INSTRUCTORS Raffi Hovasapian John Zhu

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• ## Transcription

### The Chain Rule

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

• Intro 0:00
• The Chain Rule 0:13
• Recall the Composite Functions
• Derivatives of Composite Functions
• Example I: Identify f(x) and g(x) and Differentiate 6:41
• Example II: Identify f(x) and g(x) and Differentiate 9:47
• Example III: Differentiate 11:03
• Example IV: Differentiate f(x) = -5 / (x² + 3)³ 12:15
• Example V: Differentiate f(x) = cos(x² + c²) 14:35
• Example VI: Differentiate f(x) = cos⁴x +c² 15:41
• Example VII: Differentiate 17:03
• Example VIII: Differentiate f(x) = sin(tan x²) 19:01
• Example IX: Differentiate f(x) = sin(tan² x) 21:02

### Transcription: The Chain Rule

Hello, welcome back to www.educator.com and welcome back to AP Calculus.0000

Today, we are going to discuss the chain rule.0005

Profoundly important, let us dive right on in.0008

Let us start off by recalling this idea of a composite function.0014

Recall composite functions.0018

Something like, if we had f(x) is equal to x³ and if we had g(x) is equal to, let us say, x² + 1,0030

the composite function f(g) is defined as f(g) of x.0040

We are basically, wherever we see x in here, we are just putting in the entire function.0047

It ends up being x² + 1³, that is it.0054

What we want to do, the chain rule, it allows you to find the derivative of a composite function.0060

That is it, that is all it is.0068

The majority of the functions that you are dealing with are actually going to be composite functions.0070

They are not just going to be simple functions like sin(x) or 5x or x².0073

They are going to be composite functions like x² + 1³ or sin(x)² + tan² x¹/5.0078

Most of them are going to be composites.0087

The chain rule is the one thing that you are going to be using over and over again.0088

We are going to be spending a fair amount of time with this.0093

That is all it is.0096

Let me write this down.0101

The chain rule allows us to find the derivatives of composite functions.0106

I apologize for my slightly sloppy writing.0117

Derivatives of composite functions, that is it, that is all it is.0120

Let us go ahead and go to blue.0132

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.0171

F’ is equal to f’ evaluated at g(x) × g’(x), that is the formula.0183

This is the formula for the chain rule.0195

It says that if I have a composite function, I take the derivative of the outer function.0197

I evaluate at g(x), or I put g(x) into the derivative of the function.0202

And then, I multiply it by the derivative of what is inside the derivative of g’.0210

This is why they call it the chain rule.0215

It makes more sense when you actually see the example.0217

You just keep differentiating from the outside in until you end up running out of things to differentiate.0220

That is why they call it the chain rule.0227

You do the outside function then the inside function, then the next function.0229

If you had f, g, h, you would do f’, you would do g’, then you do h’.0233

Again, this is the definition, all of this will make sense when we do the examples.0239

In the notation of this dy dx, the notation that we sometimes use, this is expressed this way.0246

Let y = some function of the variable u and u = some function of the variable x.0270

Then, dy dx is equal to, y is equal to function of u.0287

U was equal to a function of x, that means y is actually equal to a function of x.0297

When I put this u into here, I get g(x).0301

I want to find dy dx.0306

Dy dx is equal to dy du × du dx.0307

The nice thing about this particular notation is that you get the sort of see that this du and this du,0316

because we are treating them like quotients, you get to see that they cancel, leaving you with dy dx.0324

Again, it is just another way of looking at it.0332

I personally prefer this but that just a personal choice.0335

Note carefully.0341

When we say that f’(x) is equal to f’ evaluated at g(x) × g’(x), this is f’ at g(x).0349

When we have the specific x that we are talking about, we actually have to put in g(x).0368

We form f’, we find g(x), we put that in here.0373

We evaluate f’ at whatever number we get here.0378

And then, we multiply by g’ of whatever that number is.0381

This is very important.0385

I think once you see one or two examples, everything is going to make sense.0394

Let us do that.0397

Let f(x) = x³ + 5⁵, identify f(x) and g(x), then go ahead and differentiate.0403

In this particular case, f is going to be x⁵, the outer function.0413

G(x), the inner function, x³ + 5.0425

I should have put F, sorry.0433

That is okay, you will understand what is going on here.0438

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.0440

I know that there is no confusion, but just for the sake of keeping it straight.0452

