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 values of n. So there's something new in your list of rules that you can apply, of values of the derivative. That standard rule is true for negative as well as positive exponents. And that comes out of a quotient rule.
Okay, so we've done two rules. I've talked about the product rule and the quotient rule. What's next? Let's see the chain rule. So this is a composition rule. So the kind of thing that I have in mind, composition of functions is about substitution. So the kind of function that I have in mind is, for instance, y = (sin t)^10. That's a new one. We haven't seen how to differentiate that before, I think. This kind of power of a trig function happens very often. You've seen them happen, as well, I'm sure, already. And there's a little notational switch that people use. They'll write sin^10(t). But remember that when you write sin^10(t), what you mean is take the sine of t, and then take the tenth power of that. It's the meaning of sin^10(t). So the method of dealing with this kind of composition of functions is to use new variable names. What I mean is, I can think of this (sin t)^10.
I can think of it it as a two step process. First of all, I compute the sine of t. And let's call the result x. There's the new variable name. And then, I express y in terms of x. So y says take this and raise it to the tenth power. In other words, y = x^10. And then you plug x = sine of t into that, and you get the formula for what y is in terms of t. So it's good practice to introduce new letters when they're convenient, and this is one place where it's very convenient.
So let's find a rule for differentiating a composition, a function that can be expressed by doing one function and then applying another function. And here's the rule. Well, maybe I'll actually derive this rule first, and then you'll see what it is. In fact, the rule is very simple to derive. So this is a proof first, and then we'll write down the rule. I'm interested in delta y / delta t. y is a function of x. x is a function of t. And I'm interested in how