 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 multiply this way, you get (uv - u(delta v)). And I divide all this by (v delta v)u [Correction: (v delta v)v]. Okay, now the reason I like to do it this way is that you see the cancellation happening here. uv and uv occur twice and so I can cancel them. And I will, and I'll answer these questions in a minute.
Professor: Ooo, that's a v. All right. Good, anything else? That's what all hands were. Good. All right, so I cancel these and what I'm left with then is (delta u)v - u(delta v) and all this is over (v delta v)v. Ok, there's the difference. There's the change in the quotient. The change in this function is given by this formula. And now to compute the derivative, I want to divide by delta x, and take the limit. So let's write that down, delta(u/v)/delta x is this formula here divided by delta x. And again, I'm going to put the delta x under these delta u and delta v. Okay? I'm gonna put delta x in the denominator, but I can think of that as dividing into this factor and this factor. So this is ((delta u/ delta x)v) - u(delta v/delta x)). And all that is divided by the same denominator, (v delta v)v. Right? Put the delta x up in the numerator there.
Next up, take the limit as delta x goes to 0. I get, by definition, the derivative of (u/v). And on the right hand side, well, this is the derivative u(du/ dx) right? Times v. See and then u times, and here it's the derivative (dv/ dx). Now what about the denominator? So when delta x goes to 0, v stays the same, v stays the same. What happens to this delta v? It goes to 0, again, because v is continuous. So again, delta v goes to with delta x because they're continuous and you just get (v*v). I think that's the formula I wrote down over there. (du/dx)v - u(dv/dx). And all divided by the square of the old denominator. Well, that's it. That's the quotient rule. Weird formula.
Let's see an application. Let's see an example. So the example I'm going to give is pretty simple. I'm going to take the numerator to be just 1. So I'm gonna take u = 1. So now I'm differentiating 