WEBVTT mathematics/ap-calculus-ab/hovasapian
00:00:00.000 --> 00:00:05.000
Hello, welcome back to www.educator.com and welcome back to AP Calculus.
00:00:05.000 --> 00:00:08.300
Today, we are going to discuss the chain rule.
00:00:08.300 --> 00:00:14.500
Profoundly important, let us dive right on in.
00:00:14.500 --> 00:00:18.400
Let us start off by recalling this idea of a composite function.
00:00:18.400 --> 00:00:29.900
Recall composite functions.
00:00:29.900 --> 00:00:39.900
Something like, if we had f(x) is equal to x³ and if we had g(x) is equal to, let us say, x² + 1,
00:00:39.900 --> 00:00:47.100
the composite function f(g) is defined as f(g) of x.
00:00:47.100 --> 00:00:54.400
We are basically, wherever we see x in here, we are just putting in the entire function.
00:00:54.400 --> 00:01:00.300
It ends up being x² + 1³, that is it.
00:01:00.300 --> 00:01:08.100
What we want to do, the chain rule, it allows you to find the derivative of a composite function.
00:01:08.100 --> 00:01:09.800
That is it, that is all it is.
00:01:09.800 --> 00:01:13.100
The majority of the functions that you are dealing with are actually going to be composite functions.
00:01:13.100 --> 00:01:17.900
They are not just going to be simple functions like sin(x) or 5x or x².
00:01:17.900 --> 00:01:26.800
They are going to be composite functions like x² + 1³ or sin(x)² + tan² x¹/5.
00:01:26.800 --> 00:01:28.500
Most of them are going to be composites.
00:01:28.500 --> 00:01:32.700
The chain rule is the one thing that you are going to be using over and over again.
00:01:32.700 --> 00:01:36.100
We are going to be spending a fair amount of time with this.
00:01:36.100 --> 00:01:40.800
That is all it is.
00:01:40.800 --> 00:01:45.900
Let me write this down.
00:01:45.900 --> 00:01:57.000
The chain rule allows us to find the derivatives of composite functions.
00:01:57.000 --> 00:02:00.000
I apologize for my slightly sloppy writing.
00:02:00.000 --> 00:02:11.700
Derivatives of composite functions, that is it, that is all it is.
00:02:11.700 --> 00:02:51.000
Let us go ahead and go to blue.
00:02:51.000 --> 00:03:03.300
If f and g, I will leave off the x, are differentiable, and F = f(g) or F = the composite of f(g), then, the composite is differentiable.
00:03:03.300 --> 00:03:15.000
F’ is equal to f’ evaluated at g(x) × g’(x), that is the formula.
00:03:15.000 --> 00:03:17.000
This is the formula for the chain rule.
00:03:17.000 --> 00:03:22.600
It says that if I have a composite function, I take the derivative of the outer function.
00:03:22.600 --> 00:03:29.900
I evaluate at g(x), or I put g(x) into the derivative of the function.
00:03:29.900 --> 00:03:35.500
And then, I multiply it by the derivative of what is inside the derivative of g’.
00:03:35.500 --> 00:03:37.200
This is why they call it the chain rule.
00:03:37.200 --> 00:03:40.200
It makes more sense when you actually see the example.
00:03:40.200 --> 00:03:47.500
You just keep differentiating from the outside in until you end up running out of things to differentiate.
00:03:47.500 --> 00:03:49.000
That is why they call it the chain rule.
00:03:49.000 --> 00:03:52.900
You do the outside function then the inside function, then the next function.
00:03:52.900 --> 00:03:58.700
If you had f, g, h, you would do f’, you would do g’, then you do h’.
00:03:58.700 --> 00:04:06.300
Again, this is the definition, all of this will make sense when we do the examples.
00:04:06.300 --> 00:04:30.000
In the notation of this dy dx, the notation that we sometimes use, this is expressed this way.
00:04:30.000 --> 00:04:47.300
Let y = some function of the variable u and u = some function of the variable x.
00:04:47.300 --> 00:04:56.800
Then, dy dx is equal to, y is equal to function of u.
00:04:56.800 --> 00:05:01.500
U was equal to a function of x, that means y is actually equal to a function of x.
00:05:01.500 --> 00:05:05.900
When I put this u into here, I get g(x).
00:05:05.900 --> 00:05:07.300
I want to find dy dx.
00:05:07.300 --> 00:05:15.700
Dy dx is equal to dy du × du dx.
