Calculus: An Introduction to the Chain Rule

This lesson has not been reviewed.

Please purchase the lesson to review.

Computational Techniques
The Chain Rule
An Introduction to the Chain Rule Page [1 of 4]
Well, now we know how to take derivatives of functions that could be, actually, pretty complicated, being strung together with a lot of plusses and minus signs. So, you could have two X cubed minus seven X squared 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 stringing them all together.
Right now, what I want to ask you is the following challenge question. What happens if you want to take the derivative of a product? So, in fact, let me actually pose that as a sort of meta question, but I won't actually write down that particular question. Suppose you wanted to find the derivative of a product, guess a formula for taking the derivative of the product. So, for example, suppose you want to take the derivative of a product of two functions, what would you do? Remember, take the derivative, for example, of the sum of two functions, you'd just take the derivative of each of the functions and then add them up. Okay, so right now let's have you make a guess and I'll come back and see if you're right.
Okay, well, let's see how you did. I think this actually may turn out to be a trick question, but we'll see, we'll see. Let's look at a specific example to get warmed up here. Let's take a look at the function F of X equals four X cubed, plus one, multiplied by X squared, minus one. Notice, by the way, I do have a product of two separate functions. Well, here, I think, is a great, great guess. So, here 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 is a great guess. In fact, let me do the great guess in this red color, I think it's such a great guess.
So, I'll just do like we do with addition. I'll take the derivative of this, and take the derivative of that and put them together with multiplication. The derivative of that, that's not too bad. I bring the three out in front, make that a four, make that a, I'm sorry, a twelve, three times four is twelve, X raised to the power of three minus one, which is two, and then the plus one gives me a plus zero, because, remember, one 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 two out in front, so that's a two, times X to the first power, again, minus a constant, so that'd be minus zero. And, so, we see the answer is two times twelve, which is twenty-four, X raised to the X squared times X gives me an X cubed. 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. Well, how could I be certain I'm taking the derivative of this as 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, you know, maybe there's some people that don't, or didn't learn to FOIL, because I learned to FOIL things as a kid. Let me see maybe I should tell you about FOIL 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, you multiply by the first, the F, then the outside, multiply them together, the O. The inside, multiply them together, the I, and the last, multiply them together, and FOIL. So, that's like me saying FOIL. I hope you knew that, but, if not, I'm sorry. I should have told you that earlier. Well, now let me FOIL this out here. I'll do that pretty fast here. Four X cubed times X squared, is going to be four, X to the fifth power. Remember, I add exponents when you multiply the bases. And the, uh, the outside term is a minus four X cubed, the inside term is plus X squared, and the last term gives me a minus one.
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. So here's the actual, the actual answer. Remember, that red was just our guess. Now we're going to say what it really is. What if you would say, "Gee, Professor Burger, you shouldn't write down stuff like that because someone might think that, take it out of context, and think that that's actually, you're claiming the answer to be that." Well, that's their problem. They shouldn't be taking it out of context, you can either watch this thing or not, you know, that's what'd I say. So, I think this is great. Making guesses, by the way, is great.
Okay, let's take the actual answer, I'm going to take the derivative now of this thing right here. Okay, it's just a sum and difference of people, so I can, just, you 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 four X to the fifth, I bring down the five. I get a twenty, X to the five minus one, which is four, subtract off, bring the three down in front of the four. I get a twelve X to the three minus one, which is two, then I have a plus X squared, gives me a two out in front. Two X to the two minus one, which is one, derivative of a constant is zero. That is the answer, looks very different than 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 first, derivative of second and multiplying them. That doesn't work, because the answer is much more complicated. Look at that, that doesn't look like anything that has 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 each of 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, there is no way of doing it.
Now, I'm going to show you a magic trick. Now, I'm going to show you a little magic trick. And, uh, first, I'm going to make this disappear. Oh, in fact, you know I'm not through with this. So, what I'm going to do is I'll put the question over there, I'll 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's move this up, in fact I'll just keep it up here, I'll put it right there, put it over here. I keep moving it, you know, it's sort of a function that just won't stop moving. It's a mover and a shaker. In fact, it doesn't even want to go to, I want to keep it here because I've got to read it. See, I, again, my view is of that, well see, let me clear this for you. We are on the same screen, the same plane, you know, I am, as that, so I just can't stick my head out. We don't have the technology to stick my head out and look over there and see where that is. So, I'm, but I need to see the derivative, so I'm sorry, that's why -- okay, don't worry, okay, I digress.
All right, let me write the function up there. This is all for me. I know it's there, but for me, the original function was four X cubed plus one times X squared minus one, and we actually, we saw what the actual derivative is, the actual derivative we saw. We computed this, and I'm going to have to look over here, and we saw it was twenty X to the fourth minus twelve X squared, plus two X. Okay, I think we're all now caught up. And, and, and my question to you, which I think is a, is a 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 naive thing, which was the natural thing to try, by the way. Always try the easy stuff first. Derivative of this, multiply the derivative of that, it didn't work. So, now we need to think of something else.
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 twenty X to the fourth and I'm actually going to break it into two pieces. So, this actually equals, I'm not going change any thing, I'm just going to break this into two pieces. I'm going to write this as eight, eight X to the fourth, but if I write eight X to the fourth I owe you some because here I have twenty X to the fourth. So, actually, how many do I owe you? I think I owe you about twelve, so I'd better write those in, and then everything else I'm just going to keep the way it is. All I did was break this twenty X to the fourth into two pieces, uh, eight X to the fourth plus twelve X to the fourth, and the rest remains.
