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1 answer

Last reply by: Professor Hovasapian
Tue Nov 25, 2014 2:25 AM

Post by Utomo Pratama on November 24 at 10:09:07 AM

Dear Prof. Hovasapian,

Thank you for the great delivery. Now Math is becoming interesting to me.
Just some quick clarifications, if there are 3 variables, then there would be six combinations of equal mixed partial derivatives? (3!)

CMIIW

Regards,

Utomo

1 answer

Last reply by: Professor Hovasapian
Thu Feb 27, 2014 7:33 PM

Post by Chase Lottinger on February 26 at 07:10:08 PM

Hello, I was just wondering why don't you cover multivariable limits? Or is it labeled as something else in the table of contents?

1 answer

Last reply by: Professor Hovasapian
Wed Apr 3, 2013 2:30 AM

Post by Jawad Hassan on April 2, 2013

Hi raffi!

I was wondering if you know any good place to practise exam problems for multivariable calculus? once i am done with all the videos.

Or any good books you recomend that have exam problems and good solution manual.

Btw excellent video! I am realy enjoying math atm.

regards
-Jawad

2 answers

Last reply by: Matt C
Wed Mar 27, 2013 9:17 PM

Post by Matt C on March 25, 2013

In your notes you state Assume D1, D2, D1D2, D2D1, exist and are continuous then D1D2f = D2D1f. Is it ever possible that they can be not equal? Im guessing they would be discontinuous then, but that is just a guess. I have done many problems in my book and have yet to find a case where they don't equal each other, they all seem to work out.

1 answer

Last reply by: Professor Hovasapian
Tue Aug 7, 2012 4:45 PM

Post by Shahaz Shajahan on August 7, 2012

Hi, I know this is very basic but anything to take part,right?

when you was expanding D1D1 you forgot to expand out the 2 onto the second term so it should have read out 2cos(x^2+y) instead of just cos(x^2+y)

Btw you have been a real help with my exams, as i have a month left and was really worried i'd fail but you have given me renewed hope! :D

Higher and Mixed Partial Derivatives

Let f(x,y) = − 3x2 + 5xy − y2 Find [df/dx] and [df/dy]

  • [df/dx] is the partial derivative in respect to x, similarly [df/dy] is the partial derivative in respect to y.

So [df/dx] = − 6x + 5y and [df/dy] = 5x − 2y.

Let f(x,y) = − 3x2 + 5xy − y2 Find ( [d/dx] )2f and ( [d/dy] )2f

  • Note that ( [d/dx] )2f = [d/dx]( [df/dx] )f, so ( [d/dx] )2f = [d/dx]( − 6x + 5y ) = − 6.

Similarly ( [d/dy] )2f = [d/dy]( 5x − 2y ) = − 2.

Let f(x,y) = − 3x2 + 5xy − y2 Verify [d/dx][d/dy]f = [d/dy][d/dx]f

  • To compute [d/dx][d/dy]f, we find the partial derivative of f in respect to y and then take the partial derivative of f in respect to x.
  • So [d/dx][d/dy]f = [d/dx]( 5x − 2y ) = 5.

Similarly [d/dy][d/dx]f = [d/dy]( − 6x + 5y ) = 5. We note that both [d/dx][d/dy]f and [d/dy][d/dx]f yield the same result.

Let g(x,y) = cos(x2 + 1) − exsin(y) Find ( [d/dx] )2[d/dy]g

  • To compute ( [d/dx] )2[d/dy]g, we take each partial derivative to the left of g. Note that ( [d/dx] )2 = [d/dx][d/dx].

So ( [d/dx] )2[d/dy]g = [d/dx][d/dx][d/dy]g = [d/dx][d/dx]( − excos(y) ) = [d/dx]( − excos(y) ) = − excos(y).

Let g(x,y) = cos(x2 + 1) − exsin(y) Find ( [d/dy] )2[d/dx]g

  • To compute ( [d/dy] )2[d/dx]g, we take each partial derivative to the left of g. Note that ( [d/dy] )2 = [d/dy][d/dy].

