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INSTRUCTORS Raffi Hovasapian John Zhu
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Lecture Comments (5)

1 answer

Last reply by: Professor Hovasapian
Mon Apr 9, 2018 8:50 AM

Post by Patricia Xiang on April 7 at 02:05:06 AM

Hello, Prof.Hovasapian, in this video  only the introduction part has voice while the other parts are completely silent. I tried to refresh the page but it doesn’t help. All other videos have voices except this one.

2 answers

Last reply by: Venise Martinez
Wed Mar 30, 2016 8:27 PM

Post by Jerica Cui on January 29, 2016

Hello, professor!

i don't understand why the derivative of (x^2-14x) is 2x.  i thought it's (2x-14)? (in the last example)

The Product, Power & Quotient Rules

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 Product, Power and Quotient Rules 0:19
    • Differentiate Functions
    • Product Rule
    • Quotient Rule
    • Power Rule
  • Example I: Product Rule 13:48
  • Example II: Quotient Rule 16:13
  • Example III: Power Rule 18:28
  • Example IV: Find dy/dx 19:57
  • Example V: Find dy/dx 24:53
  • Example VI: Find dy/dx 28:38
  • Example VII: Find an Equation for the Tangent to the Curve 34:54
  • Example VIII: Find d²y/dx² 38:08

Transcription: The Product, Power & Quotient Rules

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

Today, we are going to continue our discussion of ways of taking the derivative of different types of functions very quickly.0005

We are going to discuss the product rule, the power rule, and the quotient rule for functions.0013

Let us jump right on in.0019

Here is what we have got, what we can do.0022

You know what, I think today I’m going to work in black.0024

What do we do when we have to differentiate functions that look like this, differentiate functions like the following.0030

Let us say f(x) = 2x² – 1³ × 3x – 4⁻³.0058

How do we differentiate something like that?0075

Or what if we had f(x) = x² × the sum of x.0077

Or what if we had f(x) = 2x² + 3x + 2/ x² - 6.0088

How do we differentiate that?0099

Or f(x) = 3x² + 1⁵.0105

How do we deal with these?0114

In the first and second cases, in other words, this one and this one, what we have is our function F(x) is actually a product of 2 separate functions of x.0118

First case, f(x) is a product of 2 independent functions of x.0142

In other words, let me go to red here.0162

We have this is one function, the 2x² – 1³.0164

This 3x – 4⁻³ is a separate function.0169

In this case, we have x² which is an entirely separate function of x and sin x which is an entirely separate function of x.0172

When they are multiplied together, how do we take the derivative of something like that?0180

In the third case, we have an independent function of x divided by an independent function of x.0185

This one over here, we have some independent function of x and it is actually raised to a power.0192

We actually have the same case here.0198

This whole thing is an independent function of x.0200

It is actually made up of a function of x which is raised to a power times a function of x raised to a power.0204

How do we handle things like this?0211

You might think to yourself, if f is the product of two functions, if I take the derivative of f, it is not just the product of the derivatives?0217

In other words, why cannot I just take the derivative of x² multiplied by the derivative of sin x.0226

It turns out that that is actually not how it happens.0233

Let me write this down.0238

You might think that, if f(x) = f(x) × g(x), then f'(x) = f'(x) × g’(x).0243

This is not true, absolutely not true.0271

It is true that the derivative of the sum is the sum of the derivatives but the derivative of a product of two functions is not the product of the derivatives.0280

It also is not true that the derivative of a quotient of two functions is equal to the quotient of the derivatives.0291

The derivative of f/g does not equal f’/g’.0309

Here are the rules for handling product of functions, powers of functions, and quotients of functions.0319

Let us go ahead and go through these.0326

I think I’m going to go back to black here.0330

Our product rule is, if f(x) and g(x) are differentiable, our presumption on the entire calculus course is that they are going to be differentiable,0334

then the derivative of fg is equal to the derivative of f × g + f × the derivative of g.0358

In other words, the derivative of this × this + the derivative of this × this.0375

You can keep them in order, essentially what you are doing is you are just taking the derivative of one function at a time, as you go down.0382

This is a shorthand notation.0389

The derivative of the product is f’ × the other function + f × g’, the derivative of the other function.0391

This can be extended to a product of 3 or more functions.0404

That is going to look like this.0439

Let us go ahead and write.0444

Let f1 f2 all the way to fn, whatever n happens to be -- 5, 10, 15, 20, however many functions.0447

F sub n, let them be n differential functions.0459

Then, the derivative of f1 × f2 × all the way to fn = f1’ × f2 × all the functions + f1 × f2’0472

