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About this Lesson
 Type: Video Tutorial
 Length: 13:11
 Media: Video/mp4
 Use: Watch Online & Download
 Access Period: Unrestricted
 Download: MP4 (iPod compatible)
 Size: 143 MB
 Posted: 11/18/2008
This lesson is part of the following series:
Calculus (279 lessons, $198.00)
Calculus Review (48 lessons, $95.04)
Calculus: Final Exam Test Prep and Review (45 lessons, $64.35)
Calculus: Techniques for Finding the Derivative (8 lessons, $15.84)
Calculus: The Product and Quotient Rules (2 lessons, $5.94)
In this lesson, we learn about taking the derivatives of quotients and applying the quotient rule of differentiation. The derivative of the quotient of two differentiable functions does not necessarily exist and is not necessarily equal to the quotient of the derivatives of both functions. Instead, you must learn and apply the quotient rule to calculate the derivative of the quotient of two functions. The quotient rule states that, when f(x) and g(x) are both differentiable and g(x) does not equal 0, the derivative of f(x)/g(x) = [g(x)* derivative of f(x)  f(x)* derivative of g(x)]/g(x)^2
Taught by Professor Edward Burger, this lesson was selected from a broader, comprehensive course, College Algebra. This course and others are available from Thinkwell, Inc. The full course can be found at http://www.thinkwell.com/student/product/calculus. The full course covers limits, derivatives, implicit differentiation, integration or antidifferentiation, L'Hôpital's Rule, functions and their inverses, improper integrals, integral calculus, differential calculus, sequences, series, differential equations, parametric equations, polar coordinates, vector calculus and a variety of other AP Calculus, College Calculus and Calculus II topics.
Edward Burger, Professor of Mathematics at Williams College, earned his Ph.D. at the University of Texas at Austin, having graduated summa cum laude with distinction in mathematics from Connecticut College.
He has also taught at UTAustin and the University of Colorado at Boulder, and he served as a fellow at the University of Waterloo in Canada and at Macquarie University in Australia. Prof. Burger has won many awards, including the 2001 Haimo Award for Distinguished Teaching of Mathematics, the 2004 Chauvenet Prize, and the 2006 Lester R. Ford Award, all from the Mathematical Association of America. In 2006, Reader's Digest named him in the "100 Best of America".
Prof. Burger is the author of over 50 articles, videos, and books, including the trade book, Coincidences, Chaos, and All That Math Jazz: Making Light of Weighty Ideas and of the textbook The Heart of Mathematics: An Invitation to Effective Thinking. He also speaks frequently to professional and public audiences, referees professional journals, and publishes articles in leading math journals, including The Journal of Number Theory and American Mathematical Monthly. His areas of specialty include number theory, Diophantine approximation, padic analysis, the geometry of numbers, and the theory of continued fractions.
Prof. Burger's unique sense of humor and his teaching expertise combine to make him the ideal presenter of Thinkwell's entertaining and informative video lectures.
About this Author
 Thinkwell
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11/14/2008
Founded in 1997, Thinkwell has succeeded in creating "nextgeneration" textbooks that help students learn and teachers teach. Capitalizing on the power of new technology, Thinkwell products prepare students more effectively for their coursework than any printed textbook can. Thinkwell has assembled a group of talented industry professionals who have shaped the company into the leading provider of technologybased textbooks. For more information about Thinkwell, please visit www.thinkwell.com or visit Thinkwell's Video Lesson Store at http://thinkwell.mindbites.com/.
Thinkwell lessons feature a starstudded cast of outstanding university professors: Edward Burger (PreAlgebra through...
More..Recent Reviews
 A nice start to the quotient rule.
 01/24/2009

There are a few things I disagree with though a bit. One of which is the chanting deal. All you really need to do for product and quotient rules is memorize 2 letters and a ' symbol with them.
product rule = fg' + gf' (where you make f = to a function and g = to a function)
quotient rule = (gf'  fg') / g^2.
Never been a huge fan of "top times bottom" chants and all that stuff. Also, it would be nice if Mr. Burger simplified some of the problems sometimes when he gets to the end. It's nice to say well we're done, but technically you're not unless multiplying everything out would be ridiculously complicated.
At any rate though, his teaching style is still very effective and helps to keep me focused when I'm drifting in my calculus book.