So ( [d/dy] )2[d/dx]g = [d/dy][d/dy][d/dx]g = [d/dy][d/dy]( − 2xsin(x2 + 1) − exsin(y) ) = [d/dy]( − excos(y) ) = exsin(y).

Let g(x,y) = cos(x2 + 1) − exsin(y) Verify [d/dy][d/dx]g = [d/dx][d/dy]g

  • Note that [d/dy][d/dx]g = − excos(y) and [d/dx][d/dy]g = − excos(y) from our previous problems.

[d/dy][d/dx]g = [d/dx][d/dy]g since [d/dy][d/dx]g and [d/dx][d/dy]g yield the same result.

Let h(x,y,z) = [1/xyz] Evaluate [d/dz][d/dx]h(1,1, − 2)

  • To compute [d/dz][d/dx]h(1,1, − 2) we first find [d/dz][d/dx]h. Note that [1/xyz] = (xyz) − 1 = x − 1y − 1z − 1.
  • Now [d/dz][d/dx]h = [d/dz]( − [1/(x2yz)] ) = [1/(x2yz2)].

Hence [d/dz][d/dx]h(1,1, − 2) = [1/((1)2(1)( − 2)2)] = [1/4].

Let h(x,y,z) = [1/xyz] Evaluate [d/dz][d/dy]h(1,1, − 2)

  • To compute [d/dz][d/dy]h(1,1, − 2) we first find [d/dz][d/dy]h. Note that [1/xyz] = (xyz) − 1 = x − 1y − 1z − 1.
  • Now [d/dz][d/dy]h = [d/dz]( − [1/(xy2z)] ) = [1/(xy2z2)].

Hence [d/dz][d/dy]h(1,1, − 2) = [1/((1)(1)2( − 2)2)] = [1/4].

Let h(x,y,z) = [1/xyz] Evaluate [d/dz][d/dy][d/dx]h(1,1, − 2)

  • To compute [d/dz][d/dy][d/dx]h(1,1, − 2) we first find [d/dz][d/dy][d/dx]h. Note that [1/xyz] = (xyz) − 1 = x − 1y − 1z − 1.
  • Now [d/dz][d/dy][d/dx]h = [d/dz][d/dy]( − [1/(x2yz)] ) = [d/dz]( [1/(x2y2z)] ) = − [1/(x2y2z2)].

Hence [d/dz][d/dy][d/dx]h(1,1, − 2) = − [1/((1)2(1)2( − 2)2)] = − [1/4].

Let h(x,y,z) = [1/xyz] Evaluate [d/dx][d/dy][d/dz]h(1,1, − 2)

  • To compute [d/dx][d/dy][d/dz]h(1,1, − 2) we first find [d/dx][d/dy][d/dz]h. Note that [1/xyz] = (xyz) − 1 = x − 1y − 1z − 1.
  • Now [d/dx][d/dy][d/dz]h = [d/dx][d/dy]( − [1/(xyz2)] ) = [d/dx]( [1/(xy2z2)] ) = − [1/(x2y2z2)].

Hence [d/dx][d/dy][d/dz]h(1,1, − 2) = − [1/((1)2(1)2( − 2)2)] = − [1/4].

*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.

Answer

Higher and Mixed Partial Derivatives

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.

Transcription: Higher and Mixed Partial Derivatives

Hello and welcome back to educator.com and welcome back to Multivariable Calculus.0000

Today, we are going to be talking about higher derivatives and mixed partial derivatives.0005

So, we introduced the notion of the partial derivative of a function of several variables.0010

You basically take the single variable derivative by holding the other variables constant, and you work through a series of partial derivatives, first derivatives. 0015

Just like for single variable, we can go ahead and take higher derivatives, second, third, fourth, and fifth, but now we can mix and match.0024

For example, we can take the first partial derivative and then we can take the derivative with respect to the other variable of that derivative, and back and forth.0030

Let us sort of just jump in with an example and see what we can do with this. 0040

Before I actually discuss that, however, I want to introduce a notion called an open set.0046

I am not going to spend a lot of time on it, I just want you to know that this idea of an open set is going to be the domain over which we are going to be defining our functions of several variables.0052

