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 gonna do is correct that little defect by adding in u(x) (v (x delta x)). Okay, now I cancelled off what was wrong with this line. But I'm still not quite there, because I haven't put this in yet. So I better subtract off uv, and then I'll be home. But I'm going to do that in a clever way, because I noticed that I already have a u here. So I'm gonna take this factor of u and make it the same as this factor. So I get u(x) times this, minus u(x) times that. That's the same thing as u times the difference. So that was a little bit strange, but when you stand back and look at it, you can see multiplied out, the middle terms cancel. And you get the right answer. Well I like that because it's involved the change in u and the change in v. So this is equal to (delta u) v (x delta x) - u(x) times the change in v. Well, I'm almost there. The next step in computing the derivative is take difference quotient, divide this by delta x. So, (delta (uv)) / (delta x) is well, I'll say (delta u / delta x) v ( x delta x).
Have I made a mistake here? This plus magically became a minus on the way down here, so I better fix that. Plus u(delta v / (delta x). This u is this u over here. So I've just divided this formula by delta x, and now I can take the limit as goes to 0, so this is as delta x goes to 0. This becomes the definition of the derivative, and on this side, I get du/dx times... now what happens to this quantity when delta x goes to 0? So this quantity is getting closer and closer to x. So what happens to the value of v? It becomes equal to x(v). That uses continuity of v. So, v (x delta x) goes to v(x) by continuity. So this gives me times v, and then I have u(n delta v/ v delta x) gives me dv/dx. And that's the formula. That's the formula as I wrote it down at the beginning over here. The derivative of a product is given by this sum. Yeah?
Student: How did you get from the first line to the second of the long equation?
Professor: From here to here?
Professor: So maybe it's easiest to work backwards and verify that what I wrote