F(x) = x⁵, g(x) = x³ + 5.0457

F = f(g) which is equal to f(g), that is that.0463

Now we go ahead and differentiate.0473

We have f’(x), we differentiate from the outside, in.0476

We said that it is equal to, let us do the formula.0483

F’ at g(x) × g’(x).0493

For this particular function, f’, this is x⁵.0500

This thing⁵, we bring it down.0506

We treat it like a regular power rule.0510

5x³ + 5⁴, this is f’ at g(x).0512

F’ is 5 × whatever it is, the 4th, the g(x) is this one right here, × g’(x) what is inside.0521

Now, I take the derivative of what is inside, 3x².0531

I simplify, 3 × 5 is 15, x² × x³ + 5⁴.0540

That is what I’m doing, I’m just differentiating from the outside, in.0550

I have x³ + 5⁵.0558

I take care of the power rule first, x³ + 5⁴.0563

And then, I take the derivative of what is inside, 3x².0568

I keep going if I need to.0573

At this point, that is it, this is the last function that is quote inside.0574

I’m fine, I’m left with 15x² × x³ + 5⁴.0579

The sin of tan(x), identify f(x) and g(x), and then differentiate.0590

In this particular case, f is going to equal the sin(x) and g is equal to the tan(x).0594

The outside function is sin, the inside function is tan.0604

F’(x), this f’, the derivative of this, I take the derivative of the sin function which is cos of tan(x).0610

That stays.0624

I multiply by the derivative of what is inside, the derivative of tan is sec² x.0625

That is my final answer, if I want, I can go ahead and bring that forward.0633

Sec² x × the cos of tan(x), either one of these is fine.0639

You are working from the outside, in.0648

You take the derivative of sin, leaving what is inside alone.0650

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.0654

Let us see what we have got here.0665

Differentiate f(x) = ∛2x³ – 4x + 7.0672

I’m going to rewrite this with a rational exponent.0678

I’m going to write 2x³ – 4x + 7¹/3.0682

I differentiate my outside function is something to the 1/3.0693

I treat it as a regular power rule, then, I differentiate what is inside.0696

F’(x) = 1/3, I will leave what is inside alone.0700

2x³ – 4x + 7, until I finish with that.0708

This is that -1, × the derivative of what is inside, × 6x² – 4.0713

I’m done, I can stop there.0724

Just start from the outside and work your way in.0729

Differentiate f(x) = -5/ x² + 3³.0737

Now we have a quotient rule and the denominator happens to be something that is a composite function.0743

The derivative of that, we are going to use the quotient rule over the entire thing.0750

The chain rule for this one.0754

F’(x), this × the derivative of that - that × the derivative of this/ this².0758

This × the derivative of that.0765

We have x² + 3³ × the derivative of -5 is 0 - -5 × the derivative of this.0767

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.0779

All of that /x² + 3³².0794

It is just 3 × 2.0800

This is going to be x² + 3⁶.0801

This is 0, I get 2 × 3 is 6, 6 × 5 is 30, - × - is +.0805

I get 30 × x² + 3²/ x² + 3⁶.0815

That goes with that, leaving 4.0825

I get 30/ x² + 3⁴.0828

That is it.0836

When I take the derivative of this, it is the outside 3, leave what is inside alone, finish that part.0838

And then, multiply by the derivative of what is inside.0846

That is the chain rule, it is a chain of functions.0849

If you have f and g, you are going to have two, one term and two terms.0853

If you have f, g, and h, you are going to have one term, two terms, three terms.0859

However many functions are nested inside of each other, that is how many terms you have in your final derivative expression.0864

Let us go to the next page, differentiate f(x) = cos (x)² + c².0875

f'(x) is equal to0881

C is a constant, c² is a constant, this is not a variable.0887

Do not let it throw you, x is the variable, f(x), x, everything else is treated as a constant.0893

F’(x), let us see, we have cos.0905

The derivative of cos is –sin.0911

We have –sin, we leave what is inside alone, x² + c² × the derivative of what is inside which is just 2x.0913