I want you to be aware of it, it is going to come up over, and over, and over again in the theorems, and the definitions -- very, very simple -- it is exactly the same as it is for single-variable calculus.0063

It is basically just a region, all the points in that region, but not including the endpoint, not including the boundaries.0072

Let us go ahead and just write the definition down and then we will move on to mixed partial derivatives.0081

So, definition. An open set in n-space -- excuse me -- is the analog, this is not really a formal definition, it is an informal definition. 0086

If you want the formal definition, you know, you can go into your books... is the analog to an open set in 1-space, which is the real number line.0108

Which is what you have been dealing with all of these years. You know that the real number line... something like this... let us say this is 0, let us say we have -1, let us say we have +1.0124

Well the open set, you remember, is the notation that has the parentheses like this instead of the closed brackets, which means that we include all the numbers between -1 and 1, but we do not include the endpoints, that is it.0134

So, basically, an open set is every point within a specified boundary, but not including the boundary points.0150

That is really all we are talking about. If you have something like... let us do a 2-space example... so this is the x,y plane, and you have some region like this.0191

Actually, you know what, let us go ahead and make it so it looks... so let us say we have some region like this, I will put dotted, something like that.0203

So this region in here, that is an open set. It does not include the boundary. That is all we are talking about.0215

Just a quick version in R3, so this is the x, the y, and the z. 0221

So we have some... let us just do a sphere. We call it a ball. Here, if you have a circle in the plane, we call it an open disc.0230

Again, the specific things that we call them are not important. The idea is the notion of an open set, or an open region.0240

Because you are not always going to get a perfect domain, you know a nice symmetrical domain like a ball or a disc.0247

In this case, let us just do that, and then maybe like that, so think of a sphere in n-space anywhere -- centered at the origin or centered at some place else -- but, not including the actual boundaries themselves, everything inside.0254

So again, these open sets... let me write this down.0268

So, these open sets form the domains over which we define our functions of several variables.0273

So, if we had some function f that is a mapping from R2 to R, let us go ahead and make sure this is a little bit more clear. 0304

So some function f, and we are taking points in R2, the vector in 2-space, a point in 2-space, we are doing something to that vector, we are spitting out a number. Let us say the domain is this.0314

Basically, all of the points in here, those are the arguments that go into the actual function. Those are the points that we operate on, if you will. That is all.0324

Just a notion that you should be aware of. It is going to come up a lot, again. All of the definitions and the theorems.0334

Okay, now let us go ahead an move on to our partial derivatives.0340

Again, we can take a second and third, fourth, fifth derivatives of functions of several variables, but now of course we are dealing with partial derivatives in the sense that we are holding one variable constant and we are differentiating with respect to the others.0345

So, let us talk about notation first. This is very, very important.0360

As you noticed from the last lesson, the notation for partial differentiation tends to become a little bit more involved because again you have more variables involved, that is what is going.0365

It is very important to be accustomed to the notation and the scientific literature and the different books that you are going to read and the different teachers that you are going to have, they tend to use notations form a lot of different sources.0373

We want to make sure that if you have not seen a particular notation, at least you can recognize was is being said.0387

We said we have... let me just write here, notation... so, we will let f be a function from R2 to R, so in this case we are going to start a function of 2 variables. 0393

So, an f(x,y) if we happen to call the variables x and y, but we do not have to it could be x1, x1, the variables themselves do not actually matter.0413

It is the notion underneath. The partial derivative of f, with respect to the first variable x is notated like that, that little modified d, df/dx.0420

We also do d1, so d1 means take the partial derivative with respect to the first variable, the first variable being x. 0433

They are written in order in the argument, that is it.0440

Again, you also have seen it as dx. Well, df/dy the partial derivative with respect to y, that is one notation, and we often write it as d2, so we are taking the partial derivative with respect to the second variable.0444

This capital D notation, that is actually very, very common in multi-variable calculus. You will this a lot also, but often times just to sort of keep the mathematics on the paper reasonably sane.0461

When you have all these df/dx's df/dy's floating around, it gets really crazy -- so, this d notation, this capital D notation - very, very, convenient.0472