Computational Techniques
The Product and Quotient Rules
The Quotient Rule Page [1 of 2]
We are now in the middle of learning these really, really fancy methods for taking derivatives. Let me just recap really fast. We have just seen the product rule, which allows us to take the derivative of complicated functions that are built up of products of other functions. And the product rule – I hope you’re going to start to get this little chant down – is the derivative of a product is equal to the first times the derivative of the second plus the second multiplied by the derivative of the first – first times the derivative of the second plus the second times the derivative of the first. It’s sort of a neat, yet sort of not intuitive formula, for finding the derivatives of products.
Now what I’d like to do is look at even more exotic ideas, and in particular, I’d like to take a look at quotients. So what that means is looking at a function that’s actually built up upon another function divided by a function. So we have two functions and then we take the quotient – we divide them – and ask what’s the derivative of that quotient?
So let’s say this as a problem first, so we can say this is a question to chew on here. So the question is, supposed that I have a function – let me called it q for quotient – and it equals a top function, f(x), and a bottom function – let me call it g(x). And the question is what is the derivative of the quotient? A great natural guess would be you take the derivative of the top and divide it by the derivative of the bottom. But you know what? I bet you that you might not be guessing that now after you saw that the product rule is a little more subtle than we first anticipated. So maybe you’re already guessing that something strange like the product rule is going to have to go on. And that intuition, or that guess, is actually a great one, because in fact, you just don’t take the derivative of the top and the derivative of the bottom. In fact, maybe I can indicate that with a very simple example where we can take a look and try that and see it won’t work.
So with a simple example – very simple example – how about the quotient q(x) equals – let’s do something real easy here. How about ? Let’s try that. So let’s try it in this sort of guess way. So here’s the guess. So the guess would be I’ll take the derivative of the bottom, take the derivative of the bottom. The derivative of the top is going to be 2x + 1. Right? Because I bring the 2 out in front, subtract 1 from the exponent. This is the +x. The derivative is then +1. And divide that by – and the derivative of x is just 1. And so in fact, that just equals 2x + 1. So that’s our guess.
Is it right? Well, we can check it by taking the derivative of this quotient in a way that we’re certain to get the right answers by using old rules. And here’s the old rule that I’m going to use. I’m going to actually just notice that, first of all, I can rewrite this a little bit and simplify it. I can factor out an x. And if I factor out an x, I see that the quotient is equal to – now factor out the x in your mind’s eye. That leaves you with an (x + 1) x. And if I now cancel those x’s, I just see x + 1. And now I can take the derivative of that. That’s actually really easy. So it turns out, this was actually a slightly harder problem slightly in disguise. The derivative of that is 1. Well, this is wrong. So we can see that the green was, once again, wrong. Green is a wrong color.
But again, please remember that wrong is not synonymous with bad. It’s just wrong, but not bad. Anyway, this does inspire that there’s more to the quotient than meets the eye. And in particular, the next thing I want to tell you, in our attempt to learn about fancy methods for taking derivatives, is called the quotient rule. So the quotient rule is a means, by answering this question, how do you take the derivative of a quotient of two functions? And just as with the product rule, it’s slightly elaborate and not obvious, because the initial guess turns out not to be a correct one.
So let me tell you what the quotient rule is. Here is the quotient rule. So suppose that we wish to take the derivative of this quotient of two functions. So f(x), some function, divided by g(x), some other function. Well, the answer is the following. It equals – I’m going to tell you exactly what it equals now. It equals the bottom multiplied by the derivative of the top. Sounding like the product rule a little bit, slight departure though, because now we subtract off the top multiplied by the derivative of the bottom. So it’s sort of like the product rule a little bit. It’s the bottom times the derivative of the top minus the top times the derivative of the bottom. But we’re not done yet, because it’s all divided by the bottom squared. I told you that’s sort of a surprise. Right? A little surprising there. That is the quotient rule.
And let’s try our hand at the quotient rule to see if this actually is at least a reasonable looking thing, to try this with the example that we just had before. So now let’s try it with finding the derivative of . So what do we do? And we know what the answer is, by the way. We know what the answer is. We just found the answer. We saw that the answer was actually 1. We saw that the answer was 1. So let’s see if we get 1 using this method.
