Preview
Buy lesson
Buy lesson
(only $3.96) 
You Might Also Like

Calculus: Derivatives of Inverse Functions 
Calculus: Logarithmic Differentiation 
Calculus: Differentiating Logarithmic Functions 
Calculus: L'Hopital's Rule, 1 to an Infinite Power 
Calculus: Basic Uses of L'Hopital's Rule 
Calculus: Derivative of the Natural Log Function 
Calculus: Derivatives of Exponential Functions 
Calculus: Derivatives of Trigonometric Functions 
Calculus: Using the Chain Rule 
Calculus: The Quotient Rule 
College Algebra: Solving for x in Log Equations 
College Algebra: Finding Log Function Values 
College Algebra: Exponential to Log Functions 
College Algebra: Using Exponent Properties 
College Algebra: Finding the Inverse of a Function 
College Algebra: Graphing Polynomial Functions 
College Algebra: Polynomial Zeros & Multiplicities 
College Algebra: PiecewiseDefined Functions 
College Algebra: Decoding the Circle Formula 
College Algebra: Rationalizing Denominators

Calculus: The Quotient Rule 
Calculus: Using the Chain Rule 
Calculus: Derivatives of Trigonometric Functions 
Calculus: Derivatives of Exponential Functions 
Calculus: Derivative of the Natural Log Function 
Calculus: Basic Uses of L'Hopital's Rule 
Calculus: L'Hopital's Rule, 1 to an Infinite Power 
Calculus: Differentiating Logarithmic Functions 
Calculus: Logarithmic Differentiation 
Calculus: Derivatives of Inverse Functions
About this Lesson
 Type: Video Tutorial
 Length: 20:44
 Media: Video/mp4
 Use: Watch Online & Download
 Access Period: Unrestricted
 Download: MP4 (iPod compatible)
 Size: 224 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 products and applying the product rule of differentiation. The derivative of the product of two differentiable functions is NOT equal to the product of the derivatives of both functions. Instead, you must learn and apply the product rule to calculate the derivative of the product of two functions. The product rule states that the derivative of f(x)*g(x) = f(x)* derivative of g(x) + g(x)* derivative of f(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 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
 2174 lessons
 Joined:
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
 Quite helpful in going over the product rule
 01/24/2009

As a follower of cramster and calcchat.com, I was pleased to come across mindbites while searching for particularly difficult concepts for me to grasp.
Professor Burger looks like he definitely knows what he's doing and it's fairly easy to follow him in what he does. I learned this a little quicker the first time in my regular Cal I class, but this refresher was nice and straightforward. The guy is definitely passionate about math, that's for sure, lol. It shows too, which is outstanding.
Computational Techniques
The Product and the Quotient Rules
The Product Rule Page [1 of 4]
Now we know how to take derivatives of functions that could be actually pretty complicated, being strung together with a lot of pluses and minus signs. So you can have 2x3  7x2 plus even the square root of x. And you can now find the derivative pretty easily by using this basic process of looking at each term separately, taking the derivatives and then string them all together. What I now want to ask you is the following challenge question. What happens if you want to take the derivative of a product? In fact, let me actually pose that as a sort of metaquestion where I won’t actually write down a particular question. Suppose you want to find the derivative of a product – guess a formula for taking the derivative of a product? So for example, suppose you wanted to take the derivative of the product of two functions, what would you do? Remember to take the derivative, for example, of the sum of two functions, you just take the derivatives of each of the functions and then add them up. So right now, let’s have you make a guess and I’ll come back and see if you’re right.
Well, let’s see how you did. I think this actually may turn out to be a trick question, but we’ll see. Let’s look at a specific example to get warmed up here. Let’s take a look at the function f(x) = (4x3 + 1)(x2  1). Notice, by the way, I do have a product of two separate functions. Here, I think, is a great guess, and maybe you made this guess. Take the derivative of this, take the derivative of that and multiply it together. So let’s do that. So here’s a great guess. In fact, let me do the great guess in this red color, because I think it’s such a great guess.
