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

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

### Implicit Differentiation

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
- Implicit Differentiation
- Example I: Find (dy)/(dx) by both Implicit Differentiation and Solving Explicitly for y
- Example II: Find (dy)/(dx) of x³ + x²y + 7y² = 14
- Example III: Find (dy)/(dx) of x³y² + y³x² = 4x
- Example IV: Find (dy)/(dx) of the Following Equation
- Example V: Find (dy)/(dx) of 6sin x cos y = 1
- Example VI: Find (dy)/(dx) of x² cos² y + y sin x = 2sin x cos y
- Example VII: Find (dy)/(dx) of √(xy) = 7 + y²e^x
- Example VIII: Find (dy)/(dx) of 4(x²+y²)² = 35(x²-y²)
- Example IX: Find (d²y)/(dx²) of x² + y² = 25
- Example X: Find (d²y)/(dx²) of sin x + cos y = sin(2x)

- Intro 0:00
- Implicit Differentiation 0:09
- Implicit Differentiation
- Example I: Find (dy)/(dx) by both Implicit Differentiation and Solving Explicitly for y 12:15
- Example II: Find (dy)/(dx) of x³ + x²y + 7y² = 14 19:18
- Example III: Find (dy)/(dx) of x³y² + y³x² = 4x 21:43
- Example IV: Find (dy)/(dx) of the Following Equation 24:13
- Example V: Find (dy)/(dx) of 6sin x cos y = 1 29:00
- Example VI: Find (dy)/(dx) of x² cos² y + y sin x = 2sin x cos y 31:02
- Example VII: Find (dy)/(dx) of √(xy) = 7 + y²e^x 37:36
- Example VIII: Find (dy)/(dx) of 4(x²+y²)² = 35(x²-y²) 41:03
- Example IX: Find (d²y)/(dx²) of x² + y² = 25 44:05
- Example X: Find (d²y)/(dx²) of sin x + cos y = sin(2x) 47:48

### AP Calculus AB Online Prep Course

### Transcription: Implicit Differentiation

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

*Today, we are going to talk about this technique called implicit differentiation.*0004

*Let us jump right on in.*0008

*Suppose, we are given the following function.*0011

*Suppose, we are given x³ + y = 7x, what is dy dx?*0019

*What is dy dx or y‘, whichever symbol you want to use.*0035

*For this one, we can solve explicitly for y and just differentiate with we have been doing all this time.*0044

*By explicitly, I mean just y on one side of the equal sign, and then, everything else in x on the other side of the equal sign.*0053

*Here we can solve explicitly for y, then differentiate as before.*0062

*Basically, I just move everything over and I end up with y is equal to 7x -3x³.*0091

*And then, y’ is 7 -, it is just x³ not 3x³, my apologies.*0099

*This just becomes -3x².*0111

*This one is easy to handle.*0114

*In other words, your function is always going to be given to you in terms of y = this.*0116

*Sometimes, you have to put it in that form, before you actually take the derivative.*0121

*What about this, what about x³ + y⁴ = -9xy.*0125

*How do we handle something like this?*0140

*Here we cannot solve explicitly for y.*0144

*We cannot rearrange this equation or manipulate it mathematically such that y = something.*0147

*Even if we could, the expression might be so complicated.*0153

*By taking the derivative of it is going to be intractable.*0157

*It is just not something you want to do.*0160

*Fortunately, there is a way around this.*0162

*This equation, it still defines, if a relation between x and y.*0165

*It is saying that if I x³, I had the y⁴, it is equal to -9xy.*0171

*There is a relation here between x and y, but the relation is implicit.*0175

*Explicit means you have one variable on one side of the equality sign.*0180

*And then, the function whatever it is on the other side, only that variable.*0184

*Y is a function of x, y = something in x.*0189

*Here it is implicit, it is implied in this.*0192

*The idea is that theoretically, you might be able to but how do we find the derivative of it, if we do not have y = something.*0195

*There is a way of doing so and that is called implicit differentiation, when you write all of these out.*0203

*Here, we cannot or do not want to express y explicitly, in terms of x.*0211

*But, the equation, it still defines a relation between x and y, an implicit relation.*0240

*In other words, the x and y are sort of mixed up in the equation, that is an implicit relation.*0269