Again, you will also see it as dy, specifically.0483

Now, if we want to take the derivative with respect to x of the first partial, in other words, d/dx of df/dx, it is like this.0488

It is going to be d/dx of df/dx, so the partial derivative with respect to x of the partial derivative with respect to x.0501

It is notated as d2f/dx2 completely analogous to single variable calculus.0516

The capital notation is something like this, just D1/D1f, in other words you have taken the first partial derivative, now take the partial derivative with respect to the first variable of what you just did.0523

This is also... you see it this way, d2f. This means take the partial derivative with respect to x, then take the partial derivative with respect to x again, or the first variable. 0539

Now, let us do d/dy of df/dy, that is going to be the same as d2f, dy2.0553

Again, you have got d2/d2f and this is often written -- sometimes -- squared f.0565

I personally do not like this notation myself, this d12f, d22f.0575

I like to see everything that I am doing, so this tells me that, it takes the first, I take the derivative with respect to the second variable, and then of that thing, I take the derivative with respect to the second variable. 0578

These are just differential operators. They tell me what to differentiate. That is all.0590

Now, let us go ahead and do, d/dx of df/dy.0596

Now, I have taken the partial derivative with respect to y, the second variable, now I am going to go ahead and take the partial derivative of that with respect to the first variable.0610

That becomes... actually, let me just do this one in reverse. This one I am going to write the capital D notation first, because I think it is a little bit better, and then I will write this modified D notation here.0623

So, d1d1, that means I have taken the derivative with respect to the first variable x, and now I differentiate that with respect to the second variable.0637

That is going to be the same as d/dy of df/dx, and that is d2f dy dx.0649

Clearly you can see that this notation is going to start to get really, really cumbersome very, very quickly.0664

I mean it is very beautiful aesthetically, and it is very nice to look at, and it is really important that when you do your math you at least like what you see, but it can be a little daunting.0669

Let us go ahead and specifically say, this is what we are doing first. We are moving from right to left.0682

This means you have a function f, do D1 to it, then do D2 on it. That is first, and this is second, order is very important here.0688

Now, of course we have D1, D2 of f, which means we have done D2 first, then done D1, and this is going to be d/dx of df/dy = d2f dx dy.0698

Wow, that is a lot.0717

Alright, so let us just do an example. That is the best way to make sense of this.0722

Again, you are reasonably comfortable with differentiation of basic functions. With multi-variable calculus just be a little bit more careful. 0727

You have to remember which variable you are differentiating with respect to so just go a little bit slower. That is all you have to do with Math in order to be correct.0736

Just go slow and be careful. So, example 1.0742

We will let f(x,y) = x3y2, so d1 the derivative with respect to x is equal 3x2y2, in other words I am treating y as a constant and just differentiating with respect to x.0751

D2 = x3y, this time I ended up holding x constant and differentiating with respect to the second variable, y.0772

Now, I am going to do d2 of d1, in other words I have done d1, now I am going to differentiate this thing with respect to y, the second variable.0791

This is going to give me, 2 × 3 is going to give me 6x2y.0801

Now, I am going to do D1 of D2, which I just did. I just did D2 that is x3y, then do D1 of that, which means I am going to differentiate this with respect to x.0809

I get 6x2y. We will stop and take a look at this.0823

D2D1, I did D1 first then I did D2, I got 6x2y.0828

here I did D2 first, then I got D1, then I did 6x2y and they ended up being the same. This is not a coincidence.0832

So, we will put MV, which means... take note of this... these ended up being equal.0843

Not a coincidence. Now, let us go ahead and write down the theorem that allows us to always that this will be the case.0863

Theorem. Let f be a function from R2 to R, defined on an open set. 0874

Now, we are going to be using... I just introduced the term open set, f, R2, this is often how you are going to see theorems in mathematics.0887

This is very precisely stated theorem, but it is nothing that we do not understand and know. It is just we want something that is formal and that is precise.0896

Defined on an open set, and assume D1 D2, D2 D1, exists. In other words assume that the partial derivatives actually exist, and are continuous.0906

These clearly exist, and these are clearly continuous functions. There is no problem here. Then, D1 D2 of f = D2 D1 of f.0935