00:05:15.700 --> 00:05:24.100
The nice thing about this particular notation is that you get the sort of see that this du and this du,
00:05:24.100 --> 00:05:32.300
because we are treating them like quotients, you get to see that they cancel, leaving you with dy dx.
00:05:32.300 --> 00:05:35.000
Again, it is just another way of looking at it.
00:05:35.000 --> 00:05:41.100
I personally prefer this but that just a personal choice.
00:05:41.100 --> 00:05:49.500
Note carefully.
00:05:49.500 --> 00:06:07.900
When we say that f’(x) is equal to f’ evaluated at g(x) × g’(x), this is f’ at g(x).
00:06:07.900 --> 00:06:13.200
When we have the specific x that we are talking about, we actually have to put in g(x).
00:06:13.200 --> 00:06:17.700
We form f’, we find g(x), we put that in here.
00:06:17.700 --> 00:06:21.600
We evaluate f’ at whatever number we get here.
00:06:21.600 --> 00:06:25.300
And then, we multiply by g’ of whatever that number is.
00:06:25.300 --> 00:06:31.100
This is very important.
00:06:31.100 --> 00:06:33.700
Let us just go ahead and start with the examples.
00:06:33.700 --> 00:06:37.000
I think once you see one or two examples, everything is going to make sense.
00:06:37.000 --> 00:06:43.200
Let us do that.
00:06:43.200 --> 00:06:53.200
Let f(x) = x³ + 5⁵, identify f(x) and g(x), then go ahead and differentiate.
00:06:53.200 --> 00:07:04.900
In this particular case, f is going to be x⁵, the outer function.
00:07:04.900 --> 00:07:13.300
G(x), the inner function, x³ + 5.
00:07:13.300 --> 00:07:17.800
I should have put F, sorry.
00:07:17.800 --> 00:07:20.600
That is okay, you will understand what is going on here.
00:07:20.600 --> 00:07:32.200
F(x) = x⁵, this should probably be, let us go ahead and make this capital because f(x) and f(x), there might be some confusion.
00:07:32.200 --> 00:07:36.900
I know that there is no confusion, but just for the sake of keeping it straight.
00:07:36.900 --> 00:07:41.000
F(x) = x⁵, g(x) = x³ + 5.
00:07:41.000 --> 00:07:43.200
You form your composite function.
00:07:43.200 --> 00:07:53.300
F = f(g) which is equal to f(g), that is that.
00:07:53.300 --> 00:07:56.100
Now we go ahead and differentiate.
00:07:56.100 --> 00:08:03.100
We have f’(x), we differentiate from the outside, in.
00:08:03.100 --> 00:08:13.300
We said that it is equal to, let us do the formula.
00:08:13.300 --> 00:08:20.500
F’ at g(x) × g’(x).
00:08:20.500 --> 00:08:26.400
For this particular function, f’, this is x⁵.
00:08:26.400 --> 00:08:30.600
This thing⁵, we bring it down.
00:08:30.600 --> 00:08:32.500
We treat it like a regular power rule.
00:08:32.500 --> 00:08:41.500
5x³ + 5⁴, this is f’ at g(x).
00:08:41.500 --> 00:08:50.700
F’ is 5 × whatever it is, the 4th, the g(x) is this one right here, × g’(x) what is inside.
00:08:50.700 --> 00:08:56.200
Now, I take the derivative of what is inside, 3x².
00:08:56.200 --> 00:09:00.300
That is my answer, f’(x).
00:09:00.300 --> 00:09:10.400
I simplify, 3 × 5 is 15, x² × x³ + 5⁴.
00:09:10.400 --> 00:09:18.500
That is what I’m doing, I’m just differentiating from the outside, in.
00:09:18.500 --> 00:09:22.700
I have x³ + 5⁵.
00:09:22.700 --> 00:09:28.200
I take care of the power rule first, x³ + 5⁴.
00:09:28.200 --> 00:09:32.700
And then, I take the derivative of what is inside, 3x².
00:09:32.700 --> 00:09:34.400
I keep going if I need to.
00:09:34.400 --> 00:09:39.400
At this point, that is it, this is the last function that is quote inside.
00:09:39.400 --> 00:09:49.700
I’m fine, I’m left with 15x² × x³ + 5⁴.
00:09:49.700 --> 00:09:53.900
The sin of tan(x), identify f(x) and g(x), and then differentiate.
00:09:53.900 --> 00:10:04.200
In this particular case, f is going to equal the sin(x) and g is equal to the tan(x).