*To find dy dx or y’ with respect to x, we use implicit differentiation.*0274

*Very important.*0294

*Here is how you do it.*0299

*Let me go to blue here.*0301

*Treat y as a function of x.*0303

*When you differentiate y, when you take the derivative of y, use the chain rule.*0318

*It does not really mean much as written.*0338

*Let us see what actually happens.*0339

*Use the chain rule then isolate the symbol dy dx.*0342

*Let us see what this means.*0359

*Let us start with letters that we had.*0362

*We had x³ + y⁴ = -9xy.*0363

*We are going to implicitly differentiate this entire thing, left side and right side.*0371

*We will start with differentiate everything.*0377

*You are differentiating with respect to x because that is what we are looking for.*0387

*We are looking for dy dx.*0390

*The variable that we are differentiating with respect to is x.*0392

*It is going to look like this.*0396

*We are going to take the ddx of x³ + the ddx of y⁴ = the ddx of -9xy.*0398

*That is all we are saying. We are saying, here we have this equation.*0413

*What we do to the left side, we do to the right side.*0417

*If we differentiate the left side, we differentiate the right side.*0419

*It retains the equality.*0422

*That is all we are doing.*0424

*You differentiate everything.*0425

*The derivative with respect to x³, that is going to be 3x².*0427

*The derivative with respect to x of y⁴, we treat y as a function of x.*0434

*Chain rules says, it becomes 4y³ dy dx.*0442

*That is what the chain rule is.*0450

*It says this is a global function, there are two things going on.*0451

*There is the y⁴, there is the power function, and then there is y itself which is a function of x.*0455

*We are presuming that it is a function of x.*0461

*It is implicit that it is a function of x.*0463

*Therefore, we have to write that dy dx, the derivative of -9xy.*0465

*This -9 is a constant, xy, we use the product rule.*0472

*It is going to be -9 × this × the derivative of that.*0477

*It is going to be x × dy dx + y × the derivative of that.*0482

*Y × 1, the derivative of x with respect to x is 1.*0488

*We have 3x² + 4y³ dy dx =, I’m going to distribute this, the 9 over this.*0497

*- 9x dy dx - 9y.*0511

*I’m going to put all the terms involving dy dx together on one side.*0522

*It is going to be 4y³ dy dx.*0527

*I’m going to bring this one over, + 9x dy dx.*0537

*Over on the other side, I’m going to put everything else.*0542

*I’m going to move this over.*0544

*It is going to be -3x² – 9y.*0546

*I have 4y³ dy dx + 9x × dy dx = this, factor out the dy dx.*0552

*I get dy dx × 4y³ + 9x = - 3x² -9y, now divide by 4y³ + 9x.*0566

*In other words, isolate the dy dx.*0586

*Dy dx = -3x² -9y/ 4y³ + 9x.*0589

*I have my derivative.*0600

*The only issue with this derivative is not really an issue.*0605

*It is just something that you have not seen, is that these derivative actually involves both variables x and y.*0607

*When a function is given explicitly like y = sin (x), the derivative is only going to involve the variable x.*0613

*Implicit derivatives, implicit differentiation expresses the derivative in terms of both variables x and y.*0621

*That is not a problem because in general, we are going to know what x and y are.*0627

*In the sense that we are going to pick some point, 5 3, -6 9, to put them in to find the derivative.*0632

*Again, the derivative is two things, it is a slope and it is a rate of change.*0640

*That is all it is.*0644

*Having it expressed with a variable itself, the y itself does not really matter.*0645

*For any value of x and y, this gives the slope at that point.*0652

*That is it, no big deal.*0670

*We are just adding that y in there.*0673

*Notice, dy dx is expressed in terms of both x and y.*0679

*It is generally going to be true.*0703

*Again, this is not a problem.*0705

*We want an expression for dy dx for y’.*0709

*This implicit differentiation gives us that expression.*0713

*Nice and straightforward.*0718

*The best thing to do is just do examples.*0720

*Once you see the examples, get used to this idea of differentiating y, treating it as of function of x using chain rule on it.*0722