Notice up here just real quickly that I just wrote, D1 D2, D2 D1, but I did not put the f. 0952

Again this is sort of a personal thing that I do. I tend to sort of minimize my notation simply because I do not like a lot of things floating around on a piece of paper.0960

The idea is that you know what you are talking about. You know that you are dealing with f, you know that D1 is the first partial derivative, you know that D2 is the second partial derivative.0969

You do not have to be that explicit. You can modify your notation depending on how much you know, how much you are comfortable with.0978

You have that freedom, do not feel that you are constrained to always write this, this, this, this. Unless you have the kind of teacher that actually demands that you write all the things out, all the x's, all the parentheses, all the y's be there.0984

Please, feel free to take some liberties with this. You are the one doing the math, you are the one that needs to be comfortable with this.0997

For me, I tend not to write the f. I know that this is not confusing at all.1004

Now, let us consider a function from R3 to R. Now let us move from 2-space to 3-space, a function of 2 variables, to a function of 3 variables.1010

Consider the function from R3 to R, or f(x,y,z).1021

That is it. Now, if we take f(y,z), so now we have D1, we will have D2, and we will have D3, partial with respect to x, partial with respect to y, partial with respect to z. 1035

Now we can sort of mix and match. Notice I can have... I can do D1, D2, D3, I can do D3, D2, D1... D2, D1, D3... D2, D3, D1... so all kinds of mixed partials are possible now.1050

Repeated applications of the theorem, which we will do an example of in just a minute, demonstrate that mixed partials are equal, mixed partial as of course 1,2,3... so no worries there... are equal regardless of order, as long as the variables with respect to which we differentiate are the same.1065

Let me just show you what that means. In other words, if I take... so D1, D2, D3 of f.1144

I have taken D3 first, then I have done D2 to that, then I have taken D1 of that.1153

Well, if I do it in different order, do... let us say, D3, D2, D1 of f... if I do D2 first, then D2, then D3... or if I do D2, D3, D1... again we are working from right to left, in other words I do D1 first, then 3 then 2. 1159

As it turns out, all of these mixed partials are the same. It does not matter which order you actually differentiate in, as long as the 1,2,3... 1,2,3... 1,2,3... as long as that is the same.1178

You can actually do it in any order. This is really, really extraordinary. I mean there is no reason to think that if you take the partial derivative of a function of several variables in... you do the mixed partial derivatives, there is no reason to believe that they should be equal and yet there it is. Really, really fantastic.1190

Okay, let us just do an example, and so it is nice to see these things sort of fall out.1209

We will let f(x,y,z) = x2y2z3.1219

Now, we are just going to do a bunch of partials. Let us do, let us start with D3, in other words z.1230

So, D3 is going to equal 3x2y2z2.1240

Now we will do D2 of D3, in other words we are going to differentiate with respect to y, this thing up here. D2 of D3.1247

That is going to equal 6x2yz2, and now we are going to do D1 of D2,D3. In other words we are going to differentiate with respect to x, this thing right here.1257

We end up getting 12xyz2.1272

So, we will do that, now let us go ahead and od it in a different order.1279

This time let us take D1, well D1 which is just Dx is just going to be 2xy2z3.1285

Now let us do D2 of D1, which is... I am going to take the derivative with respect to y of this one, of D1.1296

That is going to equal 4xyz3.1303

Now, I am going to take the derivative with respect to z of the D2 D1 that I just got.1310

I end up with 12xyz2.1317

Wow, what do you know. They are the same. Alright, now let us try another order.1322

Now let us do D1 first, we did D1 already, that is 2xy2z3.1329

Now I am going to do, instead of D2, I will do D3 next. So I will do D3 D1.1338

I am going to take the... derivative with respect to z of D1, so that is going to be 6xy2z2, I think.1344

Now, I will do D2 of the D3 D1, and I get... 12xyz2.1357

Is it the same? Yes. 12xyz2, 2xyz2, 2xyz2. I put the partial derivatives in different orders and yet it ended up being exactly the same.1368

