na 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 down is correct here. So, if you look there's a u(v (x delta x)) there. And there's also one here. And they occur with opposite signs. So they cancel. What's left is u ( x delta x) v (x delta x) - (uv). And that's just what I started with.
Student: They cancel right?
Professor: I cancelled out this term and this term, and what's left is the ends. Any other questions?
Professor: Well, I just calculated what delta uv is, and now I'm gonna divide that by delta x on my way to computing the derivative. And so I copied down the right hand side and divided delta x. I just decided to divide the delta u by delta x and delta v by delta x. Good. Anything else?
So we have the product rule here. The rule for differentiating a product of two functions. This is making us stronger. There are many more functions you can find derivatives of now. How about quotients? Let's find out how to differentiate a quotient of two functions. Well again, I'll write down what the answer is and then we'll try to verify it. So there's a quotient. Let me write this down. There's a quotient of two functions. And here's the rule for it. I always have to think about this and hope that I get it right. u'v - uv' / v^2. This may be the craziest rule you'll see in this course, but there it is. And I'll try to show you why that's true and see an example. Yeah there was a hand?
Professor: What letters look the same? u and v look the same? I'll try to make them look more different. The v's have points on the bottom. u's have little round things on the bottom. What's the new value of u? The value of u at (x delta x) is (u delta u), right? That's what delta u is. It's the change in u when x gets replaced by (x delta x). And the change in v, the new value v, is (v delta v). So this is the new value of u divided by the new value of v. That's the beginning. And then I subtract off the old values, which are - u/v. This'll be easier to work out when I write it out this way. So now, we'll cross multiply, as I said. So I get (uv (delta u)v) minus, now I cross multi