*Once you see it a couple of times, it will make sense what is going on.*0730

*Let us jump right on in.*0734

*We have our first example, xy + 4x - 3x² = 9.*0737

*Here we want you to find dy dx by both implicit differentiation and solving explicitly for y.*0746

*Here we want you to do it both ways, to see if we actually get the same answer.*0754

*Let us see what we have.*0758

*Let us do it explicitly first.*0759

*This is going to be explicit.*0761

*We have our function here.*0767

*Explicit means we want y = some function of x.*0770

*Fortunately, this one, we can do.*0773

*I have the function of xy + 4x - 3x² = 9.*0776

*That gives me, xy = 9 - 4x + 3x², y = 9 - 4x + 3x²/x.*0785

*Now we want the derivative, so y’.*0804

*I’m just going to go ahead and use quotient rule here.*0809

*This × the derivative of that - that × the derivative of this/ the square.*0810

*We are going to have x × the derivative of this which is -4 + 6x - that × the derivative of this, -9 - 4x + 3x² × 1/ x².*0815

*Let me see, this gives me -4x.*0840

*I’m going to distribute the x, + 6x².*0843

*This is going to be -9, this is going to be +4x, this is going to be -3x²/ x².*0847

*-4x and +4x go away.*0857

* 6x² and -3x², that is going to be 3x² - 9/ x².*0859

*My explicit, in this particular case, was able to separate the x and y, separate the variables.*0869

*I did my derivative the normal way.*0876

*Let us do it implicitly and see if we get the same answer.*0881

*Let us see, where are we?*0888

*Did I skip anything?*0893

*No, I did not skip anything.*0902

*Let me go ahead and do it implicitly.*0904

*3x² – 9/ x², let me just write that.*0907

*3x² – 9/ x², this was the derivative that we got.*0911

*Now let us do it implicitly.*0916

*I have xy + 4x – 3x² = 9, differentiate everything across the board.*0924

*This is differentiating everything with respect to x because we are looking for dy dx.*0940

*This is product rule, it is going to be this × the derivative of that.*0946

*I’m going to get x × the derivative of y xy‘ + y × the derivative of this + y × 1 + the derivative of 4x is 4.*0950

*The derivative of this is 6x and the derivative of 9 is 0.*0963

*I just differentiated right across the board, left side and right side.*0969

*I'm going to write this as xy’ =, I’m going to move everything over.*0975

*It is going to be 6x - y – 4.*0982

*I get y‘ is equal to 6x - y - 4/ x.*0989

*I found my y‘, my dy dx is 6x - y - 4/ x, implicit.*1000

*It involves both x and y.*1006

*Wait a minute, we are supposed to get the same thing.*1008

*When we did it explicitly, we ended up with 3x² - 9/ x², why are not these the same?*1011

*They are the same.*1021

*Let us see what we have got here.*1024

*We said, remember when we solved explicitly for this function.*1028

*We got y is equal to 9 - 4x + 3x²/ x.*1032

*We solved it explicitly.*1043

*Let us put this expression for y into here, to see what happens.*1045

*Y‘ is equal to 6x - 9 - 4x + 3x²/ x - 4/ x.*1053

*Let us simplify this out.*1072

*I’m going to get a common denominator here.*1073

*This is going to be 6x².*1076

*I’m going to distribute -9, that is a -9 + 4x - 3x².*1084

*The common denominator here is -4x.*1098

*All of this is over the common denominator x and all of this is /x.*1102

*4x - 4x, 6x² - 3x² = 3x² – 9/ x².*1107

*They are the same, the only difference is the implicit differentiation ended up expressing the derivative dy dx, in terms of both x and y.*1121

*In this particular case, because we had an explicit form, we ended up checking it by putting it in here*1131

*and realizing that we actually did get that.*1137

*Again, you are not always going to be able to do that.*1139

*In this particular case, you were able to.*1141

*But most implicitly defined relations, you are not going to be able to solve explicitly for one of the variables.*1144

*You have to leave it in terms of x and y.*1149

*This is a perfectly good derivative.*1152

*Example 2, x³ + x² y + 7y² = 14, find dy dx.*1160

*The biggest problem with calculus is the algebra.*1168

*Go slow, stay calm, cool and collected, and hopefully everything will work out right.*1170

