For more information, please see full course syllabus of Multivariable Calculus

For more information, please see full course syllabus of Multivariable Calculus

### Higher and Mixed Partial Derivatives

^{2}+ 5xy − y

^{2}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.

^{2}+ 5xy − y

^{2}Find ( [d/dx] )

^{2}f and ( [d/dy] )

^{2}f

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

^{2}f = [d/dy]( 5x − 2y ) = − 2.

^{2}+ 5xy − y

^{2}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.

^{2}+ 1) − e

^{x}sin(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].

^{2}[d/dy]g = [d/dx][d/dx][d/dy]g = [d/dx][d/dx]( − e

^{x}cos(y) ) = [d/dx]( − e

^{x}cos(y) ) = − e

^{x}cos(y).

^{2}+ 1) − e

^{x}sin(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].

^{2}[d/dx]g = [d/dy][d/dy][d/dx]g = [d/dy][d/dy]( − 2xsin(x

^{2}+ 1) − e

^{x}sin(y) ) = [d/dy]( − e

^{x}cos(y) ) = e

^{x}sin(y).

^{2}+ 1) − e

^{x}sin(y) Verify [d/dy][d/dx]g = [d/dx][d/dy]g

- Note that [d/dy][d/dx]g = − e
^{x}cos(y) and [d/dx][d/dy]g = − e^{x}cos(y) from our previous problems.

- 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^{ − 1}y^{ − 1}z^{ − 1}. - Now [d/dz][d/dx]h = [d/dz]( − [1/(x
^{2}yz)] ) = [1/(x^{2}yz^{2})].

^{2}(1)( − 2)

^{2})] = [1/4].

- 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^{ − 1}y^{ − 1}z^{ − 1}. - Now [d/dz][d/dy]h = [d/dz]( − [1/(xy
^{2}z)] ) = [1/(xy^{2}z^{2})].

^{2}( − 2)

^{2})] = [1/4].

- 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^{ − 1}y^{ − 1}z^{ − 1}. - Now [d/dz][d/dy][d/dx]h = [d/dz][d/dy]( − [1/(x
^{2}yz)] ) = [d/dz]( [1/(x^{2}y^{2}z)] ) = − [1/(x^{2}y^{2}z^{2})].

^{2}(1)

^{2}( − 2)

^{2})] = − [1/4].

- 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^{ − 1}y^{ − 1}z^{ − 1}. - Now [d/dx][d/dy][d/dz]h = [d/dx][d/dy]( − [1/(xyz
^{2})] ) = [d/dx]( [1/(xy^{2}z^{2})] ) = − [1/(x^{2}y^{2}z^{2})].

^{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.

### Multivariable Calculus

### 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 x _{1}, x_{1}, 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 d _{1}, so d_{1} 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 d _{x}. Well, df/dy the partial derivative with respect to y, that is one notation, and we often write it as d_{2}, 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 d _{y}, 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 d ^{2}f/dx^{2} completely analogous to single variable calculus.*0516

*The capital notation is something like this, just D _{1}/D_{1}f, 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, d ^{2}f. 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 d ^{2}f, dy^{2}.*0553

*Again, you have got d _{2}/d_{2}f and this is often written -- sometimes -- squared f.*0565

*I personally do not like this notation myself, this d _{1}^{2}f, d_{2}^{2}f.*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, d _{1}d_{1}, 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 d ^{2}f 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 D _{1} to it, then do D_{2} on it. That is first, and this is second, order is very important here.*0688

*Now, of course we have D _{1}, D_{2} of f, which means we have done D_{2} first, then done D_{1}, and this is going to be d/dx of df/dy = d^{2}f 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) = x ^{3}y^{2}, so d_{1} the derivative with respect to x is equal 3x^{2}y^{2}, in other words I am treating y as a constant and just differentiating with respect to x.*0751

*D _{2} = x^{3}y, this time I ended up holding x constant and differentiating with respect to the second variable, y.*0772

*Now, I am going to do d _{2} of d_{1}, in other words I have done d_{1}, 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 6x ^{2}y.*0801

*Now, I am going to do D _{1} of D_{2}, which I just did. I just did D_{2} that is x^{3}y, then do D_{1} of that, which means I am going to differentiate this with respect to x.*0809

*I get 6x ^{2}y. We will stop and take a look at this.*0823

*D _{2}D_{1}, I did D_{1} first then I did D_{2}, I got 6x^{2}y.*0828

*here I did D _{2} first, then I got D_{1}, then I did 6x_{2}y 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 D _{1} D_{2}, D_{2} D_{1}, 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, D _{1} D_{2} of f = D_{2} D_{1} of f.*0935

*Notice up here just real quickly that I just wrote, D _{1} D_{2}, D_{2} D_{1}, 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 D _{1} is the first partial derivative, you know that D_{2} 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 D _{1}, we will have D_{2}, and we will have D_{3}, 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 D _{1}, D_{2}, D_{3}, I can do D_{3}, D_{2}, D_{1}... D_{2}, D_{1}, D_{3}... D_{2}, D_{3}, D_{1}... 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 D _{1}, D_{2}, D_{3} of f.*1144