00:10:04.200 --> 00:10:09.800
The outside function is sin, the inside function is tan.
00:10:09.800 --> 00:10:24.000
F’(x), this f’, the derivative of this, I take the derivative of the sin function which is cos of tan(x).
00:10:24.000 --> 00:10:25.200
That stays.
00:10:25.200 --> 00:10:33.500
I multiply by the derivative of what is inside, the derivative of tan is sec² x.
00:10:33.500 --> 00:10:38.800
That is my final answer, if I want, I can go ahead and bring that forward.
00:10:38.800 --> 00:10:48.000
Sec² x × the cos of tan(x), either one of these is fine.
00:10:48.000 --> 00:10:50.300
You are working from the outside, in.
00:10:50.300 --> 00:10:54.200
You take the derivative of sin, leaving what is inside alone.
00:10:54.200 --> 00:11:04.800
And then, you multiply it by the derivative of what is inside and you keep going inside until you run out of things that are inside.
00:11:04.800 --> 00:11:12.300
Let us see what we have got here.
00:11:12.300 --> 00:11:17.700
Differentiate f(x) = ∛2x³ – 4x + 7.
00:11:17.700 --> 00:11:22.100
I’m going to rewrite this with a rational exponent.
00:11:22.100 --> 00:11:32.700
I’m going to write 2x³ – 4x + 7¹/3.
00:11:32.700 --> 00:11:36.100
I differentiate my outside function is something to the 1/3.
00:11:36.100 --> 00:11:40.600
I treat it as a regular power rule, then, I differentiate what is inside.
00:11:40.600 --> 00:11:48.400
F’(x) = 1/3, I will leave what is inside alone.
00:11:48.400 --> 00:11:53.400
2x³ – 4x + 7, until I finish with that.
00:11:53.400 --> 00:12:04.600
This is that -1, × the derivative of what is inside, × 6x² – 4.
00:12:04.600 --> 00:12:08.800
I’m done, I can stop there.
00:12:08.800 --> 00:12:17.400
Just start from the outside and work your way in.
00:12:17.400 --> 00:12:23.600
Differentiate f(x) = -5/ x² + 3³.
00:12:23.600 --> 00:12:29.700
Now we have a quotient rule and the denominator happens to be something that is a composite function.
00:12:29.700 --> 00:12:34.400
The derivative of that, we are going to use the quotient rule over the entire thing.
00:12:34.400 --> 00:12:38.100
The chain rule for this one.
00:12:38.100 --> 00:12:44.800
F’(x), this × the derivative of that - that × the derivative of this/ this².
00:12:44.800 --> 00:12:47.000
This × the derivative of that.
00:12:47.000 --> 00:12:59.300
We have x² + 3³ × the derivative of -5 is 0 - -5 × the derivative of this.
00:12:59.300 --> 00:13:14.600
The derivative of this is, we do power rule first, 3 × x² + 3² × the derivative of what is inside, the derivative of x² + 3 is × 2x.
00:13:14.600 --> 00:13:19.900
All of that /x² + 3³².
00:13:19.900 --> 00:13:21.500
It is just 3 × 2.
00:13:21.500 --> 00:13:25.100
This is going to be x² + 3⁶.
00:13:25.100 --> 00:13:35.200
This is 0, I get 2 × 3 is 6, 6 × 5 is 30, - × - is +.
00:13:35.200 --> 00:13:45.500
I get 30 × x² + 3²/ x² + 3⁶.
00:13:45.500 --> 00:13:48.600
That goes with that, leaving 4.
00:13:48.600 --> 00:13:56.600
I get 30/ x² + 3⁴.
00:13:56.600 --> 00:13:58.100
That is it.
00:13:58.100 --> 00:14:05.700
When I take the derivative of this, it is the outside 3, leave what is inside alone, finish that part.
00:14:05.700 --> 00:14:08.700
And then, multiply by the derivative of what is inside.
00:14:08.700 --> 00:14:13.100
That is the chain rule, it is a chain of functions.
00:14:13.100 --> 00:14:19.300
If you have f and g, you are going to have two, one term and two terms.
00:14:19.300 --> 00:14:23.700
If you have f, g, and h, you are going to have one term, two terms, three terms.
00:14:23.700 --> 00:14:35.500
However many functions are nested inside of each other, that is how many terms you have in your final derivative expression.
00:14:35.500 --> 00:14:40.900
Let us go to the next page, differentiate f(x) = cos (x)² + c².