*I say this and yet I, myself make mistakes all the time.*1176

*Hopefully we will keep our fingers crossed and we would not make any algebra mistakes.*1179

*Let us see how we deal with this one.*1183

*This is our function and we want to find dy dx.*1185

*I’m going to work with y’.*1192

*I do not want to write dy dx over and over again.*1193

*The derivative of this is 3x², x² y, this is product rule.*1197

*It is going to be + this × the derivative of that.*1205

*It is going to be x² × the derivative of y which is just y' + that × the derivative of this.*1210

*+ y × the derivative of x² is 2x.*1220

*The derivative of 7y², it is 14y y’.*1225

*Or we do the chain rule, we take care of the power and then we take dy dx.*1231

*We write the y'.*1237

*The derivative of 14 is 0.*1239

*We are going to go ahead and put things that involve the y or the dy dx together.*1244

*We have x² y' + 14y y1' =, I’m going to move this and this over to the other side.*1251

*-3x² -2xy, factor out the y'.*1263

*Y’ × x² + 14y = -3x² - 2xy, and then divide.*1270

*I get y’ is equal to -3x² - 2xy all divided by x² + 14y.*1282

*That is it, that is my derivative, that is dy dx, that is y’.*1296

*Let us move on to next example.*1305

*Here we have x³ y² + y³ x² = 4x, find dy dx.*1312

*Here same thing, this is going to be product rule.*1322

*I have this × the derivative.*1325

*I got x³ × the derivative of y² which is 2y y' + y² × the derivative of x³*1327

*which was 3x² + y³ × the derivative of x² which is 2x + x² × the derivative of y³ which is 3y² dy dx y'.*1342

*The derivative of 4x is 4, put my prime terms together.*1367

*I’m going to write this as 2x³ y y' + 3x² y² y’ = 4 - 3x² y².*1377

*I move this over to that side.*1399

*I’m going to move this over to that side, - 2xy³, factor out the y'.*1401

*Y' × 2x³ y + 3x² y² = 4 - 3x² y² - 2xy³, and then divide.*1410

*Y’ is equal to this whole thing.*1428

*-3x² y² - 2xy³ divided by 2x³ y + 3x² y².*1431

*It is exhausting, is not it?*1449

*It really is, calculus is very exhausting.*1451

*Y e ⁺x² + 14 = √2 + x² y.*1460

*Let us dive right on in.*1468

*Let me go to blue here.*1470

*This is product rule, this × the derivative of that + that × the derivative of this.*1472

*We have y × the derivative of this which is e ⁺x² × 2x + e ⁺x² × the derivative of y which is just y'.*1477

*Just writing y’, dy dx is the derivative, that is what I'm looking for.*1495

*The derivative of 14 is 0, the derivative of this, this is we know is equal to 2 + x² y ^ ½.*1504

*The derivative of that is going to be ½ × 2 + x² y⁻¹/2 × the derivative of what is inside which is 0 + this × the derivative of that.*1517

*X² y’ + this × the derivative of that, x² y’ + y × 2x.*1533

*Putting all this together, we are going to end up with 2xy e ⁺x² + e ⁺x ² y' =,*1539

*I’m going to distribute this thing over that and that.*1553

*This first one is just 0 = ½ x² y' × 2 + x² y⁻¹/2 + this × the other term.*1561

*½ × 2xy × 2 + x² y⁻¹/2.*1582

*Let me see, where am I?*1606

*Now I’m going to go ahead and collect.*1609

*Let me go ahead and do red.*1610

*This is something that involves a y' term.*1612

*This term is what involves a y’ term, I’m going to bring them together.*1615

*I’m going to get e ⁺x² y’ - ½ x² y' × 2 + x² y⁻¹/2 =, I’m going to put everything else on the other side.*1619

*I leave this over here, the 2 cancel.*1637

*I'm left with xy × 2 + x² y⁻¹/2.*1640

*I bring that over to that side, -2xy e ⁺x².*1648

*Factor out the y', I get y' × e ⁺x² - ½ x² × 2 + x² y ^-½ = xy × 2 + x² y ^ -½ -2xy e ⁺x².*1656

*We finally have y’ = xy × 2 + x² y⁻¹/2 - 2xy e ⁺x² all divided by e ⁺x² – ½ 2 + x² y – ½.*1692

