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Lecture Comments (8)

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 − 5sin(y) + [x/z] = − 5sin(y) + [y/z]

It should be
So Dy = 0 − 5sin(y) + [x/z] = − 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!

Partial Derivatives

Let f(x,y) = 10x2y − 5xy − 2x + 3y + 2. Find Dx
  • The partial derivative in respect to x keeps all other variables and numbers constant.
So Dx = 20xy − 5y − 2 + 0 + 0 = 20xy − 5y − 2.
Let f(x,y) = 10x2y − 5xy − 2x + 3y + 2.
Find Dy
  • The partial derivative in respect to y keeps all other variables and numbers constant.
So Dy = 10x2 − 5x − 0 + 3 + 0 = 10x2 − 5x + 3.
Let f(x,y,z) = 2ex + 5cos(y) + [xy/z]
Find Dx
  • The partial derivative in respect to x keeps all other variables and numbers constant.
So Dx = 2ex + 0 + [y/z] = 2ex + [y/z]
Let f(x,y,z) = 2ex + 5cos(y) + [xy/z]
Find Dy
  • The partial derivative in respect to y keeps all other variables and numbers constant.
So Dy = 0 − 5sin(y) + [x/z] = − 5sin(y) + [y/z]
Let f(x,y,z) = 2ex + 5cos(y) + [xy/z]
Find Dz
  • The partial derivative in respect to z keeps all other variables and numbers constant.
So Dz = 0 + 0 − [xy/(z2)] = − [xy/(z2)]. Note that [xy/z] = xyz − 1.
Let g(x,y) = [2sin(xy)/(4 + y2)] i) Evaluate Dx( 0,[p/2] )
  • We first take the partial derivative in respect to x, so Dx = [2ycos(xy)/(4 + y2)]
Then Dx( 0,[p/2] ) = [(2( [p/2] )cos( 0( [p/2] ) ))/(4 + ( [p/2] )2)] = [p(1)/(4 + [(p2)/4])] = [4p/(16 + p2)]
Let g(x,y) = [2sin(xy)/(4 + y2)] ii) Evaluate Dy( [p/2],0 )
  • We first take the partial derivative in respect to y, so Dy = [(2xcos(xy)(4 + y2) − 4ysin(xy))/(( 4 + y2 )2)]. Use the quotient rule.
Then Dy( [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)/(42)] = [p/4]
Let h(x,y,z) = exyz − ez Evaluate Dz( 1,1, − 1 )
  • We first take the partial derivative in respect to z, so Dz = xyexyz − ez
Then Dz( 1,1, − 1 ) = (1)(1)e(1)(1)( − 1) − e( − 1) = e − 1 − e − 1 = 0
Find ∇f of f(x,y,z) = − 4xy + cos(z).
  • The gradient of a function, denoted by ∇f, is the vector consisting of each partial derivative per component. That is ∇f = (Dx,Dy,Dz).
Now, Dx = − 4y, Dy = − 4x and Dz = − sin(z). Hence ∇f = ( − 4y, − 4x, − sin(z)).
Find ∇( − f) of f(u,v,w,x,y,z) = uv − xy + z2.
  • The gradient of a function, denoted by ∇f, is the vector consisting of each partial derivative per component. That is ∇f = (Du,Dv,Dw,Dx,Dy,Dz).
  • Now, Du = v, Dv = u, Dw = 0, Dx = − y, Dy = − x, and Dz = 2z.
Note that ∇( − f) = − ∇f, so that ∇( − f) = − (v,u,0, − y, − x,2z) = ( − v, − u,0,y,x, − 2z).
Find ∇( f + g ) given that f(x,y) = 2x2 + y and g(x,y) = xy2 − 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 = (y2 − 3,2xy).
Hence ∇( f + g ) = (4x,1) + (y2 − 3,2xy) = (y2 + 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

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 x2 + y2, 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 x3, y2, so nice basic function of 2 variables, x3, y2.0071

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 dx means you are taking the derivative of the function with respect to x.0152

You will also see it as d1. 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 da 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 d1f, or like f(x,y) if you specifically want to mention what the variables are.0178

Again, you can do something like dxf(x,y). You will also see something like this: fx or f1, 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 dx, how is that?0265

So the derivative with respect to x, that means I hold this constant so it just shows up as y2 and I just differentiate this.0274

It becomes 3x2, and since this is a constant this y2 stays. So 3x2y2, that is the derivative, partial derivative with respect to x.0280

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

Well, dx, now we are going to do the partial derivative with respect to y. Now we hold x constant, so this x3 term stays exactly as it is. 0297

We differentiate with respect to y, so this is 2y, so it becomes 2yx3. If you want you can write 2x3y, x32y, 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(x2y).0352

Well, let us see what we can do. This time we will do, I will write it as d1, so d1 means with respect to the first variable, which happens to be x. That is going to equal.. so the derivative of cos(x2y) is - sin(x2y), 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(x2y).0400

Now we will do d2, which is the same as the derivative with respect to y.0410

Now, again we are going to have -sin(x2y) and this time we are going to differentiate with respect to y which is x2 -- 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 -x2sin(x2y). 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 d1 and d2 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, d1 of (pi,1) is equal to -2 × pi × 1 × sin(pi2-1).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

d2 of (pi,1) is going to equal - pi2 × sin(pi2), 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) = x3 × sin(yz2).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 d3, let us not do all of the d's, let us just do d3.0632

So, let us go, which is going to end up being d(z), so d3 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 x3, and since z is in here it is going to be the cos(y × z2) × derivative with respect to z, which is 2yz.0668

There you go. That final answer is going to be 2x3yz × cos(yz2).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/dx1, df/2, so on... df/dxn.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) = 3x2y... 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 7xy3, 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 = 7xy3.0917

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 7y3, 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 21xy2. 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 xy2, 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