× f3 × all the functions + f1 f2 all the way up to fn-1 fn’.0501

Basically, all I'm doing is I'm leaving the functions as is and I’m just taking the derivative of the first function.0513

And then, I take the derivative of the second function, multiplied all out.0519

The derivative of the 3rd function, multiply all out, until I have taken the derivative of every single individual function.0523

Of course, I'm adding them.0530

Most of the time, we are just going to be dealing with 2 functions.0534

This one right here is sufficient.0536

That is the product rule, that is how you handle.0540

Take the derivative of the product of 2 functions.0542

What you have to do with is recognize which 2 functions you are dealing with.0545

What is f and what is g.0549

All of this will make perfect sense, once we actually see the examples.0552

Let us talk about the quotient rule.0557

As we said, the derivative of the quotient of 2 functions is not the quotient of the derivatives.0561

The quotient rule is the following.0567

Ddx (f/g) is equal to the denominator × the derivative of the numerator - the numerator × the derivative of the denominator/ g².0573

F and g are functions, these are actual functions.0592

Let us go ahead and finish off with the power rule.0600

I’m going to use u instead of f or g.0607

Let u be differentiable.0611

In other words, u(x) is an actual full function of x.0622

Let u(x) be differential.0630

The derivative of u ⁺x raised to some power, any real number is equal to,0635

It is handled the same way as a function of x.0645

It is n × the function, the exponent n -1.0648

I’m going to multiply by du dx.0656

Let us talk about what this actually looks like.0665

Before, we dealt with something like this, we said x⁵.0667

When I took the derivative, we ended up with 5x⁴.0673

We took this, brought it down, subtracted by 1.0679

X is the function, it is raised, it is variable.0682

It is raised to the 5th power.0687

What we are saying is, if you have something like this.0690

If you have 2x² + 1⁵.0692

This is your u(x).0703

It is an entire function that contains x, itself is raised to a power.0707

When you take the derivative of this, you treat it the same way.0715

It is still going to be 5 × 2x² + 1⁴, except you have to multiply by the function itself.0718

By the derivative of the function itself, du dx.0728

In this particular case, the du dx, the derivative of what is inside is, that is fine, du dx.0735

This ends up equaling 5 × 2x² + 1⁴.0747

The derivative of that is 4x.0754

It is just that extra step.0757

When the thing that you actually raising to a power is a full function of x,0759

you have to take the derivative of that function.0766

It is the same thing here.0769

What we are doing here, it is actually a special case of this x⁵.0771

When you take the derivative of this, what you are doing is you are doing 5 x⁴.0778

This is your function of x, you are taking dx of dx.0783

Dx/dx is just 1.0789

It stays 5x⁴.0793

You are still doing the same thing except your function of x just happens to be x alone.0796

It is the variable itself.0801

There is nothing else going on there.0803

In any case, I would not be able to point here.0805

It will make a lot more sense when we actually do the example problems.0807

All you have to remember is once you identify some functions that is being raised to a power,0811

you do everything the same way to do the power part.0817

And then, you just remember to multiply by the derivative of the original function itself.0820

Let us go ahead and get started with our examples.0827

Y = x² + 4 × f⁴ + 8, find dy dx.0831

There is a couple ways that we can do this that is nice.0838

Now, you have multiple paths that you can follow.0840

Do I do the product rule, do I multiply it out and just do the normal,0845

the derivative of each individual term because it is a polynomial.0851

If you want, you can just multiply this.0854

This × that, this × that, this × that, and then take the derivative.0857

Or you can treat it like a product rule.0861

Let us go ahead and do it as a product rule.0865

This is our f and this is our g.0867

The product rule says that y’ is equal to f’ × g + f × g’.0872

Let us go ahead and do that.0887

Y’ = f’ × g.0890

F’, the derivative of this is 2x × g.0893

G is x⁴ + 8, we leave g alone.0899

We leave f alone.0904

We write x² + 4 and we do g’.0906

This is 4x³.0910

That is it, you can just leave it like this, if you want to.0915

It is perfectly valid.0918

Or again, depending on what it is that your teacher or professor wants.0919

You can multiply it out.0923

If you want to multiply it out, this is just going to be y’ = 2x⁵ + 16x + 4x⁵ + 16x³.0924

2x⁵ + 4x⁵, y’ = 6x⁵ + 16x³.0946

I will do it in descending powers of x, + 16x.0955

There you go, that is it, nice and simple, product rule.0960

F’ × g + f × g’.0964

All you have to do is recognize what the f and g are.0967

That is going to be your ultimate task.0971

Y = x +1/ x-1.0975

This one, we have f here, we have g here.0979

I will go ahead and do this in red.0985

This is going to be quotient rule.0988

The quotient rule says that y’ is equal to, this is f, this is g.0995

G f’-f g’/g², denominator × the derivative of the numerator - the numerator × the derivative of the denominator/ the denominator².1006

