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Implicit Differentiation

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  • 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

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 x³1327

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 y1913

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