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

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

### Partial Derivatives

^{2}y − 5xy − 2x + 3y + 2. Find D

_{x}

- The partial derivative in respect to x keeps all other variables and numbers constant.

_{x}= 20xy − 5y − 2 + 0 + 0 = 20xy − 5y − 2.

^{2}y − 5xy − 2x + 3y + 2.

Find D

_{y}

- The partial derivative in respect to y keeps all other variables and numbers constant.

_{y}= 10x

^{2}− 5x − 0 + 3 + 0 = 10x

^{2}− 5x + 3.

^{x}+ 5cos(y) + [xy/z]

Find D

_{x}

- The partial derivative in respect to x keeps all other variables and numbers constant.

_{x}= 2e

^{x}+ 0 + [y/z] = 2e

^{x}+ [y/z]

^{x}+ 5cos(y) + [xy/z]

Find D

_{y}

- The partial derivative in respect to y keeps all other variables and numbers constant.

_{y}= 0 − 5sin(y) + [x/z] = − 5sin(y) + [y/z]

^{x}+ 5cos(y) + [xy/z]

Find D

_{z}

- The partial derivative in respect to z keeps all other variables and numbers constant.

_{z}= 0 + 0 − [xy/(z

^{2})] = − [xy/(z

^{2})]. Note that [xy/z] = xyz

^{ − 1}.

^{2})] i) Evaluate D

_{x}( 0,[p/2] )

- We first take the partial derivative in respect to x, so D
_{x}= [2ycos(xy)/(4 + y^{2})]

_{x}( 0,[p/2] ) = [(2( [p/2] )cos( 0( [p/2] ) ))/(4 + ( [p/2] )

^{2})] = [p(1)/(4 + [(p

^{2})/4])] = [4p/(16 + p

^{2})]

^{2})] ii) Evaluate D

_{y}( [p/2],0 )

- We first take the partial derivative in respect to y, so D
_{y}= [(2xcos(xy)(4 + y^{2}) − 4ysin(xy))/(( 4 + y^{2})^{2})]. Use the quotient rule.

_{y}( [p/2],0 ) = [(2( [p/2] )cos( [p/2](0) )(4 + (0)

^{2}) − 4(0)sin( [p/2](0) ))/(( 4 + (0)

^{2})

^{2})] = [(p(1)(4) − 0)/(4

^{2})] = [p/4]

^{xyz}− e

^{z}Evaluate D

_{z}( 1,1, − 1 )

- We first take the partial derivative in respect to z, so D
_{z}= xye^{xyz}− e^{z}

_{z}( 1,1, − 1 ) = (1)(1)e

^{(1)(1)( − 1)}− e

^{( − 1)}= e

^{ − 1}− e

^{ − 1}= 0

- The gradient of a function, denoted by ∇f, is the vector consisting of each partial derivative per component. That is ∇f = (D
_{x},D_{y},D_{z}).

_{x}= − 4y, D

_{y}= − 4x and D

_{z}= − sin(z). Hence ∇f = ( − 4y, − 4x, − sin(z)).

^{2}.

- The gradient of a function, denoted by ∇f, is the vector consisting of each partial derivative per component. That is ∇f = (D
_{u},D_{v},D_{w},D_{x},D_{y},D_{z}). - Now, D
_{u}= v, D_{v}= u, D_{w}= 0, D_{x}= − y, D_{y}= − x, and D_{z}= 2z.

^{2}+ y and g(x,y) = xy

^{2}− 3x.

- Note that ∇( f + g ) = ∇f + ∇g, it suffices to find each individual gradient of f and g and add them.
- Now ∇f = (4x,1) and ∇g = (y
^{2}− 3,2xy).

^{2}− 3,2xy) = (y

^{2}+ 4x − 3,2xy + 1). Note that this is vector addition.