So I’ll just do like we did with addition. I’ll take the derivative of this and take the derivative of that and put them together – multiplication. The derivative of that is not too bad. I bring the 3 out in front and make that a 12. x raised to the power, 3  1, which is 2. And then the plus 1 gives me a plus zero, because remember 1 is a constant. The derivative of a constant is zero. I could write the plus zero there, but I won’t. I’ll just keep that out. And then I’ll multiply that by the derivative of this term. Bring the 2 out in front, so that’s a 2 times x1, again minus a constant, so that would be minus zero. And so we see the answer is 2 x 12, which is 24x raised to the – x2 times x gives me an x3.
So that is, I think, a great guess. How can we determine if that’s the right answer or not? How can we, in general, determine if that recipe – that rule that we’ve created, the rule of taking the derivative of this and multiplying by the derivative of that – is right or not? Well, one way is to take the derivative of this in a manner in which we’re certain of the answer and then compare it to this guess. How could I be certain of taking the derivative of this is an answer? Well, one thing I could do is sort of multiply that all out – FOIL it – and then take the derivative of each term using the procedure we know to be correct. So let’s try that right now. So let me actually FOIL all this out for you. So I’m going to untangle all that.
By the way, I hope you all know what FOIL means. Maybe there’s some people that didn’t learn those FOILs. And I learned this FOIL thing as a kid. Let me tell you about FOILs just for a second here. FOIL stands for “First, Outside, Inside and Last.” And the idea behind FOILing is that if you have two things like this and you want to multiply them together, you first take the first – multiply by the first, the F; then the outside, multiply them together, the O; the inside and multiply them together, the I; and the last, multiply them together and FOIL. So that’s why I keep saying FOIL. I hope you knew that, but if not, I’m sorry. I should have told you that earlier.
But now let me FOIL this out here. I’ll do that pretty fast here. 4x3(x2) is going to be 4x5. Remember I add exponents when you multiply the bases. And the outside term is a minus 4x3. The inside term is plus x2. And the last terms give me a minus 1. So there, I just FOILed it out and now I can take the derivative. So I take the derivative of this, so here’s the actual answer. Remember that red was just our guess. Now, we’re going to say what it really is.
I wonder if people would say, “Gee, Professor Berger, you shouldn’t write down stuff like that, because someone might take it out of context and think that you’re actually claiming the answer to be that. Well, that’s their problem. They shouldn’t be taking it out of context. You’re either going to watch this thing or nor, that’s what I say, so I think this is great. Making guesses, by the way, is great.
Let’s take the actual answer. I’m going to take the derivative now of this thing right here. Okay? It’s just the sum and difference of people, so I know how to do that. I just take each one individually, take its derivative and string them all together. So the actual retail value is the derivative of 4x5. I bring down the 5. I get a 20x to the 5  1, which is 4. Subtract off. Bring the 3 down in front of the 4. I get a 12x to the 3  1, which is 2. Then I have a plus x2. It gives me a 2 out in front – 2x to the 2  1, which is 1. The derivative of a constant is zero. That is the answer. It looks very different from my guess. My guess must be wrong.
Now some people would be saddened by this, but I’m not, because I learned a lesson. I learned that, in fact, you can’t take derivatives of products by just taking derivative of the first and the derivative of the second and multiplying them. That doesn’t work, because the answer is much more complicated. Look at that. That doesn’t look like anything to have to do with these two pieces right here. This is really just a complicated looking thing.
In fact, I don’t see much to do with this at all, really. I mean, how could you just look at these terms individually – looking at each of these terms individually and somehow getting this answer? I don’t see how to do it. Maybe there is no way of doing it. Now I’m going to show you a little magic trick, and first I’m going to make this disappear. In fact, I’ll just put it over there.
So what I’m going to do is I’ll put the question over there – I put the original function. So there’s that function way over there. And now I’m going to put this answer way over there. Let me write the function up there. This is all for me. I know it’s there, but for me, the original function was (4x3 + 1)(x2  1). And we actually saw what the actual derivative is. The actual derivative we saw – we computed this, and I’m going to have to look over here. We saw it was 20x4  12x2 + 2x. I think we’re all now caught up. And my question to you, which I think is real serious challenge now, is how do we just take that function and this function separately, do something to it, and then get the derivative to look like this? You see we tried the naïve thing, which was a natural thing to try, by the way. Always try the easy stuff first – derivative of this, multiply it by the derivative of that. It didn’t work, so now we have to think of something else. Well, now I’m going to show you a magic trick. So watch me, because this is real magic. I’m searching for a pattern. I’m going to take this 20x4 and I’m actually going to break it into two pieces.
