Quotient rule from product & chain rules

We already know that the product rule tells us that if we have the product of two functions-- so let's say f of x and g of x-- and we want to take the derivative of this business, that this is just going to be equal to the derivative of the first function, f prime of x, times the second function, times g of x, plus the first function, so not even taking its derivative, so plus f of x times the derivative of the second function. So two terms, in each term we take the derivative of one of the functions and not the other, and then we switch. So over here is the derivative of f, not of g. Here it's the derivative of g, not of f. This is hopefully a little bit of review. This is the product rule. Now what we're essentially going to do is reapply the product rule to do what many of your calculus books might call the quotient rule. I have mixed feelings about the quotient rule. If you know it, it might make some operations a little bit faster, but it really comes straight out of the product rule. And I frankly always forget the quotient rule, and I just rederive it from the product rule. So let's see what we're talking about. So let's imagine if we had an expression that could be written as f of x divided by g of x. And we want to take the derivative of this business, the derivative of f of x over g of x. The key realization is to just recognize that this is the same thing as the derivative of-- instead of writing f of x over g of x, we could write this as f of x times g of x to the negative 1 power. And now we can use the product rule with a little bit of the chain rule. What is this going to be equal to? Well, we just use the product rule. It's the derivative of the first function right over here-- so it's going to be f prime of x-- times just the second function, which is just g of x to the negative 1 power plus the first function, which is just f of x, times the derivative of the second function. And here we're going to have to use a little bit of the chain rule. The derivative of the outside, which we could kind of view as something to the negative 1 power with respect to that something, is going to be negative 1 times that something, which in this case is g of x to the negative 2 power. And then we have to take the derivative of the inside function with respect to x, which is just g prime of x. And there you have it. We have found the derivative of this using the product rule and the chain rule. Now, this is not the form that you might see when people are talking about the quotient rule in your math book. So let's see if we can simplify this a little bit. All of this is going to be equal to-- we can write this term right over here as f prime of x over g of x. And we could write all of this as-- we could put this negative sign out front. We have negative f of x times g prime of x. And then all of that over g of x squared. Let me write this a little bit neater. All of that over g of x squared. And it still isn't in the form that you typically see in your calculus book. To do that, we just have to add these two fractions. So let's multiply the numerator and the denominator here by g of x so that we have everything in the form of g of x squared in the denominator. So if we multiply the numerator by g of x, we'll get g of x right over here and then the denominator will be g of x squared. And now we're ready to add. And so we get the derivative of f of x over g of x is equal to the derivative of f of x times g of x minus-- not plus anymore-- let me write it in white-- f of x times g prime of x, all of that over g of x squared. So once again, you can always derive this from the product rule and the chain rule. Sometimes this might be convenient to remember in order to work through some problems of this form a little bit faster. And if you wanted to kind of see the pattern between the product rule and the quotient rule, the derivative of one function just times the other function. And instead of adding the derivative of the second function times the first function, we now subtract it. And all that is over the second function squared. Whatever was in the denominator, it's all of that squared. So when we're taking the derivative of the function in the denominator up here, there's a subtraction, and then we are also putting everything over the second function squared.