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Lecture Comments (5)

3 answers

Last reply by: Professor Hovasapian
Wed Mar 13, 2013 5:34 AM

Post by Yujin Jung on March 11, 2013

Hello!

Do you by any chance teach the 'Jacobian matrix' of a function?
Our lecturer referred to it as he was teaching the chain rule and was wondering how they relate to each other..

Thank you :)

0 answers

Post by Caleb Lear on October 11, 2012

I like your method for using the chain rule, I wonder if there's a way to adapt it for implicit differentiation? Right now I'm just deriving as before and multiplying by a partial where I need to.

The Chain Rule

Let C(t) = ( t,[1/2]t ). Find the composition F(C(t)) given F(x,y) = 2xy.
  • The composition F(C(t)) is formed by taking the components of C and subsituting them for x and y.
Then with x = t and y = [1/2]t, we have F(C(t)) = 2(t)( [1/2]t ) = t2.
Let C(t) = ( sin(t),cos(t) ). Find the composition F(C(t)) given F(x,y) = x2 + y2.
  • The composition F(C(t)) is formed by taking the components of C and subsituting them for x and y.
Then with x = sin(t) and y = cos(t), we have F(C(t)) = (sin(t))2 + (cos(t))2 = 1.
Let C(t) = ( et,t,[1/t] ). Find the composition F(C(t)) given F(x,y,z) = 2x2 − xyz + y2.
  • The composition F(C(t)) is formed by taking the components of C and subsituting them for x, y and z.
Then with x = et, y = t and y = [1/t], we have F(C(t)) = 2( et )2 − (et)(t)( [1/t] ) + (t)2 = 2e2t − et + t2.
Find the derivative of F(C(t)) = 5t2 + e3t − 1.
  • Since our function is of one variable, we find F′(C(t)) by taking the derivative in respect to t.
Thus F′(C(t)) = 10t + 3e3t.
Let F(x,y) = cos(xy) and C(t) = ( √t , − √t ).
Evaluate F′(C(p)).
  • First we find F(C(t)), let x = √t and y = ( √t , − √t ).
  • Then F(C(t)) = cos( √t ( − √t ) ) = cos( − t) = cos(t).
So F′(C(t)) = − sin(t) and F′(C(p)) = − sin(p) = 0.
Let F(x,y,z) = x2y + yz and C(t) = ( t,cos(t),etsin(t) ).
Evaluate F′(C(t))
  • Since our functions are complex, we use F′(C(t)) = ∇F(C(t)) ×C′(t) to find the derivative.
  • Computing ∇F(x,y,z) yields ∇F(x,y,z) = (2xy,x2 + z, − y), so ∇F(C(t)) = (2tcos(t),t2 + etsin(t), − cos(t)).
  • Computing C′(t) yields C′(t) = (1, − sin(t),etcos(t) + etsin(t)). Note that to compute [ etsin(t) ]′ we apply the product rule.
  • Thus F′(C(t)) = ∇F(C(t)) ×C′(t) = (2tcos(t),t2 + etsin(t), − cos(t)) ×(1, − sin(t),etcos(t) + etsin(t)) = 2tcos(t) − t2sin(t) − etsin2(t) − etcos2(t) − etcos(t)sin(t).
Simplifying results in F′(C(t)) = 2tcos(t) + t2sin(t) − etcos(t)sin(t) − et.
Let F(x,y) = x2ln(2x + y) and C(t) = ( sin(t),cos(t) ).
Evaluate F′(C( [p/2] ))
  • First we find F′(C(t)), using F′(C(t)) = ∇F(C(t)) ×C′(t) to find the derivative.
  • Computing ∇F(x,y) yields ∇F(x,y) = ( [(2x2)/(2x + y)] + 2xln(2x + y),[(x2)/(2x + y)] ), so ∇F(C(t)) = ( [(2sin2(t))/(2sin(t) + cos(t))] + 2sin(t)ln(2sin(t) + cos(t)),[(sin2(t))/(2sin(t) + cos(t))] ). Note that to compute [ x2ln(2x + y) ]′ we use the product rule.
  • Computing C′(t) yields C′(t) = (cos(t), − sin(t)).
  • Thus F′(C(t)) = ∇F(C(t)) ×C′(t) = ( [(2sin2(t))/(2sin(t) + cos(t))] + 2sin(t)ln(2sin(t) + cos(t)),[(sin2(t))/(2sin(t) + cos(t))] ) ×(cos(t), − sin(t)).
  • We can simplify our scalar product by inputing t = [p/2], so F′(C( [p/2] )) = ( [(2sin2( [p/2] ))/(2sin( [p/2] ) + cos( [p/2] ))] + 2sin( [p/2] )ln(2sin( [p/2] ) + cos( [p/2] )),[(sin2( [p/2] ))/(2sin( [p/2] ) + cos( [p/2] ))] ) ×(cos( [p/2] ), − sin( [p/2] )) = ( 1 + 2ln(2),[1/2] ) ×(0, − 1).
Thus F′(C( [p/2] )) = − [1/2].
Let F(x,y,z) = zcos(x)sin(y) and C(t) = ( pt2,√t ,t ). Evaluate F′(C(t))
  • We let x = pt2, y = √t and z = t and find F(C(t)).
  • So F(C(t)) = tcos(pt2)sin( √t ). To find F′(C(t)) we apply the product rule twice.
  • So F′(C(t)) = [ tcos(pt2) ]( − cos( √t )( [1/(2√t )] ) ) + [ t( sin(pt2)(2pt) ) + cos(pt2) ]sin( √t ).
Simplifying yields F′(C(t)) = [( − √t cos(pt2)cos( √t ))/2] + 2pt2sin(pt2)sin( √t ) + cos(pt2)sin( √t ).
Let F(x,y) = √{x2 + y2} and C(t) = ( e2t,e − 2t ). Evaluate F′(C(0))
  • We let x = e2t, y = e − 2t and find F(C(t)).
  • So F(C(t)) = √{( e2t )2 + ( e − 2t )2} = √{e4t + e − 4t} . We can now compute F′(C(t)). Note that √{e4t + e − 4t} = ( e4t + e − 4t )[1/2].
  • Thus F′(C(t)) = [1/2]( e4t + e − 4t ) − [1/2]( 4e4t − 4e − 4t ).
Hence F′(C(0)) = [1/2]( e4(0) + e − 4(0) ) − [1/2]( 4e4(0) − 4e − 4(0) ) = 0.
Let F(x,y) = √{x2 + 1} + √{y2 − 1} and C(t) = (t,t).Evaluate F′(C(2))
  • We let x = t, y = t and find F(C(t)).
  • So F(C(t)) = √{t2 + 1} + √{t2 − 1} . We can now compute F′(C(t)). Note that √{t2 + 1} + √{t2 − 1} = ( t2 + 1 )[1/2] + ( t2 − 1 )[1/2]
  • Thus F′(C(t)) = [2t/2]( t2 + 1 ) − [1/2] + [2t/2]( t2 − 1 ) − [1/2] = [t/(√{t2 + 1} )] + [t/(√{t2 − 1} )].
Hence F′(C(2)) = [2/(√{22 + 1} )] + [2/(√{22 − 1} )] = [2/(√5 )] + [2/(√3 )].