Let us do it.1021

Y = g which is x-1 × the derivative of the numerator which is 1 – f,1024

which is x + 1 × the derivative of the denominator which is 1, the derivative of x – 1 is 1, all over the denominator², x-1².1034

If I distribute the negative sign, i on top, I get x -1 - x-1/ x-1².1050

X cancels, I’m left with -2/x-1².1062

That is my y’, that is my dy dx.1069

At a given point, x, if I choose a point and I put it in here, this is going to be the slope of my tangent line to that curve.1076

This represents the rate of change.1087

For a unit change, for a small change, I’m making xy, at that point changes by that much.1091

Rate of change, slope.1103

Let us try this one.1109

We have this thing raised to the 5th power.1116

In this particular case, this is going to be our u(x) and it is a whole function raised to the 5th power.1120

We are going to use our power rule.1129

Our power rule says, f⁵ = 5 × f⁴ × f’.1136

Let us go ahead and do it.1154

Y’ =, that comes down, 5 ×, I will leave that alone, 4x³ – 3x² +6x + 9⁴ × the derivative × f’.1156

Now I take the derivative of what is inside, × 12x² - 6x + 6.1175

I’m done, I’m not going to simplify this anymore.1186

I’m not going to multiply it out, that is it.1188

I’m done, I just have to run through the rule.1192

Nice and straightforward.1196

Example number 4, y = 5/3x – 5⁴.1199

It is a quotient, in the sense that you have 5 over that.1206

You can certainly use the quotient rule.1210

Another way to handle this is actually doing just the power rule, believe it or not.1212

Because I can actually write this as 5 × 3x – 5⁻⁴.1218

You have multiple paths that you can take.1229

You can do quotient rule or you can do power rule.1230

There is always going to be the case like this, with problems like this.1232

You are going to have a choice and the choice does not matter.1235

Sometimes it is the same amount of effort.1237

No matter if you go this path or this path.1239

Sometimes, one of them is going to be a lot quicker than the other.1241

Sometimes one of them is going to be messy, the other one is going to be not so messy.1244

It just depends and this is a question of experience as to which one is going to be better.1251

Oftentimes, you do not know which one is going to be better.1256

You just have to jump in, go down the path.1259

If it looks to be too complicated, you change path and you go down another path, very simple.1261

Let us go ahead and do y’ here.1267

Y’ = 5, this is a constant.1270

This is going to be, I’m going to bring this down.1274

This is going to be -4 × 3x-5.1277

I’m going to take the derivative of this to be -4, -4 -1⁻⁵ × the derivative of this which is 3.1282

Let me do it over here because I need some room. 5 × -4 is -16.1294

-16 × 3, 5 × -4 is -20.1304

-20 × 3 is -60.1310

Let us not jump to going here.1313

-60 × 3x -5⁵, this is our answer.1314

Let us also do this via the quotient rule to see what it looks like.1326

Let us also do this via the quotient rule.1333

Remember, we said what the quotient rule is.1349

If this is f, this is g, it is g f’- f g’/ g².1352

Y’ equals this, 3x -5⁴ × the derivative of that f’ which is 0, -5 × g’ the denominator,1362

× the derivative of this which is the derivative of the denominator is 4 × 3x – 5³.1380

4-1 is 3 × the derivative of what is inside which is 3/ this², 3x – 5⁸.1392

This is equal to, this is just 0.1407

We have -60 to be 3x -5³/3x – 5⁸.1411

That leaves 5, you are left with y’ = -60/3x-5⁵.1424

There you go, this is the same as that.1438

It is just expressed with a negative exponent.1440

The only thing that you have to watch out for is watch this exponent very carefully.1443

If it is negative that -1 becomes more negative.1448

Notice, we went from -4 to -5.1452

Here, we left it as is, the denominator.1456

When we took this g‘, f4 came here.1461

4 -1 is 3, you just have to watch the signs.1466

If it is positive, it goes down.1471

If it is negative, it still goes down, becomes further negative.1473

That is the only thing you have to watch out for.1476

Again, you are going to make a mistake.1478

I made a mistake, I still make a mistake after all these years.1480

It is just the nature of the game.1484

We just have to be as vigilant as possible.1486

Let us go ahead and go on here.1491

Let us see, now we have y = this thing.1495

It looks like we do have just to do the quotient rule.1503

We will just do quotient rule.1506

Once again, recall quotient rule.1508

Let me go back to red here.1514

Quotient rule, y’, if this is f and this is g, we have g f’- f g’/ g².1516

Y’ or dy dx is equal to this × the derivative of that.1532

We have x³ – 3x – 2 × the derivative of this which is 2x – this x² -5 × the derivative of this1539