*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

### 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.

- Intro 0:00
- Partial Derivatives 0:19
- Example 1
- Example 2
- Example 3
- Example 4
- Definition 1
- Example 5
- Example 6
- Notation and Properties for Gradient

### Multivariable Calculus

### Transcription: Partial Derivatives

*Hello and welcome back to educator.com and welcome back to multivariable calculus. *0000

*In the last lesson we introduced the notion of a function of several variables.*0005

*In today's lesson what we are going to do is we are going to actually talk about doing Calculus on those functions, the idea of a derivative, so let us just jump right on in.*0009

*So, when differentiating a function of several variables, really it is the exact same thing as a function of single variable, except what you do is you hold every other variable, you just treat it like a constant.*0018

*You do exactly what you have always done before. Nothing is different. *0033

*So if you have x ^{2} + y^{2}, let us pretend that we are just taking the derivative with respect to x, and we will get to the notation in just a minute.*0037

*You just hold y, you just treat it like a constant, and you just differentiate with respect to x.*0045

*Then if you want to take the derivative of y, now this time you hold the x constant and you differentiate with respect to y.*0050

*Let us just do some examples, and I think it is the best way to make sense of it.*0058

*So, we will start with a 2-variable example. Example 1.*0062

*We will let f(x,y) equal x ^{3}, y^{2}, so nice basic function of 2 variables, x^{3}, y^{2}.*0071

*Before I go ahead and actually do the derivatives, I just want to introduce the notation that you are going to see. *0085

*You are used to seeing... you know... dx, or you are used to seeing f'(x)... those are the standard notations for single variable calculus. *0091

*In multi-variable calculus you are going to see lots of different notations. There are some that are standard.*0100

*I am going to introduce several of them, and as it turns out, we are actually going to be using different variations of it.*0106

*We want to make sure that you actually recognize the different notations that are used, because again we are dealing with more variables now, so the notation becomes a little bit more cumbersome, a little bit more complicated.*0116

*But we try to keep it as close to this as possible.*0125

*If we want to differentiate with respect to x, so we might see something like this, we will see df/dx, something that looks kind of like a d, but is a little more curvy.*0130

*We also use, we tend to use capital letters for the derivative of functions of several variables. *0145

*So d _{x} means you are taking the derivative of the function with respect to x.*0152

*You will also see it as d _{1}. Let us say that you have a function of 19 variables, well you cannot.. x,y,z,... you cannot just come up with a bunch of different, so we just say the first variable, the second variable, the third variable, the fourth variable...*0157

*Things like that, so d _{a} means I am differentiating the function with respect to the first variable. Everything else constant.*0169

*These are the shorter notations that we use. If you want to be a little bit more explicit, you can do something like d _{1}f, or like f(x,y) if you specifically want to mention what the variables are.*0178

*Again, you can do something like d _{x}f(x,y). You will also see something like this: f_{x} or f_{1}, something like that.*0196

*These are all different notations that are used to notate the partial derivative of a function of several variables with respect to whichever variable you are talking about.*0209

*Now, quick thing, remember when we did something like dy/dx in single variable calculus? We can actually treat this like a fraction.*0220

*In other words, we can move things around, move the denominator around. This notation which is analogous to this notation, this is not a fraction.*0228

*Symbolically we can sometimes think about it, especially later when we talk about the chain rule, but it is very, very important to remember this is not the same as that.*0240

*This is just a notation talking about the derivative of the function with respect to the variable x, but you cannot do the same thing with it that you do with this.*0249

*It is not going to be a problem though. Let us go ahead and jump right on in and see what we can do.*0259

*I am going to go ahead and say, I am going to go ahead and differentiate this with respect to x, so I will do d _{x}, how is that?*0265

*So the derivative with respect to x, that means I hold this constant so it just shows up as y ^{2} and I just differentiate this.*0274

*It becomes 3x ^{2}, and since this is a constant this y^{2} stays. So 3x^{2}y^{2}, that is the derivative, partial derivative with respect to x.*0280

*Again, we say partial because we are holding everything else constant.*0292

*Well, d _{x}, now we are going to do the partial derivative with respect to y. Now we hold x constant, so this x^{3} term stays exactly as it is. *0297

