For more information, please see full course syllabus of AP Calculus AB

For more information, please see full course syllabus of AP Calculus AB

### 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

### AP Calculus AB Online Prep Course

### 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 this *1539

*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-7 *2522

*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 this*2644

*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 – 2x³ *2745

*+ 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

2 answers

Last reply by: Venise Martinez

Wed Mar 30, 2016 8:27 PM

Post by Jerica Cui on January 29 at 01:50:49 AM

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)