 In this video I wanna revisit the quotient rule and talk about how it behaves with respect to the chain rule which we've been learning about in lecture 22 right here. So just as a reminder, we saw previously that if f and g are two differentiable functions such that g of x doesn't go to zero, then the derivative of f of x divided by g of x is gonna be g of x times f prime of x minus f, or f of x times g prime of x over g of x squared for which as the order of the functions in numerator matters, I taught you of course the poem, low d high minus high d low square of the bottom. Here we go to try to help us remember the quotient rule. We've seen examples of computing things with the quotient rule, but we didn't actually see the proof. We didn't see why the quotient rule works. And I had promised you that we'd see it later on. We're now at that moment. It turns out we can prove the quotient rule by combining the chain rule which we've proven with the product rule which we proved earlier. And so I wanna show you how that's gonna work out here. So consider we wanna calculate the derivative of f of x divided by g of x, okay? Well, the first thing to observe is if you're dividing by g of x, that's the same thing as multiplying by g to the negative one power. Now be careful here, in this context, this negative one superscript does not mean the inverse function. This actually means the negative one exponent. That is, this is a power right here. So you'll notice I've turned my quotient into a product. This suggests that the product rule is gonna come and play as we try to take the derivative. So by the product rule, we're gonna get f prime of x times g of x, g of x to the negative one power. And then you're gonna get f of x times the derivative of g of x to the negative one power. So that's why the product rule was necessary. But how do we deal with this part right here? Cause when we look at this expression g of x to the negative one power, we should think of this in terms of function decompositions. There's two functions in play. There's the reciprocal function u to the negative one. And then there's the inner function of g of x. So we put g of x inside of u. That's where this g of x to the negative one came from. So if we take the derivative using the chain rule here, we're gonna take the derivative of u to the negative one. We'll buy the power rule that should look like negative u to the negative two power. And then if we take the derivative of the inner function, that should just be a g prime of x. You get something like this. And that's exactly what translates over to this expression right here. We get a negative g of x to the negative two power times g prime of x. Great. Now let's go back to fraction form. So if you have a g of x to the negative one, that means you're dividing by g of x. If you have a g of x to the negative two, that means you're dividing by g squared. Now these fractions have, they have uncommon denominators. So to find a common denominator, we need to take the first fraction times top and bottom by g of x, as you can see right here, for which case then the new numerator will become f prime of x times g of x. The second denominator is already f prime of, excuse me, f of x times g prime of x. And so combine those together, you get the usual quotient rule. So voila, there's the proof of the quotient rule. But it turns out that this proof also illustrates another interesting observation. It shows us how we can calculate the derivative of a quotient without using the quotient rule. If you just combine the product rule with the chain rule, you can always calculate the derivative of a function. Let's take a look at such a thing. Let's find the derivative of f of x, which is the function x squared plus one over three x plus two with this idea of the chain rule. So we can write this as a product. We get x squared plus one times three x plus two to the negative one power. So if we take the derivative using the product rule, f prime of x would look like, well, we take the derivative of x squared plus one, that's gonna get two x, then you're gonna get a three x plus two to the negative one power. Then the next one, we're gonna get the x squared plus one, and then taking the derivative using the chain rule, we're gonna get negative three x plus two to the negative two power times that by three, like so. And so if writing these as fractions, you're gonna get two x over three x plus two. The next one, you're gonna get subtract x squared plus one times that by three over three x plus two squared. In order to have a common denominator, we need another three x plus two here. So we have to do that to the numerator as well. For which case then, if we just pause for a moment at, just at this moment, what we get is we're gonna get a low d high minus high d low square the bottom. And here we go. So I want you to see here that if you rewrite every fraction as a product of the reciprocal, you can always calculate the derivative using the chain rule and the product rule, which are a lot easier to remember because with the quotient rule, the location, the minus sign and such, the order of operations matters in that context. If you mess it up, you can mess up the whole derivative. So you actually don't need the quotient rule, but why don't we have the quotient rule? Because of the quotient rule, you can skip all of this garbage, right? All of this stuff goes away. If you just apply the quotient rule directly, you end up right here. So the point of memorizing the quotient rule is to simplify this calculation so that we can just start at this moment and we end up with a six x squared plus four x minus three x squared minus three all above three x plus two squared. If we combine like terms, we get a three x squared plus four x minus three all on top of three x plus two quantity squared, which is the same derivative. So I would definitely recommend the quotient rule. It'll simplify your calculations in the long run, but also be aware that you don't necessarily need it. If you're in some type of desert island scenario, you're playing crashes in the South Pacific and only you and your volleyball Wilson survive and you have to calculate the derivative of a quotient in order to get back home. Be aware that if your quotient rule didn't make it out of the crash, it's okay. If you with the scraps called the product rule and chain rule, you can reproduce the quotient rule if necessary, although I think memorizing it really would be a time saver for you in the long run.