*We differentiate with respect to y, so this is 2y, so it becomes 2yx ^{3}. If you want you can write 2x^{3}y, x^{3}2y, it does not matter, multiplication is commutative.*0307

*There are certain standards -- they generally tend to put the x before the y, but we are not concerned with cosmetics, we are concerned with the mathematics.*0324

*Let us do another example, there is nothing here that you have not done before. Just treat the other variable as a constant, that is it. *0332

*You just have to be sort of careful when you do this, because again the functions are getting a little bit more complicated, so you just have to be careful with what is what.*0339

*It is perfectly common to make errors, it is not a big deal, you just check it carefully.*0347

*So example 2, let us go ahead and let f(x,y) equal to... this time let us do a trigonometric function, cos(x ^{2}y).*0352

*Well, let us see what we can do. This time we will do, I will write it as d _{1}, so d_{1} means with respect to the first variable, which happens to be x. That is going to equal.. so the derivative of cos(x^{2}y) is - sin(x^{2}y), and of course this is a chain rule because we have a function of x,y, in that thing.*0363

*So now we take the derivative of what is inside with respect to x, so that is going to be × 2xy.*0387

*So the derivative of this, we hold the y constant, we differentiate, we get 2xy.*0394

*We are going to write that as -2xy × sin(x ^{2}y).*0400

*Now we will do d _{2}, which is the same as the derivative with respect to y.*0410

*Now, again we are going to have -sin(x ^{2}y) and this time we are going to differentiate with respect to y which is x^{2} -- differentiating with respect to that, we are going to hold this as constant, so some constant × y, the derivative is just the constant.*0415

*Therefore, you end up with -x ^{2}sin(x^{2}y). That is it. Nothing going on, nothing strange.*0437

*Now let us do one more example where we actually will evaluate a derivative at a certain point. *0449

*You remember the derivative gives you a function. Well, that function, if you actually have a specific point, a specific point (x,y) that you put in there, it gives you a number. Same as single-variable calculus.*0455

*So, let us see, we have got... okay... example 3.*0470

*This time, we want to evaluate d _{1} and d_{2} from the previous example, so we want to evaluate this at the point (x,y) = pi,1. *0478

*So, we want to know what the derivative is at the point (pi,1).*0504

*Well, d _{1} of (pi,1) is equal to -2 × pi × 1 × sin(pi^{2}-1).*0510

*When you just evaluate it, put all the numbers in, the number you get is 2.7037, so you see that it is exactly like it is in single-variable calculus. *0529

*d _{2} of (pi,1) is going to equal - pi^{2} × sin(pi^{2}), and you end up with 4.2469.*0540

*That is it. Let us go ahead and do a function of 3 variables.*0555

*We have example 4, now we will let f(x) = x ^{3} × sin(yz^{2}).*0565

*Now, notice the notation that I used, slightly different notation. Up before in the previous examples, I specifically wrote out f(x,y).*0583

*In this case, in the argument for the function, I put a vector, a vector x.*0592

*Again, a vector is just a point, it is a shorthand notation for the point, in this case, (x,y,z).*0598

*I could have written f(x,y,z), but I want to be using notation that you will be seeing. Various different types of notation. We need to recognize them.*0604

*Single-variable calculus is reasonably uniform in its notation. Different books, different fields, physics, engineering, mathematics, chemistry, they tend to use a whole bunch of different notations and it is good to be familiar with them. *0612

*We do not want to just stick with one notation, we want to be able to recognize others.*0626

*Now, let us go ahead and do, let us just do like d _{3}, let us not do all of the d's, let us just do d_{3}.*0632

*So, let us go, which is going to end up being d(z), so d _{3} which is equal to d(z), which is equal to df/dz, all of those notations.*0643

*That is going to equal, well, let us see what we have got. With respect to z, so we are differentiating with respect to z which means that we are holding everything else constant.*0657

*So this is going to be x ^{3}, and since z is in here it is going to be the cos(y × z^{2}) × derivative with respect to z, which is 2yz.*0668

