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### More Chain Rule Example Problems

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
• Example I: Differentiate f(x) = sin(cos(tanx)) 0:38
• Example II: Find an Equation for the Line Tangent to the Given Curve at the Given Point 2:25
• Example III: F(x) = f(g(x)), Find F' (6) 4:22
• Example IV: Differentiate & Graph both the Function & the Derivative in the Same Window 5:35
• Example V: Differentiate f(x) = ( (x-8)/(x+3) )⁴ 10:18
• Example VI: Differentiate f(x) = sec²(12x) 12:28
• Example VII: Differentiate 14:41
• Example VIII: Differentiate 19:25
• Example IX: Find an Expression for the Rate of Change of the Volume of the Balloon with Respect to Time 21:13

### Transcription: More Chain Rule Example Problems

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

Today, I thought we would do some more chain rule example problems.0005

The chain rule was one of those things that you are pretty much going to be using all the time in your differentiation.0008

You are never going to be dealing with straight simple functions.0015

It can be a little bit deceiving, in the sense that you think you have a grasp on it.0019

And yet somehow things tend to slip away because if you forget to take that last derivative or whatever it is.0025

In any case, I just thought it would be nice to do some more example problems.0030

Let us jump right on in.0033

Nothing particularly difficult, just more practice, something nice to see.0036

Differentiate f(x) = sin of cos(tan(x)).0041

We have the sin, the cos, the tan.0049

We have 3 things, we are going to end up with 3 terms in our derivatives.0051

That is how it works.0054

The number of things that you actually have, the number of actual functions,0055

that is how many terms you end up with in your derivative.0058

A good way to think about it if you lose your way.0061

I will stick with black or blue, we will stick with blue.0066

F’(x) = we are going to take the sin, the derivative of the sin.0071

We are working outside in.0075

It is going to be the cos of what is inside.0078

What is inside is the cos(tan(x)).0082

Now we take the derivative of what is inside.0087

The derivative of the cos(tan(x)) is –sin tan(x).0091

Now we take the derivative of the tan(x) which is sec² x.0098

That is it, we are done.0104

F’(x), if you want to go and bring this negative sign out front, you are going to end up with f’(x) = -cos of the cos(tan(x)) × sin(tan(x)) × sec² x.0106

Is there a way to simplify that, maybe or maybe not?0128

That it is not really worth it.0131

You are just going to go ahead and use the function as is.0134

Nothing particularly strange, you just have to follow through and make sure you go down the line, go down the chain.0137

Example number 2, find an equation for the line tangent to the given curve at the given point.0145

Our function is sin(x)² π/2.62438.0153

Let us go ahead and find the derivative.0159

F’(x) is equal to, the derivative of sin is cos.0162

It is going to be cos(x)² × the derivative of what is inside, 2x.0169

Therefore, if I want I can that 2, it does not really matter.0178

Now we are going to go ahead and put π/2 into the x, to actually go ahead and find what this slope is.0184

The f’(π/2) = cos(π/2)² × 2 × π/2.0194

You end up with, these two cancel here.0210

You end up with cos.0214

Some number I went ahead and converted it to degrees, just for the heck of it.0216

141.7°, it is just going to be π²/ 4.0221

The cos(π²)/ 4 is just going to give you some number, and then × π.0225

When we solve that, we end up with -0.78 π.0230

This is our slope, we want the equation of the line, that is just y - y1 which is 0.6243 = m,0237

the slope -0.785 × x - x1, which is –π/2.0248

There you go, you can leave it like that, not a problem.0256

Example number 3, F(x) is equal to f(g(x)).0264

They give a certain values f(4), f(6), f’(4), g(6), g’(6), find f’(6).0269

F(x) is f(g(x)), therefore, f’(x) by the chain rule is f’ at g(x) × g’(x).0277

Therefore, f’ at 6 = f’ at g(6) × g’(6).0293

This equals, g(6) is equal to 4.0305

This is going to be f’(4) × g’(6).0312

This is f’(4) is equal to 2 and g’(6) is equal to 9.0318

That is it, just working it out, plugging it in.0330

Nice and straightforward.0333

Differentiate this function and then graph both the function of the derivative in the same window.0337

I just thought that I will throw that extra graphing thing in there.0343

Let us see what we have got.0348

We have f(x) = this.0350

It looks like we are going to be doing the quotient rule.0353

This × the derivative of that - that × the derivative of this divided by this².0358

F’(x) is going to equal this × the derivative of that.0367

That is going to be x × ½ × 1 – x³⁻¹/2 × the derivative of this thing, we still have to take the derivative of that.0372

That is going to be × -3x², this × the derivative of that.0387

Yes, very good, - that × the derivative of that.0404

The derivative of this is 1.0409

It just stays 1 – x³¹/2.0412

And then, × the derivative of that which is 1.0419

We will take all of that /x².0422

I’m going to go ahead and rewrite this.0429

I will do -3x².0433

I’m going to put this and I'm going to go ahead and bring this down to the bottom.0438