Now again, you can memorize this in formulaic, but what I do personally is I remember it as a little chant. And I really mean chant, because when I’m in the middle of this complicated looking thing, I’ve got to stop, regroup and chant to see where I am. And the way I chant is, the derivative of the top divided by the bottom is the bottom times the derivative of the top minus the top times the derivative of the bottom, all over the bottom squared. I chant it just like that, with tops and bottoms.
Also let me point out the fact that there is a negative sign there. You can’t flipflop the order of these things. Otherwise, you’re going to be off. So you always have to start with the bottom times the derivative of the top then subtract the top multiplied by the derivative of the bottom, all over the bottom squared.
Okay, let’s try it with this example here. So we have the bottom multiplied by the derivative of the top, the quantity 2x + 1, minus the top, x2 + x, times the derivative of the bottom – and the derivative of x is just 1 – all divided by the bottom, x2. Well, I’ve got to tell you this doesn’t look like 1 to me. This looks pretty complicated. Let’s simplify it and see what it looks like after we simplify. Well, if I distribute that x, I see 2x2 + x. So that’s just distributing that. And then I have the .
A mistake has been made. I once again forgot to subtract both of these terms. I need parentheses around here, so I have to subtract off that x. In fact, that’s a pivotal fact, because if I distribute that negative sign, what I do see is the following. I see a 2x2 minus an x2. If I have two x2’s and subtract off one x2, I’m just left with an x2. And then I’m left with x and then I have a x. Wait a minute, x  x is zero, so all I’m left with is x2. This term remains and these drop out. I divide by x2. That equals 1. Well that was what we knew the derivative to be when we computed it. So in fact, at least this seems reasonable for this particular problem. And in fact, it’s not only reasonable for that problem, it’s reasonable for all problems where, in fact, you have a quotient.
Let me try to illustrate this with one example. We’ll use the quotient rule. And in fact, I will, once again, so that we can share the fun. There’s no point in me having all the fun, you just watching enviously along as I come over to you on the information superhighway here. I want you to find the derivative of this. So let me write down the function here. q(x) = . I want you to try this one now on your own – come up with a guess – and feel free to use the quotient rule. Feel free to use the quotient rule. Give this one a shot right now.
How’d you make out? Well, let’s see how we did. Let me use the quotient rule because this is the quotient of two functions. And so what do I have to do? Well, I’m going to chant my chant. So q’, the derivative, is – by the way, look how complicated that is. Will you just stop for a second and look how complicated? We are now in the position of taking derivatives of these complicated functions. Could you imagine if you had to take the derivative of this function using the definition? Wow, that would be some work, and yet, now not that big of a deal. I’d miss a long formula, but better this than the definition. So no backtalk, otherwise, I’ll bring the definition back out here, so watch it. Just joking.
So let’s take the derivative of this using the quotient rule. So I chant it. The bottom – so I write out the bottom – 3x2  6, all multiplied by the derivative of the top. And the derivative of the top I bring down to 3 x 6 is 18x31, which is 2, plus 2 x 2 is 4x to the – and 2  1 is 1. Minus – and then the derivative of 14x is just 14. And now I lost my place, so I’ve got to start all over again. Bottom times the derivative of the top minus the top – I’m going to continue over here. 6x3 + 2x2  14x all multiplied by the derivative of the bottom. So now you take the derivative of the bottom, which is 6x and the derivative of 6 – that’s a constant – is zero. Am I all done? Yes. Bottom times the derivative of the top minus the top times the derivative of the bottom, all that over the bottom squared. Wow.
You could simplify that if you wanted to, but guess what? I have no desire to simplify this. Anyway, just a matter of getting down these facts for how to take derivatives of quotients, and then just extending it and using it very carefully. And I really do recommend the chant method so you don’t get confused where you are during this process. Why don’t you try some on your own? And I’ll see you in a bit. Bye.
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There are a few things I disagree with though a bit. One of which is the chanting deal. All you really need to do for product and quotient rules is memorize 2 letters and a ' symbol with them.
product rule = fg' + gf' (where you make f = to a function and g = to a function)
quotient rule = (gf'  fg') / g^2.
Never been a huge fan of "top times bottom" chants and all that stuff. Also, it would be nice if Mr. Burger simplified some of the problems sometimes when he gets to the end. It's nice to say well we're done, but technically you're not unless multiplying everything out would be ridiculously complicated.
At any rate though, his teaching style is still very effective and helps to keep me focused when I'm drifting in my calculus book.