00:14:40.900 --> 00:14:47.200
f'(x) is equal to
00:14:47.200 --> 00:14:53.100
C is a constant, c² is a constant, this is not a variable.
00:14:53.100 --> 00:15:05.100
Do not let it throw you, x is the variable, f(x), x, everything else is treated as a constant.
00:15:05.100 --> 00:15:11.000
F’(x), let us see, we have cos.
00:15:11.000 --> 00:15:12.900
The derivative of cos is –sin.
00:15:12.900 --> 00:15:23.000
We have –sin, we leave what is inside alone, x² + c² × the derivative of what is inside which is just 2x.
00:15:23.000 --> 00:15:26.300
The derivative of c² is 0, it is constant.
00:15:26.300 --> 00:15:29.700
Our final answer is, I will go ahead and bring this forward.
00:15:29.700 --> 00:15:37.300
-2x × sin(x)² + c².
00:15:37.300 --> 00:15:44.200
There you go.
00:15:44.200 --> 00:15:51.500
F(x) = cos⁴ x + c².
00:15:51.500 --> 00:16:01.000
F’(x) here, this, I’m going to write, cos⁴ x.
00:16:01.000 --> 00:16:04.900
If you are comfortable with the notion of the 4 staying there, that is fine.
00:16:04.900 --> 00:16:13.600
If you want, you can go ahead and rewrite this as cos(x), the whole thing⁴ +,
00:16:13.600 --> 00:16:19.200
sorry, this is f(x), I’m rewriting it, + c².
00:16:19.200 --> 00:16:21.800
Now we can do f’(x).
00:16:21.800 --> 00:16:30.400
The derivative of this is 4 × cos(x)³.
00:16:30.400 --> 00:16:39.400
Now, × the derivative of what is inside, the derivative of the cos is –sin, + the derivative of this is just 0.
00:16:39.400 --> 00:16:48.300
Our final answer is -4 sin x cos³ x.
00:16:48.300 --> 00:17:06.000
I went ahead and put the 3 back there.
00:17:06.000 --> 00:17:09.700
X³ × e ⁻x³.
00:17:09.700 --> 00:17:15.000
This is going to be product rule and chain rule.
00:17:15.000 --> 00:17:19.600
Our chain rule here because this is e ⁺u, u is –x³.
00:17:19.600 --> 00:17:22.400
It is a function of x itself.
00:17:22.400 --> 00:17:29.100
F’(x), the derivative of this × this.
00:17:29.100 --> 00:17:39.200
3x² × e ⁻x³ + the derivative of this × this.
00:17:39.200 --> 00:17:52.300
The derivative of this is going to be e ⁻x³ × the derivative of this.
00:17:52.300 --> 00:17:59.000
The derivative of this thing is e ⁺u du dx.
00:17:59.000 --> 00:18:03.500
In other words, due to differentiation, this is inside.
00:18:03.500 --> 00:18:06.500
The outside function is this.
00:18:06.500 --> 00:18:07.600
This is the inside function.
00:18:07.600 --> 00:18:14.000
The derivative of this is -3x².
00:18:14.000 --> 00:18:22.400
This is the derivative of that part, now we have to multiply it by x³.
00:18:22.400 --> 00:18:27.200
The derivative of this × this + the derivative of this × this.
00:18:27.200 --> 00:18:29.100
The derivative of this is this whole thing.
00:18:29.100 --> 00:18:43.100
Our final answer is 3x² e ⁻x³ – 3x⁵ e ⁻x³.
00:18:43.100 --> 00:18:45.300
If you want to factor out the e ⁻x³, that is fine, you can.
00:18:45.300 --> 00:18:51.200
But you are absolutely welcome to leave it as is.
00:18:51.200 --> 00:18:57.900
Let us see, where are we?
00:18:57.900 --> 00:19:01.300
I think that is correct, good.
00:19:01.300 --> 00:19:06.900
Let us go to example 8, differentiate f(x) = sin of tan(x²).
00:19:06.900 --> 00:19:09.600
Now we have three functions.
00:19:09.600 --> 00:19:15.700
We have the sin, we have the tan, and we have x², which means we are going to have three terms.
00:19:15.700 --> 00:19:22.700
We are going to work from outside, in, in our chain.
00:19:22.700 --> 00:19:29.700
Do not worry, we are going to do plain more examples after this lesson is complete.
00:19:29.700 --> 00:19:33.400
The derivative of the outside function, the derivative of sin is cos.