I’m going to put these, it is going to be -3x³ 3x² x divided by, I will leave the 2.0442

I’m going to bring this down below.0451

I’m going to actually rewrite it as a radical.0453

This is going to be √1 – x³.0456

This is going to be –√1 – x³.0460

All of that is going to be /x².0470

I get myself a common denominator here.0475

The common denominator is going to be 2 × 1- √x³.0482

We get -3x³ – 2 × 1 – x³.0488

Because when I multiply this × that, the radical sign goes away.0498

All over 2 × √1 – x³/ x².0505

Here we are going to have – 3x³.0512

A – and – x³, this is going to be -2x³ – 2/ 2 × √1-x³/ x².0518

I will just go ahead and bring this x² and put it in the denominator.0542

I’m left with -2x³ – 2.0544

If I did my arithmetic correctly, which is always the task up.0548

2x², 1 – x³.0552

There you go, this is our derivative.0559

If I did the algebra correct.0561

As far as what it looks like, it basically just looks like this.0564

This red one, this is the original f(x).0570

This little light blue one, this is f’(x).0574

F(x), this is f(x), f’(x) and f’(x), there you go.0579

This way, you see the slope is positive.0587

If it is positive, it is above the x axis.0591

At some point, it actually hits 0, becomes horizontal.0594

And then, the slope becomes negative, that is the derivative of the line.0596

Here the slope is negative, starts to become a little positive.0599

It zeros out again and it actually drops back down.0605

It starts to get a little positive but then it drops back down.0608

That is it, straight looking graph but these are the functions that we are going to be dealing with in the real world.0611

Differentiate f(x) = x – 8/ x + 3⁴.0621

Here, f’(x), chain rule.0628

We will do 4 × x – 8/ x + 3³ ×, the derivative of what is inside.0633

The derivative of what is inside is going to be quotient rule.0645

It is going to be x + 3 × the derivative of the top which is 1, -x – 8 × the derivative of the bottom which is 1, all over x + 3².0649

This is going to equal, let me bring it down here.0666

This is going to equal 4 ×, I‘m going to separate this one out.0672

It is going to be 4 × x³/ x + 3³ ×,0680

Here we have x + 3 – x – 8.0690

I will just write it out, it is not a problem.0699

It is going to be x + 3 – x + 8.0701

Sorry about that, it is going to be –x.0706

A – and -8 is going to be +8.0708

This is going to be x + 3, it is going to be².0712

This is going to equal, the x – x, they go away.0716

We have 3 and we have 8.0720

3 + 8 is 11, 11 × 4 is 44.0726

On the top we have 44 and we have x – 8³.0731

The bottom x + 3³, x + 3².0735

We get x + 3⁵, there we go.0738

Just have to follow the chain.0746

F(x) = sec² 12x.0751

This one is really nice and simple.0754

Let us see, sec² 12x.0761

We have to deal with this one first.0766

This is the same as, the sec(12x)².0768

We have to do chain rule first.0776

It is going to be 2 sec 12x.0779

Now we take the derivative of the sec which is going to be sec 12x tan 12x.0793

And then, × the derivative of the 12x, there is that.0805

Our final answer is going to be, I will take the 2 and the 12 together, it is going to be 24.0809

Sec 12x, I will put those together.0815

Sec² 12x tan 12x, that is our final answer.0819

We have three things going on.0831

We have the power function, we have the sec function, and we have the 12x function.0832

There are 3 three things so we are going to have 3 terms in our derivative.0838

Here is the first term, that takes care of the second term.0842

This takes care of the third term.0850

Just keep it that way.0852

If you want, you can draw the lines in between to separate your terms.0855

Identify how many actual functions you have, and then that the terms.0859

If you have counted 1, 2, 3, and then you only ended up with two, you know that you are going to need one more.0864

That one more is going to be the derivative of that.0869

As many functions as you have in your main function, that is how many terms0871

you are going to have in your derivative of the function.0876

Differentiate this, e ⁺x² + 3x × sin(x²) + 3x/ cos x.0884

This is going to be long and painful.0894

Maybe not too bad, let us see what is going on here.0899

Let us go back to blue.0902

We have f’(x), probably going to run out of room but it is not a problem.0904

It is going to be this × the derivative of that – that × the derivative of this/ that².0910

This × the derivative of this.0916

We have cos x × the derivative of this.0919

The derivative of this, the top is going to be product rule.0924

It is going to be this × the derivative of that + that × the derivative of this.0927

Here is what we end up getting.0934

It is going to be e ⁺x² + 3x × the derivative of that.0942

The derivative of that is going to be cos(x²) + 3x × the derivative of what is inside0949

which is going to be 2x + 3 + that × the derivative of this.0959

It is going to be + sin(x²) + 3 × the derivative of this.0966

The derivative of that is equal to e ⁺x² + 3x ×, I will just bring it down here, the derivative of that which is 2x + 3.0977

