lecture Notes (PDF)
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Professor: I am Haynes Miller, I am substituting for David Jerison today. So you have a substitute teacher today. So I haven't been here in this class with you so I'm not completely sure where you are. I think just been talking about differentiation and you've got some examples of differentiation like these basic examples: the derivative of x^n is nx^(x-1). But I think maybe you've spent some time computing the derivative of the sine function as well, recently. And I think you have some rules for extending these calculations as well. For instance, I think you know that if you differentiate a constant times a function, what do you get?
Professor: The constant comes outside like this. Or I could write (cu)' = cu'. That's this rule, multiplying by a constant, and I think you also know about differentiating a sum. Or I could write this as (u v)' = u' v'. So I'm going to be using those but today I'll talk about a collection of other rules about how to deal with a product of functions, a quotient of functions, and, best of all, composition of functions. And then at the end, I'll have something to say about higher derivatives. So that's the story for today. That's the program.
So let's begin by talking about the product rule. So the product rule tells you how to differentiate a product of functions, and I'll just give you the rule, first of all. The rule is it's u'v uv'. It's a little bit funny. Differentiating a product gives you a sum. But let's see how that works out in a particular example. For example, suppose that I wanted to differentiate the product. Well, the product of these two basic examples that we just talked about. I'm going to use the same variable in both cases instead of different ones like I did here. So the derivative of (x^n)sin x. So this is a new thing. We couldn't do this without using the product rule. So the first function is x^n and the second one is sin x. And we're going to apply this rule. So u is x^n. u' is, according to the rule, nx^(n - 1). And then I take v and write it down the way it is, sine of x. And then I do it the other way. I take u the way it is, that's x^n, and multiply it by the derivative of v, v'. We just saw v' is cosine of x. So that's it. Obviously, you can differentiate longer products, products of more things by doing it one at a time.
Let's see why this is true. I want to try to show you why the product rule holds. So you have a standard way of trying to understand this, and it involves looking at the change in the function that you're interested in differentiating. So I should look at how much the product uv changes when x changes a little bit. Well, so how do compute the change? Well, I write down the value of the function at some new value of x, (x delta x). Well, I better write down the whole new value of the function, and the function is uv. So the whole new value looks like this. It's u (x delta x)v(x delta x). That's the new value. But what's the change in the product? Well, I better subtract off what the old value was, which is u(x) v(x).
Okay, according to the rule we're trying to prove, I have to get u' involved. So I want to involve the change in u alone, by itself. Let's just try that. I see part of the formula for the change in u right there. Let's see if we can get the rest of it in place. So the change in x = ((u(x) delta x) - u(x)). That's the change in x [Correction: change in u]. This part of it occurs up here, multiplied by v (x delta x), so let's put that in too. Now this equality sign isn't very good right now. I've got this product here so far, but I've introduced something I don't like. I've introduced u (v ( x delta x)), right? Minus that. So the next thing I'm gon