which is 3x²/ x³ -3x – 2².1557

We have to multiply all of this out on top.1570

Y’ = 2x and x³ that should be 2x⁴ – 6x² – 4x, this minus sign, x².1574

3x² is 3x⁴, the minus sign becomes -3x⁴ x² -3.1593

This is -3x², - and – is +, 3x².1603

This × this is -15x², it is going to be a +15x².1609

This is -5 × -3 is +15.1617

It is -15/ x³ – 3x – 2².1622

When we put it together, 2x⁴- 3x⁴, that takes care of these.1635

We get a –x⁴ -6x² + 3x² is -3x² + 15x⁵.1640

We get a +12x² and then we have -4x.1653

And then, we have our -15/ x³ - 3x -2.1660

That is y’ or dy dx.1669

There you go.1676

Whenever you are dealing with quotient rule, as we get them to more complex functions, trigonometric functions, things like that, exponentials,1677

you are going to notice that the quotient rule, the biggest problem is going to be where do I stop simplifying.1686

Again, it is up to your teacher what they want.1693

Sometimes, they will just ask you to stop right there and just leave it like that.1696

Other times, they will ask you to simplify as much as possible, personal choice.1700

The mathematics itself, the important part is being able to recognize the f and g, and to keep the order straight, that is what matters.1705

Suppose that u and x and v(x) are functions that are differentiable at x = 0.1720

I think I will actually work in blue here.1728

U(0) = 7, U’ (0) = -2, v(0) =-3, v’(0) = 2.1732

Find the values of the following at x = 0.1740

Here we are going to form the derivatives and then just use these values, plug them in.1744

Let us start with ddx(uv).1753

Ddx(uv) that is equal to u’ v + u v’.1758

U’ is -2, v is -3, this is all at 0.1771

+u at 0 which is 7 × v’ which is 2.1784

This is 6 + 14 = 20.1793

I do not even need to know what the functions are.1803

I know what the values of the functions are and their derivative.1805

The rest is very simple, good.1808

Let us do the ddx(u/v).1813

The ddx(u/v) is v u‘ – u v’/ v².1823

V is -3, u’ is -2, - u which is 7, × v’ which is 2/ v².1840

V is -3².1857

You get 6 -14/9.1862

6 – 14, I think is -8.1868

You get -8/9, good.1871

We have ddx of, this time we have v/u.1882

This is the denominator × the derivative of the numerator – the numerator × the derivative of the denominator / the denominator².1889

Plug the values in, we are going to get 7 × 2 -3 and -2/7².1902

That is going to equal 14 – 6/ 49, that equals 8/49.1914

We have ddx (5v – 4u).1928

That is 5v’, the derivative of this is just the constant × the derivative of this, that is just v’ – 4 × u’ = 5 × 2 – 4 × -2 = 10 + 8 = 18.1937

The final one, ddx (u⁴), u is a function of x.1969

It is equal to 4 × u³ × du dx which is u’.1983

We have 4 × 7³ × -2.1994

We end up with -2744.2003

We actually have our final one, ddx(7v⁻³).2013

V is a function of x, this becomes 7 × -3 × v -4.2022

-3 – 1 is -4, × dv dx which is v’.2033

V is a function of x, we have to multiply by v’.2042

That is the rule for powers.2047

This becomes -21/v⁴, I just went ahead and put the -4 down there, × v’.2049

That equals -21/-3⁴ × 2.2064

You end up getting a final answer of -42/81, if I have done my arithmetic correctly.2073

Hope that made sense.2085

Let us see, example number 7.2091

Find the equation for the tangent to the curve x²/ x³ + 3 at x = 2.2097

Very simple, we find the derivative, we put the value 2 into that derivative, that gives us a slope.2103

We find the point, the y value, and then we just form the equation.2112

Let us go ahead and start with y’.2118

Let us see, y’ = this is f and this is g.2122

This × the derivative of that – that × the derivative of this/ this².2130

We have x³ + 3 × the derivative of this which is 2x – x² × the derivative of this which is 3x²/ x³ + 3².2135

Y’ is equal to 2x⁴ + 6x² – 3x⁴/x³ + 3², y’ at 2.2158

When I put 2 into this, I end up with -16 + 24 / 121.2188

I end up with 8/121.2201

This is going to be our slope.2206

It is going to be a slope of our line.2207

The derivative at the given point is the slope of the tangent line through that point, through the x value of that point.2209