The derivative of c² is 0, it is constant.0923

Our final answer is, I will go ahead and bring this forward.0926

-2x × sin(x)² + c².0930

There you go.0937

F(x) = cos⁴ x + c².0944

F’(x) here, this, I’m going to write, cos⁴ x.0951

If you are comfortable with the notion of the 4 staying there, that is fine.0961

If you want, you can go ahead and rewrite this as cos(x), the whole thing⁴ +,0965

sorry, this is f(x), I’m rewriting it, + c².0973

Now we can do f’(x).0979

The derivative of this is 4 × cos(x)³.0982

Now, × the derivative of what is inside, the derivative of the cos is –sin, + the derivative of this is just 0.0990

Our final answer is -4 sin x cos³ x.0999

I went ahead and put the 3 back there.1008

X³ × e ⁻x³.1026

This is going to be product rule and chain rule.1030

Our chain rule here because this is e ⁺u, u is –x³.1035

It is a function of x itself.1039

F’(x), the derivative of this × this.1042

3x² × e ⁻x³ + the derivative of this × this.1049

The derivative of this is going to be e ⁻x³ × the derivative of this.1059

The derivative of this thing is e ⁺u du dx.1072

In other words, due to differentiation, this is inside.1079

The outside function is this.1083

This is the inside function.1086

The derivative of this is -3x².1087

This is the derivative of that part, now we have to multiply it by x³.1094

The derivative of this × this + the derivative of this × this.1102

The derivative of this is this whole thing.1107

Our final answer is 3x² e ⁻x³ – 3x⁵ e ⁻x³.1109

If you want to factor out the e ⁻x³, that is fine, you can.1123

But you are absolutely welcome to leave it as is.1125

Let us see, where are we?1131

I think that is correct, good.1138

Let us go to example 8, differentiate f(x) = sin of tan(x²).1141

Now we have three functions.1147

We have the sin, we have the tan, and we have x², which means we are going to have three terms.1149

We are going to work from outside, in, in our chain.1156

Do not worry, we are going to do plain more examples after this lesson is complete.1163

The derivative of the outside function, the derivative of sin is cos.1170

This is cos of tan(x)² × the derivative of tan(x)² is equal to sec² (x)².1173

This is still a function of x.1191

I take the derivative of that × 2x.1195

There you go.1200

My final answer is going to be 2x, cos of tan(x)² × sec² (x)².1204

Either one of these, it is absolutely fine.1220

Let us see what we have got.1225

See how that happen, there are three terms.1230

There are three functions, the sin, the tan, and the x².1232

These are all functions of x.1235

I have one term, two terms, three terms.1238

Keep going until you ran out.1241

In this particular case, my function of x is just x.1246

The derivative of x is just 1.1249

It is true, when we do keep going, that final derivative is just 1 so we do not put it there.1253

Now we have sin of tan² (x).1263

F’(x) = the derivative of sin is cos.1267

This is the cos of tan² (x).1274

Let me rewrite this again until we are comfortable with these ideas.1284

Let me rewrite f(x) = sin of tan(x)², three functions.1291

I have the sin function, I have the tan function, and I have the squared function.1311

Sin function, tan function, squared function.1319

Three functions, three terms.1324

F’(x) = the derivative of sin is cos.1328

I leave what is inside alone, the tan(x)².1332

Now I multiply by the derivative of what is inside.1338

The tan(x)² is equal to 2 × tan(x)¹ × the derivative of tan x which is sec² x.1341

My final answer, if I were to put it together, I can leave it like this.1359

I can bring the 2 to front, I guess.1364

2 × cos of tan² x × tan x × sec² x.1366

I hope that make sense.1383

Just the last one, at least for this particular lesson.1391

Here we have a product rule and we are going to have chain rule here and chain rule here.1395

F’(x) is equal to the derivative of this × this.1402

The derivative of this is 4 × x² – 2³ × the derivative of what is inside × 2x × this,1408

which is 5x² – 2x – 2⁸, + the derivative of this which is 8 × 5x² – 2x – 2⁷.1423

We are still taking the derivative of this × the derivative of what is inside which is tan x – 2 × this x² – 2⁴.1442

We have two functions, the outer function and the inner function, there are two terms.1463

The derivative of this × this + the derivative of this which is two terms × this.1468

There you go, you can stop there.1476

You do not really need to simplify this.1478

It is going to be too complex to simplify.1479

That is all, that is the chain rule.1482