That is the first part, this × the derivative of that.0997

From that, this is still the numerator here, -this × the derivative of this.1001

It is going to be –e ⁺x² + 3x × sin(x²) + 3x × the derivative of this × -sin x.1012

It is going to be all of this /that².1028

What are we going to do with this?1039

Probably, you do not want to simplify it too much.1040

I’m going to go ahead and pull out a, 2x + 3, I have a 2x + 3, I’m going to pull that out.1043

I have an e ⁺x² + 3x, that is one term.1055

I have an e ⁺x² + 3x, that is another term.1058

I have an e ⁺x² + 3x, that is another term.1063

Let us see, let me just take this first part here.1067

This first part, let me deal with that.1076

I’m going to go ahead and factor out a 2x + 3.1079

I’m going to factor out an e ⁺x² + 3x.1085

I’m also going to factor out a cos(x).1092

That is going to be cos(x)² + 3x + sin(x)² + 3x.1096

I’m going to deal with this term right here.1116

I’m just going to actually leave it alone, - and – becomes +.1118

It becomes + e ⁺x² + 3x × sin(x)² + 3x × sin(x).1121

All of that is the numerator, and then, cos² x.1138

I think it is best to just leave it at that point, just one little step of simplification.1144

You could have left it at that, it is not a big deal.1150

Just make it a little bit cleaner, if nothing else.1152

There you go, very complicated as we would expect.1156

We have a very complicated looking function.1159

Exponential, trigonometric, there is going to be a lot going on.1162

Differentiate f(x) = 6 ⁺sin(π/2x).1167

Let us recall, anytime we have a constant raised to some power,1175

the derivative of that = a ⁺u and then the nat-log of the base which is a.1182

F’(x) that is going to equal 6 ⁺sin(π/2x) × ln (6) × du dx, chain rule.1189

Now we take the derivative of this.1209

The derivative of sin(π/2x) is cos(π/2x) × the derivative of π/2x which is π/2.1213

We can put some things together here.1226

Let me go back to blue.1230

Ln (6) is a constant, π/2 is a constant.1232

Let us write it as π ln 6/ 2 × this thing × this thing ×,1236

Let me go ahead and put a parenthesis.1253

6 ⁺sin(π/2x) × cos(π/2x).1255

There we go, that is our final answer.1266

This one right here, the rate of change of the radius of a spherical balloon with respect to time as we inflate it is 1.5 cm/s,1274

find an expression for the rate of change of the volume of the balloon with respect to time.1285

The rate of change of the radius of the spherical balloon with respect to time, that is dr dt, that is equal to 1.5 cm/s.1290

Basically, you have this balloon, you are inflating it, it is getting bigger and bigger.1306

This radius is changing, for every second it is growing by 1.5 cm.1313

That is all this says.1320

It is the cm/s, radius is in centimeters, time is in second.1321

This is a rate of change, that is how we express the rate of change.1326

Dy dx is the rate of change of y for every unit change in x.1329

For every unit interval of time, 1 second, the radius changes by 1.5 cm.1333

That is all this is, it is just the rate.1341

They want to know, find an expression for the rate of change of the volume with respect to time.1344

What they want is dv dt.1348

We have a relationship between v and r.1352

Let us go ahead and deal with that.1355

We know that the volume of a sphere is 4/3 π r³.1358

We have dr dt, we want dv dt.1369

Dv dt, we will just differentiate everything with respect to t.1370

Dv dt = 4/3 π ×, r is a function of t.1376

We treat it chain rule, it is going to be × 3 r² dr dt.1386

The 3 and the 3 cancel, therefore, I get dv dt = 4 π r² dr dt.1395

Dr dt is 1.5, I plug that in here.1410

I get 4 π r² × 1.5.1415

The final answer is, the rate of change of volume with respect to time is equal to 6 π r².1422

Whatever the radius happens to be at moment, I put that into this equation.1435

That will tell me, at that moment, how fast the volume is changing.1439

Notice, here, the rate of change of the radius is constant.1443

That means the radius is growing by 1.5 cm every second, it is not changing.1448

But every second it is 1.5 cm, another second 1.5 cm, another second 1.5 cm.1453

The volume is not linear, it is not constant.1458

It actually depends on what r is.1462

As r increases, the rate at which the volume is changing actually keeps getting more and more.1465

Because now, the rate of change of volume depends on r.1474

As r changes, the square of r, the number is going to change.1480

If r is 4, this is going to be 4 × 4 is 16, 6 × 16 π.1483

When the radius is 4, the volume is changing at 6 × 16 π.1488

If the radius is 10, 10² is 100, 6 × 100.1494

When radius is 100, the volume is changing per second at 600 π.1500

The rate of change of the radius is constant but the rate of change of volume is not constant.1508

It actually depends on r.1515

This is the general procedure.1518

You will be seeing a lot of this when we do related rates.1519

I think that is actually the last example for today.1526

Thank you so much for joining us here at www.educator.com.1528

We will see you next time, bye.1531