The slope is equal to 8/ 121.2221

Let us find f(2), we need to find the y value.2224

The f(2) that is just going to be 2²/ 2³ + 3 which is going to be 4/11.2228

Our point is 2 and 4/11.2243

Therefore, our equation is y – 4/11 = 8/121 × x – 2.2253

There you go, we want the equation of the tangent to the curve at x = 2.2265

When x = 2, the y value is 4/11.2273

Another line passes through that point.2275

The slope of the line is the derivative at that point.2279

Y -y1 = m × x - x1.2282

Let us see here.2291

X²/x – 7, find d² y dx².2294

We need to differentiate this twice.2299

Let us start with y’.2302

This is f and this is g, this × the derivative of that, which is going to be x -7 × 2x – x² × the derivative of that which is 1/ x – 7².2306

This is going to equal 2x² - 14x – x²/ x – 7².2327

Therefore, we have a final answer of y’ is equal to 2x² - x², that is just equal to x² – 14x/ x – 7².2341

Now we need to find y” which means we differentiate this.2365

Let me rewrite this as, y’ is equal to x² - 14x.2370

I can do this via the quotient rule again or I can do the product rule.2382

I’m just going to go ahead and rewrite this as, x-7⁻².2386

I just brought it up, I’m going to do it as a product rule instead, y” which is the derivative of this.2392

Let us go to black here.2402

This is our f and this is our g.2404

Our g is actually this whole thing.2410

But notice that g itself is made up of a function raised to a power.2415

You just have to be really careful here.2421

This is the product rule.2423

This is f × g.2427

We are going to have f’g + f g’.2429

The derivative of this × that + the derivative of this × that.2437

F‘ is equal to 2x × g which is x – 7⁻² + f.2445

F is equal to x² -14x × the derivative of g’.2460

The derivative of this whole thing is -2 × x-7⁻³ × the derivative of what is in here which is just 1.2469

Now I simplify, I’m going to write this as 2x/ x – 7² +, this is -, I’m sorry.2485

This is going to end up being a -.2499

-2 × x² – 14x.2506

I’m going to bring this thing down to the denominator.2510

I’m going to write it as x – 7³, that makes sense, -2.2513

This thing stay on top, I just brought this down.2519

I’m going to find a common denominator here by multiplying the top and bottom by x-72522

because I want the denominator to be x -7³.2527

This equals 2x × x – 7 – 2 × x² – 14x/ the common denominator which is x – 7³.2533

2x² – 14x -2x² + 28x, x-7³.2558

2x² – 2x², they go away.2579

-14 + 28, I’m left with 14x on top.2585

I’m left with x – 7³ in the bottom.2590

This is my y”.2597

Here is my y’, my first derivative.2600

Here is my second derivative.2603

I’m going to do this again but this time I’m going to use the quotient rule instead of the power rule,2608

just for the sake of seeing another path.2614

Let us go ahead and write y’ again.2619

Y’ that was equal to x² - 14x/ x -7².2623

I’m going to the derivative of this.2633

Y” = this × the derivative of that.2638

X – 7² × the derivative of this which is 2x – that × the derivative of that x² – 14x × the derivative of this2644

which is 2 × x – 7 × the derivative of what is inside which is 1/ x -7².2659

That is going to equal, I’m going to multiply it all out.2676

X² – 14x + 49 × 2x – 2 ×, I’m going to multiply this all out.2681

The x² -14x × x – 7.2697

I pull the 2 out, brought it here.2700

This is going to be x³ – 7x² -14x² + 98x/ x – 7².2702

I’m sorry, x – 7²², this is going to be⁴, my apologies.2720

There you go, I’m telling you.2726

Vigilance is the most important thing in calculus.2728

Let me see, what do we got.2734

Let us go ahead and multiply this out.2737

We got 2x³ -28x² + 98x – 2x³2745

+ 14x² +28x² – 196x/ x – 7⁴.2764

2x³ goes with 2x³ – 28x² + 28x².2783

I end up with 14x², 98x -, this is going to be -98x/ x – 7⁴.2793

I’m going to pull out a 14x and I’m going to write that as x – 7.2809

This is x -7⁴.2815

This one cancels one of this.2818

I’m left with 14x/ x – 7³.2820

It is absolutely your choice.2825

You can either go product rule, you can go quotient rule.2826

You are going to end up with the same answer.2831

One of them is more tedious, usually than the other.2832

In this particular case, it looks like the quotient rule is the one that is more tedious.2836

I hope that helped a little bit.2842

Thank you so much for joining us here at www.educator.com.2843

We will see you next time, bye.2845