Now I want to rearrange them, of course, when you add numbers, seven plus three is the same thing three plus seven, 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, uh, eight X to the fourth, plus two X. I'm just writing those two terms down here, and I'll write the rest of the terms down here now. Twelve X to the fourth, minus twelve X squared. And, now, you'll notice, actually, I can factor some things out of here. Here I could actually factor out a common factor of two X. Let me actually do that. Let me factor out the common factor of two X. What am I left with? Well, if I take a two X out of here I'm left with a four X cubed. If I take a two X out of there, I'm just left with a one. So, all I did there was look at this term and factor out the common factor of two X. Distribute that back in and you'll see it's exactly that.
Okay, now, what should I do here? Well, here let me factor out the common factor of twelve X squared. If I take a twelve X squared out of this piece I'm left with a, an X squared. If I take a twelve X squared out of there, I'm left with just a minus one; times that twelve X squared. 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, as the second thing there. So, I've taken this mysterious looking answer and I've magically, inside of that, voila, uncovered 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 a derivative of that piece. And this is a derivative of that piece. So, what have we discovered? We discovered that, in fact, there is a formula, there is a, a system of taking derivatives of products. It's not the naive one; it's the more elaborate one. It's one where you write down the first term; not it's 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, that always works.
So, in fact, we are now in a position to really pick up and look at fancy methods for taking derivatives - fancy methods. 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 know 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, now how can you remember this? Well, one, one, way is to actually memorize this formula, F times G, the derivative of that is F times derivative of G plus G times derivative of F. That is honestly not the way I remember it. I'll tell you, you might not like this method or not, but I'll tell you how I remember it. I remember it by saying it in terms of the first and second. So, in my mind, whenever I do a product of a 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 uh, P of X. Suppose it equals five X cubed, plus six X squared, minus one, times three X to the nine, minus one half X, plus seven. You want to find the derivative of that. Well, you could multiply it all out like we did before and get the correct actual answer, or, we're now empowered, using this fancy method of the, the Product Rule to actually just use as fact. So, the derivative, look at this -- really intense functions we can now take the derivative of. The first, so I write down the first, five X cubed, plus six X squared, minus one, 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 nine. Nine times three is twenty-seven. X to the nine minus one, which is eight, minus, and then, well, I've got X to the first power, bring that down. I just see then, a half times one, which is a half. X to the one, minus one is zero. X to the zero is just one, so I won't write anything in there, plus, and a derivative of a constant is zero, and that's where I am.
Now, this is the place where I always get so confused because this is a multi-step process. I've got to step back and I've got to re-chant the, the Product Rule again to see where I am. And, I think it's a great idea, by the way, of course, that's just because I do it, but you might want to do this too. Whenever you get to this stage, go back and re-chant the Product Rule and see where you are. First times the derivative of the second - oh, okay, plus the second, so I write down the second. So three X to the ninth, minus one half X plus seven 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, I'll bring the three out in front. That's fifteen X to the three, minus one, which is two plus. Bring the two down, times six is twelve, X to the two minus one, which is one 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, 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 two X to the fourth, minus seven X, minus three, times, and I'm just going to keep writing over here, square root X minus one over X, plus one. Now, let me just tell you, please, please, uh, I want to remind you of the little chant here. First times derivative of the second, plus second times 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'd you make out? Are you getting it? It takes a while, but I think once you get it, you'll feel really good. We're going to use the Product Rule here. The Product Rule -- I'll put it way down on the bottom here, in case you want to take a look at it. 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 longie, folks. It's the first, so that's two X to the fourth, minus seven X, minus three, times the derivative of the second. Well, now I've got to take the derivative of that, and that's going to be a bit of work. So, I've got to chew this up carefully. By the way, notice how I'm putting parenthesis here, because I've got to multiply those pieces together.
Well, I've got to take the derivative of the square root of X, so you might actually want to run off and, 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 square root of X, that equals X to the one half power, and now I know how to take the derivative. I bring the one half in front, X to the, and then one half minus one is minus a half power. I could rewrite that. The minus sign means it's underneath the square, the one half means square root, that two remains. 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, the little green problem there. So the moral of that part of the story is that here I could just write down one over two square root of X. Then I've got to subtract off the derivative of one over X.
Well, that actually might be another little problem you might want to do off on the sidelines here. In fact, I'll do that one for you on the sidelines really fast. How would you do the derivative of one over X? I'd first write that as X to a power, and that would be X to the minus one power, because it's underneath. The derivative of that, I'd bring the minus one out in front, X, and then the minus one, minus one is minus two, and so, this equals a minus one, and that negative exponent means downstairs X squared. So, the derivative of one over X, I now see is negative one over X squared. If I insert that in right now, I see a negative one, so that negative and this negative produce a positive one over X squared, and the derivative of a one 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, plus the second, square root of X minus one over X plus one, multiplied by the derivative of the first. The derivative of the first, let's do that. Four times two is eight, X to the four minus one is just three, minus seven X. The derivative of seven X is just seven. The derivative of a constant is zero, and so there we have it. There is the derivative of that complicated looking 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.