*There you go. That final answer is going to be 2x ^{3}yz × cos(yz^{2}).*0689

*That is it. You just have to be really, really careful when you are doing these partial derivatives to make sure that you are holding the proper thing constant, make sure that you are taking care of the chain rule... that is it, that is all that is going on here.*0700

*So, now we have come to a point, I am going to introduce a definition, something called a gradient of a function of several variables, the gradient vector. *0715

*This is a profoundly important definition. if you do not take anything else away from this course, by all means you want to be able to understand both algebraically, and later geometrically, what the gradient means. *0723

*It comes up all the time and it is a profound importance.*0735

*So, definition... in fact, let us go ahead and change the color of our ink here. I am going to go ahead and do this one in blue.*0740

*So, definition, let f be a function from RN to R, so a function of several variables.*0751

*Then, the gradient of f, notated grad(f) -- we will give another notation of it, one that is a little more common, later on in this lesson -- notated grad(f) is the vector, very important, the gradient is a vector, not a number... oops, getting all kinds of crazy lines here.*0764

*The vector... grad(f) is equal to df/dx _{1}, df/_{2}, so on... df/dx_{n}.*0805

*Let me tell you what this means. If I have a function of let us say 5 variables, the gradient is the vector where the components of that vectors, the first component is the derivative of f, the partial derivative of f with respect to the first variable.*0828

*The second component is the partial derivative with respect to the second variable, and so on.*0845

*The third, the 4th, the 5th component is the partial derivative with respect to the 5th variable, very important.*0851

*The derivative is not a number, it is a vector. It is called the gradient vector. Let us go ahead and just do an example. I think that is the best way to make sense of this.*0859

*I am going to go back to black here. Let us go, example 5.*0870

*Let f(x) = 3x ^{2}y... no, wait, what happened here, function of several variables, nope, what a minute, no let us go ahead and do it like this.*0878

*Yeah, that is fine, let us go ahead and do 7xy ^{3}, in fact, you know what, I tend to prefer the other notation personally.*0905

*I am going to go ahead and write f(x,y) explicitly = 7xy ^{3}.*0917

*Now, the gradient of f is equal to, well, we take the partial derivative with respect to x, so when we hold everything else constant and take the derivative we get 7y ^{3}, that is the first component.*0925

*The second component is the partial derivative with respect to y.*0943

*So, we go ahead and differentiate with respect to y holding the x constant. That is going to be 3 × 7 is 21, so it is going to be 21xy ^{2}. That is the gradient vector. *0947

*What you are taking is the function of several variables. You are differentiating with respect to each variable, and the derivatives that you get respectively end up being the components of the gradient vector.*0961

*Let us go ahead and do another one, so example 6, let us do 3 variables now, f(x,y,z) = x × y × sin(yz).*0975

*Be very, very careful here, again look at the function here before you just dive right on in.*0995

*So, let us do df/dx, so df/dx, the derivative with respect to x is going to be... well, notice we said you hold everything else constant, so you hold the y and z constant.*1000

*This sine function, sin(yz), x is not in this argument. Because x is not in this argument, this whole sin(yz) is a constant and y is a constant.*1013

*So, the derivative of this is just plain old y × sin(yz).*1025

*Make sure you do not just automatically start taking the derivative and say x × cos(yz) because there is no x in the argument for the sine function. It is just a constant. *1032

*You treat it like a constant. Let me make my z a little bit clearer here.*1045

*Now, let us do df/dy. Okay, df/dy is going to be... okay, so now you notice we have a y and we have a sin(yz) so this is going to be a product function.*1050

*We are going to have to differentiate the way we differentiate a product, 1 × the derivative of the other + the derivative of that one × the other. That is what it looks like.*1067

*So we are going to take xy, the derivative of this which is cos(yz) × derivative of what is inside there because this is a chain rule, × z.*1078

*Again, you just have to be very, very careful and it is a very common error, it is not a problem, that happens, that is life.*1094

*Plus, now let us differentiate with respect to... it's the other derivative, so it is this × the derivative of that + that × the derivative of this.*1101

*The other things that are constant here, remember this is our function, this is the RY, so it is these 2, so it is going to be + x sin(yz) × the derivative of y which is 1.*1110