So this actually equals – I’m not going to change anything. I’m just going to break this into two pieces. I’m going to write this as 8x4. But if I write 8x4, I owe you some, because here I have 20x4. So actually, how many do I owe you? I think I owe you about 12, so I’d better write those in. And then everything else, I’m just going to keep the way it is. So all I did was break this 20x4 into two pieces – 8x4 + 12x4 – and then the rest remains.
Now I want to rearrange them. Of course, when you add numbers, 7 + 3 is the same thing as 3 + 7, so you can do any order you want. Let me take these two people and put them together. So I put these two people together, then I see 8x4 + 2x. I’m just writing those two terms down here, and I’ll write the rest of the terms down here now: 12x4  12x2. And now you’ll notice, actually, I can factor some things out of here. Here I can actually factor out a common factor of 2x. So let me actually do that. Let me factor out the common factor of 2x. What am I left with? Well, if I take a 2x out of here, I’m left with a 4x3. If I take a 2x out of there, I’m just left with a 1. So all I did there was look at this term and factor out the common factor of 2x. Distribute that back in and you’ll see exactly that.
Now what should I do here? Well, here, let me factor out the common factor of 12x2. If I take a 12x2 out of this piece, I’m left with an x2. If I take a 12x2 out of there, I’m left with just the minus 1 times that 12x2. You can check again. But something magical has happened. Look down. Look at this. This is exactly the same as the first thing here. This is the exact same thing as the second thing there. So I’ve taken this mysteriouslooking answer and I’ve magically, inside of that – voila! I’m covered – bits of the original thing. Of course there’s all this other stuff left over.
But look at that other stuff. That’s actually familiar too. This is the derivative of that piece. And this is the derivative of that piece. So what have we discovered? We discovered that, in fact, there is a formula – there is a system – of taking derivatives of products. If not the naïve one, it’s a more elaborate one. It’s one where you write down the first term – not its derivative, but the first term – and multiply it by the derivative of the second term. Then you’ve got to add to it the second term multiplied by the derivative of the first term. And you know what? That always works.
So in fact, we are now in the position to really pick up and look up at fancy methods for taking derivatives. And the first thing we’re going to look at, the one we just discovered here. The one we just are discovering is what’s known as the product rule. And what the product rule says is the following. If you want to take the derivative of a product – that is something that’s made up of the product of two functions – what you do is you take the first and you multiply it by the derivative of the second. You add to it the second multiplied by the derivative of the first. That is the product rule.
Let me write that down for you. This is really fun. So if you want to take the derivative of a product – look how I have here – I want to take the derivative of the product, this function multiplied by that function, f times g. What is it? It turns out it equals the following, and this is called the product rule. It’s the first multiplied by the derivative of the second plus the second multiplied by the derivative of the first. That is called the product rule.
By the way, how can you remember this? One way is to actually memorize this formula, f times g. The derivative of that is f times the derivative of g plus g times the derivative of f. That is honestly not the way I remember it. You might not like this method or not, but I’ll tell you the way I remember it. I remember it by saying it in terms of first and second. So in my mind, whenever I do a product rule problem, even by myself or whatever it is, I think first times the derivative of the second – this is actually what I say in my mind – the first times the derivative of the second plus the second times the derivative of the first. That’s how I think about the product rule.
Let’s do an example. Let’s find the derivative. Let’s call it p(x). Suppose it equals (5x3 + 6x2  1)(3x9  x + 7). You want to find the derivative of that. Well, you could multiply it all out like we did before and get the correct actually answer. Or we’re now empowered, using this fancy method of the product rule, to actually just use this fact. And so the derivative – look at these really intense functions we can now take the derivative of. So I write down the first. 5x3 + 6x2  1 multiplied by the derivative of the second. So I take this and multiply it by the derivative of that. So now I go off and figure out the derivative. Well, I know how to do that. I take each piece separately – bring down the 9. 9 x 3 = 27. x to the 9  1, which is 8, minus – and then I’ve got x1. Bring that down. I just see then x 1, which is . x11 is zero. x0 is just 1. So I’m won’t write anything in there. Plus – and the derivative of a constant is zero, and that’s where I am.