This is very, very, very deep and extraordinary. I am just going to write that. I am just going to write "pretty amazing".1380

Now, let us see what else we can do here.1392

Regarding notation, there is just one caution that I am going to throw out there.1400

So, let us write caution: do not confuse the following.1405

Much of the problem with higher mathematics is notation. Notation, it is just one of those things. We have to have some way of representing what it is that we are actually doing.1418

As things become more complicated... well, things get more complicated, so we just want to make sure we know how we are operating.1427

What kind of math we are actually doing. Do not confuse the following.1436

If you see d/dx2 of f, that is the same as d2fdx2.1441

This is D1(D1) of f. In other words take the derivative with respect to the first variable, then take the derivative again with respect to the same variable.1455

It is not the same as this one. df/dx2, which is D1 of f squared.1470

This one says take the derivative, take the first partial derivative, or the derivative with respect to the first variable and square it as a number, or a variable, or a function.1486

This one notice, it is the d/dx, the differential operator that is squared. 1499

Here, f is actually inside, here f is outside. This one right here, this right here, not the same. This one is also written this way.1504

Remember we said often times we will write the square, D12f, so D12f is not the same as D1f2.1514

This one says differentiate with respect to f, then differentiate again with respect to the same variable.1524

This one says differentiate with respect to f, then square that number or function.1530

Two very, very different things. Another reason why I actually prefer that notation.1533

This one is reasonably clear, this one can be a bit confusing.1538

Okay, let us just do another example here. So, example 3.1543

We will let f(x,y) = sin(x2+y), and we will do... well, let us just run through them.1551

Let us run through all of them. We have D1.1569

The derivative with respect to x. We have cos(x2+y) × 2x, which we will write again... I mean you leave it like this it is not a problem but it is traditional to sort of bring all the functions forward and leave the trigonometric function to the end.1575

So, 2x × cos(x2+y).1599

D2 = cos(x2+y) × 1, which equals... because this is the derivative of the inside, right? chain rule... cos(x2+y).1607

Now, let us do this for the heck of it, let us do D12 which is the same as D1, D1.1625

So, we have D1 already, now we are going to differentiate this with respect to x. So this is going to be... I tend to pull my constants out, so I am going to pull this out, and it is going to be a power rule. x × cos(x2+y).1636

So it is going to be this × the derivative of that, and end up being -x × sin(x2+y) × 2x + that × the derivative of that, which is this × the derivative of this which is just 1. 1652

Again, I have pulled the 2 out. So, it is cos(x2+y), so again, the idea is just be really, really careful and go really slowly.1674

There is no hurry, the idea is to be right, not to finish quickly.1681

When we put this together, we end up with, so the 2, and the 2... this is a × not a minus sign.1687

You end up with -4x2 × sin(x2+y) + cos(x2+y). That is the derivative of the derivative.1696

Derivative with respect to x of the derivative with respect to x. D12, D1 D1. I really love this capital notation, and I love not writing the f's.1710

Now, let us go ahead and do D22.1722

So, D22, which is D2 of D2, when we do that, we end up with -sin(x2+y) × 1.1726

So, equals -sin(x2+y). I hope you guys are checking this.1742

Now let us do D1 D2. When we do D1 D2, we end up with, in other words, we have taken D2, now we are going to take D1 of D2.1750

-sin(x2+y) × 2x is going to be -2xsin(x2+y).1761

Now, we will do D2 of D1, in other words we have done D1 first, now we are going to do D2 of that.1783

We get -2x × sin(x2+y) × 1, which equals -2xsin(x2+y).1789

Well, what do you know, this ends up being the same as that. This mixed partial is the same as that mixed partial.1803

The order is irrelevant, but it is 1,2... 2,1.1810

That is it. So mixed partial derivatives of higher order are the same for functions of several variables provided... so the order does not matter... provided that all of the particular variables with respect to which you are differentiating are there.1815

So 2,3,1... 3,2,1... 1,2,3... 1,3,2... all of those will actually end up being equal, and I personally think that is absolutely extraordinary.1833

Thank you for joining us here at educator.com, we will see you next time for some more Multivariable Calculus. Take care, bye-bye.1842