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we already know that the product rule tells us that if we have the comp 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 we want to take the derivative of this business now 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 plus the first function so not even taking its derivative so plus f of X times the derivative of the second function times the derivative of the second function so two terms and 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 the of F naught of G here it's the derivative of G naught of F that's this is hopefully a little bit of review this is the product rule now 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 read arrive it from the product rule so let's see what we're talking about so let's imagine if we had a 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 f of X times G of X to the negative 1 power 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 F prime of x times the just the Reg the second function which is just G of X to the negative 1 power G of X to the negative 1 power plus plus the first function which is just f of X plus 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 use something to the negative one power with respect to that something is going to be negative one times that something which in this case is G of X to the negative two 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 bath 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 as f prime of x over G of X F prime of x over G of X and we can write all of this as put this negative sign out front we have negative f of X negative f of X times G prime of x times G prime of X G prime of X and then all of that over G of x squared all of that over write this a little bit neater all of that over G of x squared G of x squared and it still isn't 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 in 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 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 times G of X minus not plus anymore - let me write it in white - minus f of X minus f of X times G prime of X time G prime of X all of that over all of that over G of x squared 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 derivative of the second function times the first function we now subtract it and all of 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