*That ends up being xy, oops, let us do xyz, so that is xyz × cos(yz) + x × sin(yz).*1129

*I certainly hope that you are going to double check this for me. *1145

*Now let us do the third partial derivative df/dz, so now we hold the x and y constant, so we end up getting... let us see, what happens... so if we hold the x and y constant, that means these are constant, so we end up with xy cos(yz) × derivative of what is inside with respect to z which is y.*1150

*That will equal xy ^{2}, if I am not mistaken, cos(yz), there you go.*1182

*This is df/dx, this is df/dy, this is df/dz. The gradient vector, the gradient of f, is equal to the first component is df/dx, the second component is df/dy, and the third component is df/dz.*1190

*This, that is the second component, and that is the third component. It is very, very important, the gradient is a vector.*1214

*Now, let us go ahead and finish off by discussing the alternate notation, the more common notation that you are going to see in especially in physics and engineering.*1225

*We will finish off with a couple of properties for the gradient, really simple properties, exactly as they are for first variable calculus.*1239

*So, let us say there is another notation for gradient, which is very common.*1247

*The symbol is this upside down triangle, an upside down delta, f(x,y) or just this, and it is called ∇f.*1276

*It is called the del operator, basically when it says you do del to f, it means you take the gradient of f, that is it.*1292

*It is just an alternate notation instead of writing G R A D I E N T, it is just a... again, it is just a symbol.*1300

*So, let us finish off some properties and they are exactly what you expect.*1308

*The gradient, if I have 2 functions f and g, functions of several variables, if I add the functions then take the gradient, it is the same as taking the gradient and then adding those.*1314

*So, gradient of f + g = gradient of f + gradient of g. That is it.*1326

*The gradient of a constant times a function = constant × the gradient of the function... ∇f, so if you like the del notation, I personally do not like the del notation, but that is just me, so it is ∇f + ∇g = c × ∇f.*1337

*This right here, when you add 2 things, it is the same as taking... so this is again, we are operating, we are operating on this function.*1366

*Operating on the sum of a function is the sum of the operation on that function.*1377

*In this case taking the gradient of a constant × a function, is the same as a constant... this is defined, is actually the definition of linearity.*1382

*For those of you who have done linear algebra, you would recognize this as the algebraic definition of linearity.*1391

*It turns out the gradient operator is a linear operator.*1395

*Okay, so we have introduced the gradient, again a profoundly, profoundly important notion in multi-variable calculus.*1399

*We will go ahead and stop the lesson here, thank you for joining us at educator.com.*1406

*We will see you next time, bye-bye.*1409

0 answers

Post by Arwa Shayaa on October 18, 2015

videos take time just talking without heading to the point directly -__-

0 answers

Post by I M on November 15, 2014

Simply not enough example questions. Disappointing

1 answer

Last reply by: Professor Hovasapian

Fri Aug 22, 2014 8:19 PM

Post by Denny Yang on August 14, 2014

Let f(x,y,z) = 2ex + 5cos(y) + [xy/z]

Find Dy

You wrote the answer as:

So Dy = 0 âˆ’ 5sin(y) + [x/z] = âˆ’ 5sin(y) + [y/z]

It should be

So Dy = 0 âˆ’ 5sin(y) + [x/z] = âˆ’ 5sin(y) + [x/z]

1 answer

Last reply by: Professor Hovasapian

Fri Mar 22, 2013 4:14 PM

Post by Abdelrahman Megahed on March 20, 2013

5.27 you are the best

1 answer

Last reply by: Professor Hovasapian

Tue Feb 5, 2013 12:54 AM

Post by Josh Winfield on February 4, 2013

Example #3; For D-1= -2xysin(x^(2)y) therefore D-1 should be -2Ï€sin(Ï€^2) not -2Ï€sin(Ï€^(2)-1) but your numerical answer of 2.7037 is correct.

I make note of your mistakes only to help others become less confused and to reaffirm what they might be thinking (hmm is that right???). I am loving this course and i think you are a great teacher!