ply 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 1 / v, the reciprocal of a function; 1 over a function. Here's a copy of my rule. What's du/ dx in that case? u is a constant, so that term is in this rule. I don't have to worry about this. I get a minus. And then u = 1, and dv/ dx. Well, v is whatever v is. I'll write dv/dx as v'. And then I get a v^2 in the denominator. So that's the rule. I could write it as (v^-2)v'. (-v'/v^2). That's the derivative of 1 / v.
How about sub-example of that? I'm going to take the special case where u = 1 again. And v = x^n. And I'm gonna use the rule that we developed earlier about the derivative of x^n. So what do I get here? d / dx (1/x^n) is, I'm plugging into this formula here with v = x^n. So I get minus, uh, v^-2. If v = x^n, v^-2 is, by the rule of exponents, x^-2n. And then v' is the derivative of x^n, which is (nx)^(n-1).
Ok, so let's put these together. There's several powers of x here. I can put them together. I get -nx ^ ((- 2n) (n - 1)). One of these n's cancels. And what I'm left with is ((-n) - 1). So we've computed the derivative of 1 / x^n, which I could also write as x^-n, right? So I've computed the derivative of negative powers of x. And this is the formula that I get. If you think of this - n as a unit, as a thing to itself, it occurs here in the exponent. It occurs here, and it occurs here. So how does that compare with the formula that we had up here? The derivative of a power of x is that power times x to one less than that power. That's exactly the same as the rule that I wrote down here. But the power here happens to be a negative number, and the same negative number shows up as a coefficient and there in the exponent. Yeah?
Professor: How did I do this?
Professor: Where did that x^-2n come from? So I'm applying this rule. So the denominator in the quotient rule is v^2. And v was x^n, so the denominator is x^2n. And I decided to write it as x^-2n. So the green comments there... What they say is that I can enlarge this rule. This exact same rule is true for negative values of n, as well as positive val