*There you go, that is our final answer.*1725

*Very complicated looking, but again, once you have x and y, you just plug them in and you solve it.*1729

*Let us see what we have got.*1738

*That was example 4.*1740

*Example 5, 6 sin x cos y = 1.*1744

*Let me go back to blue here.*1748

*Let me do it down here.*1756

*6 ×, this is product rule, sin x × cos y.*1761

*It is going to be this × the derivative of that + that × the derivative of this.*1765

*Sin x × the derivative of cos y which is –sin y × y' + cos y × the derivative of sin x which is cos x.*1771

*The derivative of 1 = 0.*1786

*We end up with, 6 goes away, we end up with - sin x sin y × y' + cos x cos y = 0.*1791

*We have -sin x sin y y' = -cos x cos y.*1812

*Therefore, y' is equal to cos x cos y, I will leave the minus sign, that is fine, divided by -sin x.*1832

*Sin y, I can go ahead and cancel.*1845

*I can just leave it as cos x cos y/ sin x sin y.*1848

*The negatives cancel.*1851

*Or I can write it as cot x cot y, either one of these is absolutely fine.*1852

*Hope that makes sense.*1862

*Example 6, same thing, trigonometric, just more complicated.*1866

*Product rule, it is going to be a long differential.*1877

*This × the derivative of that.*1884

*X² × 2 × cos(y) y'.*1886

*The derivative of cos² is 2 cos y × the derivative of what y is, y’.*1893

*Wait, hold on a second, let me do this right.*1906

*This × the derivative of this.*1911

*X² × the derivative of cos² y, that is going to be 2 cos y × the derivative of the cos y*1913

*which is × -sin y × the derivative of y which is y’.*1926

*There we go, that is better. This × the derivative of that + cos² y × the derivative of x² which is 2x.*1931

*Now, this one, + y × the derivative of sin x which is cos x + sin x × the derivative of y which is y’.*1944

*All of this is equal to 2 × this × the derivative of that which is sin x × -sin y y' + cos y × cos x.*1958

*I do not really need to simplify it, I can just go ahead and jump right on into what it is.*1997

*That is fine, I will just go ahead and rewrite it.*2002

*Let me do it in red, I’m making this a little bit simpler.*2010

*-2x² cos y sin y y’*2018

*+ 2x cos² y + y cos(x) + sin x y’.*2030

*Let me make sure I have everything right here.*2056

*This × the derivative of that, cos x.*2068

*Wait a minute, this is 2 sin x.*2086

*Okay, everything looks good.*2092

*+ sin xy, good, all of that is equal to -2 sin x sin y y’ + cos y cos x.*2096

*I’m going to bring everything, I’m going back the blue.*2118

*This term has a y' in it.*2123

*This term has a y' in it.*2128

*This term has a y' in it.*2130

*I’m going to bring all of those over to one side.*2132

*That is going to be -2x² cos y sin y y' + sin x y' + 2 sin x sin y y' and all of that is going to equal,*2136

*I’m going to put them this term and this term, I’m going to bring those over that side.*2162

*= cos y cos x - 2x cos² y – y × cos(x).*2166

*Let me go back to red.*2186

*When I factor out the y’ and I divide, I end up with y’ = this thing on the right.*2187

*Cos y cos(x) – 2x cos² y – y × cos(x) divided by this -2x² cos(y) sin y + sin(x) + 2 sin x sin y.*2199

*Keeping track of it all, that is it.*2233

*The fundamental part is differentiating this very carefully.*2236

*There we are, this is calculus, welcome to calculus.*2244

*I have an extra page for that too.*2251

*Great, I’m squeezing everything into one page.*2252

*Example 7, the √xy = 7 + y² e ⁺x.*2257

*We just do the same thing that we always do.*2264

*We know that this is equal to xy¹/2.*2268

*Let me differentiate.*2274

*We end up with, it is going to be ½ xy ^-½ × the derivative of what is inside the xy,*2275

*which is going to be this × the derivative of that xy' + y × the derivative of the x which is 1 = 0 + this × the derivative of that,*2297