*I have taken D _{3} first, then I have done D_{2} to that, then I have taken D_{1} of that.*1153

*Well, if I do it in different order, do... let us say, D _{3}, D_{2}, D_{1} of f... if I do D_{2} first, then D_{2}, then D_{3}... or if I do D_{2}, D_{3}, D_{1}... again we are working from right to left, in other words I do D_{1} 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) = x ^{2}y^{2}z^{3}.*1219

*Now, we are just going to do a bunch of partials. Let us do, let us start with D _{3}, in other words z.*1230

*So, D _{3} is going to equal 3x^{2}y^{2}z^{2}.*1240

*Now we will do D _{2} of D_{3}, in other words we are going to differentiate with respect to y, this thing up here. D_{2} of D_{3}.*1247

*That is going to equal 6x ^{2}yz^{2}, and now we are going to do D_{1} of D_{2},D_{3}. In other words we are going to differentiate with respect to x, this thing right here.*1257

*We end up getting 12xyz ^{2}.*1272

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

*This time let us take D _{1}, well D_{1} which is just D_{x} is just going to be 2xy^{2}z^{3}.*1285

*Now let us do D _{2} of D_{1}, which is... I am going to take the derivative with respect to y of this one, of D_{1}.*1296

*That is going to equal 4xyz ^{3}.*1303

*Now, I am going to take the derivative with respect to z of the D _{2} D_{1} that I just got.*1310

*I end up with 12xyz ^{2}.*1317

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

*Now let us do D _{1} first, we did D_{1} already, that is 2xy^{2}z^{3}.*1329

*Now I am going to do, instead of D _{2}, I will do D_{3} next. So I will do D_{3} D_{1}.*1338

*I am going to take the... derivative with respect to z of D _{1}, so that is going to be 6xy^{2}z^{2}, I think.*1344

*Now, I will do D _{2} of the D_{3} D_{1}, and I get... 12xyz^{2}.*1357

*Is it the same? Yes. 12xyz ^{2}, 2xyz^{2}, 2xyz^{2}. 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/dx ^{2} of f, that is the same as d^{2}fdx^{2}.*1441

*This is D _{1}(D_{1}) 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/dx ^{2}, which is D_{1} 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, D _{1}^{2}f, so D_{1}^{2}f is not the same as D_{1}f^{2}.*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(x ^{2}+y), and we will do... well, let us just run through them.*1551

*Let us run through all of them. We have D _{1}.*1569

*The derivative with respect to x. We have cos(x ^{2}+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(x ^{2}+y).*1599

*D _{2} = cos(x^{2}+y) × 1, which equals... because this is the derivative of the inside, right? chain rule... cos(x^{2}+y).*1607

*Now, let us do this for the heck of it, let us do D _{1}^{2} which is the same as D_{1}, D_{1}.*1625

*So, we have D _{1} 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(x^{2}+y).*1636

*So it is going to be this × the derivative of that, and end up being -x × sin(x ^{2}+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(x ^{2}+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 -4x ^{2} × sin(x^{2}+y) + cos(x^{2}+y). That is the derivative of the derivative.*1696

*Derivative with respect to x of the derivative with respect to x. D _{1}^{2}, D_{1} D_{1}. I really love this capital notation, and I love not writing the f's.*1710

*Now, let us go ahead and do D _{2}^{2}.*1722

*So, D _{2}^{2}, which is D_{2} of D_{2}, when we do that, we end up with -sin(x^{2}+y) × 1.*1726

*So, equals -sin(x ^{2}+y). I hope you guys are checking this.*1742

*Now let us do D _{1} D_{2}. When we do D_{1} D_{2}, we end up with, in other words, we have taken D_{2}, now we are going to take D_{1} of D_{2}.*1750

*-sin(x ^{2}+y) × 2x is going to be -2xsin(x^{2}+y).*1761

*Now, we will do D _{2} of D_{1}, in other words we have done D_{1} first, now we are going to do D_{2} of that.*1783

*We get -2x × sin(x ^{2}+y) × 1, which equals -2xsin(x^{2}+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

1 answer

Last reply by: Professor Hovasapian

Tue Sep 8, 2015 8:58 PM

Post by Jim Tang on September 7 at 04:45:11 PM

Wait I'm confused. The partial derivative of x^2+y with respect to x is 2x and not 2x+y? I think in some other lecture you wrote 2x+y but I might be mistaken.

1 answer

Last reply by: Professor Hovasapian

Tue Jul 28, 2015 11:27 PM

Post by Jonathan Snow on July 25 at 08:27:10 PM

Hey,

Thanks for the great lecture, I was wondering why you said "power rule" at around 27:30, did you mean product rule?

1 answer

Last reply by: Professor Hovasapian

Tue Nov 25, 2014 2:25 AM

Post by Utomo Pratama on November 24, 2014

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, 2014

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