Now this is a place where I always get so confused, because this is a multistep process. I’ve got to step back and I’ve got to rechant the product rule again to see where I am. And I think it’s a great idea, by the way. Of course, that’s because I do it, but you might want to do this too. Whenever you get to this stage, go back and rechant the product rule and see where you are. The first times the derivative of the second. Oh, okay. Plus the second – so I write down the second. So 3x9  x + 7. And I multiply that by the derivative of the first. So now I’ve got to go back and I’ve got to compute the derivative of that. Well, bring the 3 out in front. That’s 15x31, which is 2, plus bring the 2 down, times 6 is 12x to the 2  1, which is 1, minus – and the derivative of a constant is zero. So I get that. And that really long answer is the derivative.
So in fact, look at these functions we can now take the derivative of. Using the product rule, we can actually take the derivative of even really complicated functions that, if we were to untangle it, would require too much work, to unFOIL all that. But now, the product rule, no problem.
Let me do one last example using the product rule and then I’ll let you try a whole bunch. In fact, you know what? I’ll make up this example right now and I’ll have you try it first. Why should I be having all the fun here? You should be having some of the fun here too. This is for both of us.
Here we go. How about this one? Have some fun with this. How about (2x4  7x  3)( . Now let me just tell you, I want to remind you of the little chant here: first times the derivative of the second plus second times the derivative of the first. And I invite you to actually take on that chant when you’re in the middle of the problem. Whenever you get stuck or you forget or you lose your way, don’t panic. Just chant the little mantra. Okay, see how you do and then we’ll do it together.
Well, how did you make out? Are you getting it? It takes awhile, but I think once you get it, you’ll feel really good. So we’re going to use the product rule here. The product rule, I’ll put it way down in the bottom here, in case you want to take a look at it. It’s the first times the derivative of the second plus the second times the derivative of the first. And I’m going to keep chanting that throughout this problem because this is going to be a longy, folks. It’s the first – so that’s 2x4  7x  3 times the derivative of the second. Well now, I’m going to take the derivative of that, and that’s going to be a bit of work. So I’ve got to chew this off carefully.
By the way, notice I’m putting parentheses here because I’ve got to multiply those pieces together. Well, I’ve got to take the derivative of . So you might actually want to run off and do that separately here. In fact, maybe I’ll just run off and do that separately as well. Take a little piece of paper here and do that separately really fast for us. So how would I do that? I’d say , that equals . And now I know how to take the derivative. I bring the in front and then x to the – and then  1 is  power. I could rewrite that. The minus sign means it’s underneath. The means square root. That 2 remains and so I could write it like that. So in fact, that’s the derivative of that piece. I had to go off and do that. It’s a little green problem there. So the moral of that part of the story is there here, I could just write down . Then I’ve got to subtract off the derivative of .
Well, that actually might be another little problem you might want to do off on the sidelines here. So in fact, I’ll do that one for you on the sidelines really fast. How would you do the derivative of ? I’d first write that as x to a power, and that would be x1, because it’s underneath. So the derivative of that, I bring the 1 out in front – x – and then the 1  1 is 2. And so this equals the 1. And that negative exponent means downstairs x2.
So the derivative of I now see is . If I insert that in right now, I see a 1. So that negative and this negative produce a + . And the derivative of a 1 is zero, so I just add zero. And okay, where am I? Well, I don’t know. I lost track of where I am. I’ve got to go back to my little chant. So I’ve got the first multiplied by the derivative of the second – oh, okay, now I know where I am – plus the second, , multiplied by the derivative of the first. So the derivative of the first, let’s do that. 4 x 2 is 8x to the 4  1, which is 3, minus 7x. The derivative of 7x is just 7. The derivative of a constant is zero. And so there we have it. There is the derivative of that complicatedlooking function and I just used the product rule. It’s just this chant. So I want you to think about this chant and try some on your own. Have fun and I’ll see you soon.
Get it Now and Start Learning
Embed this video on your site
Copy and paste the following snippet:
Link to this page
Copy and paste the following snippet:
As a follower of cramster and calcchat.com, I was pleased to come across mindbites while searching for particularly difficult concepts for me to grasp.
Professor Burger looks like he definitely knows what he's doing and it's fairly easy to follow him in what he does. I learned this a little quicker the first time in my regular Cal I class, but this refresher was nice and straightforward. The guy is definitely passionate about math, that's for sure, lol. It shows too, which is outstanding.