*y² e ⁺x + e ⁺x × the derivative of that which is 2y y'.*2310

*Let me distribute this part.*2319

*I’m going to distribute this over that and this over that.*2330

*I'm going to get ½ x y' × xy ^-½ + ½ y × xy⁻¹/2 = y² e ⁺x + 2y e ⁺x y'.*2334

*Here is a y’ term, let me go back to blue.*2371

*Here is y’ term, here is a y’ term.*2375

*Let me put those together.*2378

*I have ½ x y’ xy⁻¹/2 – 2y y’ e ⁺x.*2380

*Let me move this over to that side.*2392

*It equals y² e ⁺x – ½ y xy⁻¹/2.*2395

*Let me go back to red.*2406

*When I factor out the y’, I’m going to get y’ × ½ x × xy⁻¹/2 – 2y e ⁺x = y² e ⁺x – ½ y xy⁻¹/2.*2407

*I will go ahead and divide.*2435

*I’m left with y’ = y² e ⁺x – ½ y × xy⁻¹/2 divided by ½ x × xy⁻¹/2 – 2y e ⁺x.*2436

*There you go.*2457

*Example number 8, same thing, just equation, any other equation that we have to deal with.*2467

*Let us go ahead and differentiate this.*2476

*This is going to be 4 × this thing.*2477

*The derivative of this thing is going to be 2 × this x² + y² × the derivative of what is inside.*2483

*That is going to be 2x + 2y y' = 35 × 2x – 2y y.*2492

*I get 8 ×, I’m going to multiply this.*2513

*This is a binomial × a binomial.*2519

*I’m going to get 2x³.*2521

*This × this is going to be + 2x² y y' + y² × 2x is going to be 2x y² + 2y³ y' = 70x-70y y'.*2525

*Here I have a y' term, here I have a y' term, here I have a y’ term.*2550

*Let me go back to red, put a parentheses around that.*2556

*Let me multiply through, all of these becomes 16, that stays 70.*2559

*I’m going to go ahead and just multiply, and move things around.*2563

*On this slide, I'm going to have 16.*2568

*Let me go to blue.*2572

*I’m going to have 16x² y y' + 16 y³ y'.*2575

*I’m going to bring the 70 y’ over + 70y y' = 70x - 16 x³, that is this one.*2586

*And then, - 16xy².*2601

*Go back to red, there is my y’.*2608

*Y’ and y’, factor out, divided by what is left over.*2610

*I’m going to be left with y' is equal to 70x -16x³ - 16xy² divided by 16x² y + 16y³ + 70 y.*2614

*There is my derivative.*2638

*We have x² + y² = 25.*2647

*This time they want us have to find d² y dx².*2649

*They want us to find y”, the 2nd derivative.*2652

*I will just do it in black.*2656

*X² + y² = 25.*2660

*The derivative of x² is 2x.*2662

*The derivative of y² is 2y dy dx, y’ = 0.*2665

*I have 2y y' = -2x.*2674

*Therefore, y’ = -x/y, that is my y'.*2682

*I want y”.*2690

*Now I take the derivative of this.*2694

*Again, I differentiate implicitly.*2699

*Here, this is going to come down.*2702

*This is going to end up becoming, it is going to be this × the derivative of that - that × the derivative of this/ this².*2705

*Let me go ahead and just keep my negative sign here.*2715

*Let me write y” =, the negative sign that is this negative sign.*2720

*It is going to be this × the derivative of that, y × 1 - this × the derivative of that – x y'/ y².*2726

*That is equal to distribute the negative sign.*2739

*We get x y' - y/ y².*2741

*However, notice the second derivative, not only does it have x and y but it also has the y'.*2751

*I already know what y' is, I already found the first derivative.*2760

*Y' is equal to -x/y.*2762

*I can take this -x/y, stick it into y'.*2765

*Let me do it down here.*2774

*This =, y” = x × y' which is - x/y - y/ y²*2775

*which is equal to -x²/ y - y/ y² = -x² - y²/ y/ y²,*2791

*which becomes -x² - y²/ y³.*2813

*When you take the second derivative, the second derivative is actually going to involve the first derivative.*2824

*But you already found the first derivative, you can put that back in.*2830

*Something like this, reasonably simple and straightforward.*2834

*Not going to be so reasonable and straightforward, when you actually do some of the longer problems like we are going to do in a second.*2840