00:19:33.400 --> 00:19:50.700
This is cos of tan(x)² × the derivative of tan(x)² is equal to sec² (x)².
00:19:50.700 --> 00:19:55.100
This is still a function of x.
00:19:55.100 --> 00:19:59.900
I take the derivative of that × 2x.
00:19:59.900 --> 00:20:03.800
There you go.
00:20:03.800 --> 00:20:19.700
My final answer is going to be 2x, cos of tan(x)² × sec² (x)².
00:20:19.700 --> 00:20:25.600
Either one of these, it is absolutely fine.
00:20:25.600 --> 00:20:29.800
Let us see what we have got.
00:20:29.800 --> 00:20:32.000
See how that happen, there are three terms.
00:20:32.000 --> 00:20:35.600
There are three functions, the sin, the tan, and the x².
00:20:35.600 --> 00:20:38.400
These are all functions of x.
00:20:38.400 --> 00:20:41.500
I have one term, two terms, three terms.
00:20:41.500 --> 00:20:45.700
Keep going until you ran out.
00:20:45.700 --> 00:20:49.300
In this particular case, my function of x is just x.
00:20:49.300 --> 00:20:53.200
The derivative of x is just 1.
00:20:53.200 --> 00:21:02.700
It is true, when we do keep going, that final derivative is just 1 so we do not put it there.
00:21:02.700 --> 00:21:06.700
Now we have sin of tan² (x).
00:21:06.700 --> 00:21:13.900
F’(x) = the derivative of sin is cos.
00:21:13.900 --> 00:21:23.700
This is the cos of tan² (x).
00:21:23.700 --> 00:21:31.200
Let me rewrite this again until we are comfortable with these ideas.
00:21:31.200 --> 00:21:50.700
Let me rewrite f(x) = sin of tan(x)², three functions.
00:21:50.700 --> 00:21:59.200
I have the sin function, I have the tan function, and I have the squared function.
00:21:59.200 --> 00:22:04.200
Sin function, tan function, squared function.
00:22:04.200 --> 00:22:08.100
Three functions, three terms.
00:22:08.100 --> 00:22:12.600
F’(x) = the derivative of sin is cos.
00:22:12.600 --> 00:22:18.100
I leave what is inside alone, the tan(x)².
00:22:18.100 --> 00:22:21.200
Now I multiply by the derivative of what is inside.
00:22:21.200 --> 00:22:39.600
The tan(x)² is equal to 2 × tan(x)¹ × the derivative of tan x which is sec² x.
00:22:39.600 --> 00:22:44.400
My final answer, if I were to put it together, I can leave it like this.
00:22:44.400 --> 00:22:46.600
I can bring the 2 to front, I guess.
00:22:46.600 --> 00:23:03.600
2 × cos of tan² x × tan x × sec² x.
00:23:03.600 --> 00:23:11.500
I hope that make sense.
00:23:11.500 --> 00:23:15.400
Just the last one, at least for this particular lesson.
00:23:15.400 --> 00:23:22.600
Here we have a product rule and we are going to have chain rule here and chain rule here.
00:23:22.600 --> 00:23:27.900
F’(x) is equal to the derivative of this × this.
00:23:27.900 --> 00:23:43.300
The derivative of this is 4 × x² – 2³ × the derivative of what is inside × 2x × this,
00:23:43.300 --> 00:24:01.900
which is 5x² – 2x – 2⁸, + the derivative of this which is 8 × 5x² – 2x – 2⁷.
00:24:01.900 --> 00:24:23.300
We are still taking the derivative of this × the derivative of what is inside which is tan x – 2 × this x² – 2⁴.
00:24:23.300 --> 00:24:28.000
We have two functions, the outer function and the inner function, there are two terms.
00:24:28.000 --> 00:24:35.900
The derivative of this × this + the derivative of this which is two terms × this.
00:24:35.900 --> 00:24:37.800
There you go, you can stop there.
00:24:37.800 --> 00:24:38.900
You do not really need to simplify this.
00:24:38.900 --> 00:24:42.000
It is going to be too complex to simplify.
00:24:42.000 --> 00:24:44.200
That is all, that is the chain rule.
00:24:44.200 --> 00:24:48.100
Do not worry about this, the chain rule is very important.
00:24:48.100 --> 00:24:53.400
In the next lesson, we are going to be continuing our example problems for the chain rule.
00:24:53.400 --> 00:24:55.700
Thank you so much for joining us here at www.educator.com.
00:24:55.700 --> 00:24:56.000
We will see you next time, bye.