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Calculus: Using the Chain Rule

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About this Lesson

  • Type: Video Tutorial
  • Length: 12:54
  • Media: Video/mp4
  • Use: Watch Online & Download
  • Access Period: Unrestricted
  • Download: MP4 (iPod compatible)
  • Size: 139 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 Chain Rule (3 lessons, $6.93)

The chain rule is a technique used for differentiating composite functions. In this lesson, you will learn the chain rule as well has how to apply it. Professor Burger will carefully walk through mistakes to avoid when using the chain rule as well as the correct way to use the chain rule in conjunction with the product rule for differentiation. The chain rule states that the derivative of g(h(x)), where g(x) and h(x) are differentiable functions, is the derivative of g(h(x)) times the derivative of h(x).

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 UT-Austin 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, p-adic 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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Founded in 1997, Thinkwell has succeeded in creating "next-generation" 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 technology-based textbooks. For more information about Thinkwell, please visit www.thinkwell.com or visit Thinkwell's Video Lesson Store at http://thinkwell.mindbites.com/.

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Nopic_orng
thank you!
02/23/2010
~ Aisha1

thank you! saved me on my midterm!!!!

Nopic_orng
thank you!
02/23/2010
~ Aisha1

thank you! saved me on my midterm!!!!

Computational Techniques
The Chain Rule
Using the Chain Rule Page [1 of 2]

Well, we are now in the middle of learning these really, really fancy methods for taking derivatives. Let me just recap really fast. We had just see 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 are 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 second, plus second times derivative of the first. Sort of neat yet, sort of, not intuitive, not intuitive formula for finding the derivatives of products.

Okay, what now 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 function. So, we have two functions and then we take the quotient, we divide them and ask what’s the, what’s the derivative of that quotient? So, let’s try -- let me say this as a problem first, so this is sort of a question to sort of chew on here. The question is, suppose that I have a function, let me call it a Q for quotient, and it equals – well, a top function, f of x, and a bottom function, let me call it g of 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, a very simple example, so a simple example – how about the quotient, Q of f equals, let’s do something really easy here. How about x squared, plus x all divided by x. Let’s try that. So, let’s, let’s try it in this sort of guess way. Here’s the guess. So the guess would be – well, I’ll take the derivative of the top, take the derivative of the bottom. The derivative of the top is going to be two x plus one, right. Then I bring the two out in front, subtract one from the exponent, which is a plus x. The derivative is then plus one, and divide that by, and the derivative of x is just one, so, in fact, that just equals two x plus one.

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 answer by using old rules. And here’s the old rule that I’m going to use. I’m going to actually -- 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 plus one, all multiplied by the x. And if I now cancel those x’s, I just see x plus one. And now I can take the derivative of that, that’s actually real easy. So it turns out this is actually a certainly harder problem, slightly in disguise. The derivative of that is one. Well, this is wrong. 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.

Okay. 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, 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, not obvious, because the initial guess turns out not to be, not to be a correct one.

Let me tell you what the Quotient Rule is. Here is the Quotient Rule. Here is the Quotient Rule. Now suppose that we wish to take the derivative, take the derivative of this quotient of two functions. So, f of x, some function, divided by g of x, some other function. Well, the answer is the following. It equals; here 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, the quotient, the Product Rule a little bit. It’s the bottom times the derivative of the top minus the top times derivative of the bottom, but we’re not done yet, it’s all divided by the bottom squared. I told you that’s sort of a surprise – little surprising there. That is the Quotient Rule and let’s try our hands at the Quotient Rule to see that this is actually a, is at least a reasonable looking thing, to try this with the example that we just had before.

Now, let’s try it, try it with the finding the derivative of x squared plus x, all over x. So, what do we do? Well, I’m going to – 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 one. We saw that the answer was one. So, let’s see if we get one using this method. Well, 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, re-group and chant to see where I am. And the way I chant this, the derivative of top divided by a 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, the tops and bottoms. Also, let me point out the fact that there is a negative sine there. You can’t flip flop the order of these things; otherwise, you’re going to be off. 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, multiply it by the derivative of the top, the quantity two x plus one, minus the top, x squared plus x, times the derivative of the bottom, and the derivative of x is just one, all divided by the bottom x squared. Well, I’ve got to tell you this doesn’t look like one to me. This looks pretty complicated.

Let’s simplify it; see what it looks like after we simplify it. Well, if I distribute that x, I see two x squared plus x, so that’s just distributing that, and then I have to minus x squared plus x, all divided by x squared. A mistake has been made. I, once again, forgot to subtract both of these terms. I need parenthesis around here, so I have to subtract off that x. In fact, that’s a pivotal fact, because if I distribute that negative sine what I do see is the following. I see a two x squared, minus an x squared, if I have two x squared and I subtract off one x squared, I’m just left with an x squared. And then I’m left with x, and then I have a minus x. Wait a minute, x minus x is zero. So, all I’m left with is x squared. This term remains and these drop out. I divide by x squared. Huh! That equals one. 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, with -- use the Quotient Rule. And, in fact, I will, once again – so that we can share the fun – there’s no, no point in me having all the fun. And you’re just watching enviously along as I come over to you on the information super highway here. I want you to find the derivative of this. So, let me write down the function here. Q of x equals six x cubed, plus two x squared, minus fourteen x, all divided by three x squared minus six. I want you to try this one now on your own to 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, of two functions, and so, what do I have to do? Well, I’m going to chant my chant. So, Q prime the derivative is – look how funny, look how complicated that is – can we 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 taken the derivative of this function using the definition? Wow! That would be some work. And, yet, now, it’s not that big of a deal. I admit that’s a long formula, but better this than the definition. So no back talk, otherwise I’ll bring my definition back out here, so watch it. Okay, just joking, we’re all done.

Okay, so let’s take the derivative of this using the Quotient Rule. So, I chant it, the bottom, so I write out the bottom, three x squared, minus six, all multiplied by the derivative of the top. And the derivative of the top, I bring down the three time six is eighteen, x to the three minus one, which is two, plus two times two is four, x to the, and two minus one is one, minus, and then, uh, the derivative of minus fourteen x is just fourteen, and now I lost my place. So, I’ve to start all over again.

Bottom times the derivative of the top, okay, minus the top, and I’m going to continue over here, six x cubed plus two x squared minus fourteen x, all multiplied by the derivative of the bottom. So, now you take the derivative of the bottom, which is six x, and the derivative of minus six, that’s a constant of zero. Am I all done? Yes. Bottom times the derivative of the top, minus the top times the derivative of the bottom, all over, all that, all that, 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, it’s just a matter of getting down this fact 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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