*You would have to decide the extent to which you want to actually put in.*2845

*The extent to which you want to simplify it, things like that.*2853

*As long as you realize that, y” is going to also contain the first derivative and you would already have the 1st derivative.*2856

*Those are going to be the important parts.*2864

*Example 10, let us see what we can do.*2870

*Example 10, let me go ahead and go back to blue.*2878

*The derivative of sin x is the cos(x).*2886

*The derivative of cos y =, it is negative, + -sin y y1'.*2890

*The derivative of sin 2x is cos 2x × the derivative of 2 which is 2, it is 2 × cos(2x).*2903

*Let us solve, -sin y y’ is equal to 2 cos 2x – cos x.*2916

*Therefore, we get y' is equal to 2 cos 2x - cos x/ -sin y.*2928

*I’m going to go ahead and take this negative sign, bring it up top.*2942

*Flip these two and I’m going to write this as cos(x) -2 cos(2x)/ sin y.*2945

*I hope that made sense.*2958

*A negative sign actually can go top or bottom.*2958

*It does not really matter.*2960

*I went ahead and brought it up here.*2961

*When I distributed over this, the negative of this and this, this one becomes positive, I put it first.*2964

*This one is negative, I put it second.*2969

*That is all I have done.*2971

*There we go, we have y’.*2973

*Now, y”.*2977

*Y”, let me do this one in red.*2982

*Y”, I'm actually going to be differentiating this one implicitly.*2988

*It is going to be this × the derivative of that - that × the derivative of this/ this².*2995

*That = sin y, this × the derivative of this.*3001

*The derivative of cos x is -sin x.*3010

*The derivative of this cos 2x is going to be -2 sin 2x.*3017

*-2 × 2 is 4, it is going to be +, it is going to be +4, × sin(2x).*3025

*That is this × the derivative of that - this cos x - 2 cos 2x × the derivative of this which is cos y y’/ sin² y.*3032

*Let us take a look at this, before we do anything.*3059

*It involves sin x, it involves sin y, or cos x and cos y.*3065

*It looks like the only y' term that we have is here.*3070

*That is this one.*3075

*If you were going to simplify it, you do not have to.*3077

*If you actually have to, for y', this is y'.*3080

*You take this expression, you put it into here.*3087

*And then, you simplify the expression as much as possible.*3091

*I’m not going to go ahead and do this.*3094

*For this particular problem, something that is going to end up looking like this, I would just leave it alone.*3096

*If your teacher is going to have you do this, where you actually have to substitute in,*3103

*only express things in terms of x and y, then you have to do something simple.*3106

*Otherwise, you will be here for 4 days doing this.*3110

*But for something like this, you are going to deal with it at some point in your career.*3113

*You have y', you have y”, if you need to, you go ahead and put it in.*3119

*Again, most of the time, when you are working,*3123

*you are going to be working with mathematical software so it is going to be doing this for you.*3126

*That is it, straight differentiation.*3130

*You differentiate the first one implicitly.*3134

*You get this, you differentiate the second one implicitly.*3135

*It is going to get progressively more complicated, simply by virtue of the fact that this happens to be a quotient rule.*3139

*I hope that that made sense.*3147

*Thank you so much for joining us here at www.educator.com.*3148

*We will see you next time, bye.*3150

1 answer

Last reply by: Professor Hovasapian

Fri Aug 18, 2017 3:17 AM

Post by Kevin Wang on August 8 at 02:07:10 PM

Hi Professor Hovasapian,

In Example 6, at 35:16, how is it that there isn't a coefficient of 2 in front of the term cos(y)cos(x) ?

1 answer

Last reply by: Professor Hovasapian

Tue Jan 31, 2017 6:46 AM

Post by Efrain Loeza Martinez on January 27 at 11:46:18 AM

Lesson : Implicit Differentiation.

Hi professor. In Example V ( 30:00)

¿Why do you cancel the 6?

2 answers

Last reply by: Professor Hovasapian

Fri Nov 18, 2016 8:26 PM

Post by Muhammad Ziad on November 15, 2016

Hi Professor Hovasapian,

In Example 2, why is the derivative of 7y^2, written as 14yy', instead of just 14y.

Thank you!