*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

The Chain Rule

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
  • The Chain Rule 0:45
    • Conceptual Example
    • Example 1
    • The Chain Rule
    • Example 2: Part 1
    • Example 2: Part 2 - Solving Directly

Transcription: The Chain Rule

Hello and welcome back to educator.com and multi-variable calculus.0000

Today we are going to talk about the chain rule, and you remember from single-variable calculus the chain rule just allowed you to differentiate functions that were composite functions.0005

Composite functions where something like sin(x3), something like that, you took the derivative of the sine, then you took the derivative of what was inside the argument, the x3, that became 3x2.0013

Now that we are dealing with these vector value functions, these functions of several variables and we have introduced the gradient, we can actually bring those tools to bare on differentiating a function, a composite function that involves functions of several variables.0027

So let us just jump right on in, and let me give a quick description of what it is that is going on. 0045

Again, what we want to do is not just jump right into the mathematics, we do not just want to write symbols on a page, we want to be able to understand what is happening. 0052

When there is understanding, when you can see what is going on, you can use the intuition that you have already developed to decide what goes next and where to go. That is the whole idea.0060

We want you to understand what is happening mathematically before you actually do the mathematics.0071

That is the easy part, the mathematics will come, that is just symbolic manipulation but we want to see what is going on.0076

Let us start in R2, let us start in the plane, again we are using our geometric intuition to help us guide our mathematics.0082

So, let us say we have something like this. Now let us say we have some region in... you know, here... so now, let us suppose the following things.0090

So suppose f is a function from R2 to R, so a function of 2 variables, defined on the open set u. This is u by the way. Open set u.0104

Now, also suppose that C, which is a function from R to R2 -- we are dealing with R2, okay -- R to R2, is a curve in 2-space, that passes through u.0125

Let us say this is just something like that. So passes through u. Now, here is what is great.0163

The points along the curve, the function itself is defined on this open set.0172

In other words the points in this set can be used for this function f.0177

Well, the curve happens to pass through u, so the point on the curve can be used in f.0182

That is what we are doing, so what we can do is we can actually form the composite function, what we have is some curve that has nothing to do with u and yet it happens to pass through u.0191

We also have a function that is defined on u because there is this overlap we can use the... in the function f, we can use the points along the curve.0200

That is what is really, really great here.0211

Let us go ahead and write this down, and then the points along c(t) can be used by the function f.0215

In other words, we can form the composite function of c, which is f(c).0240

Let me write it with the actual t... oh this is a capital F by the way, sorry about that... small f's big F's, hm.0270

F(c(t)), that is it, it is a composite function. Except now, instead of a composite function of single variables, it is a composite function of multi-variables.0278

In this case, we are dealing with 2. Let us just do a quick example, and then we will discuss it just a little bit more. Just to make sure we understand this concept.0289

This is a profoundly important concept, so like the gradient we definitely want to have a good sense of what is going on here.0298

We will take the time to make sure that that is the case.0307

Let us just do a quick example. Just so that we can see... so example 1.0311

Now, let c(t), let the curve equal the following: let go t3 then t2, so t3 and t2.0319

Now we will let our f, capital F of (x,y), of two variables, let us call it ln(y) and let us do cos(xy). How is that?0332

Just a... you know... nice random function. Now, f(c(t)). Notice. F is a function of two variables x and y, so its argument contains 2 things. 0345

c(t) has two things, for x we put in the t3, for y we put in the t2, into here and we form the function F as a function of t.0366

Watch what happens. F(c(t)) = f(t3,t2), right?0380

f(x,y) that just means these 2 things, so wherever I see a y I put in whatever is in here, wherever I see an x I put in what is in here.0392

Well F(c(t)) is this, it actually spits out 2 values, those 2 values, this is x, this is y.0400

We end up getting the logarithm of t2 × cos... natural logarithm × the cosine(t3t2, which equals ln(t2) × cos(t5). 0409

That is it, that is all I have done here. I have formed this composite function now, with 2 different functions that mapped to different spaces, but the space where one maps to is exactly the space that the next function needs as its domain in order to take the next step.0432

Let me actually write down this whole idea of the functions again, so C(t) is a map from R to R2.0450

In other words, it takes numbers, real numbers, and it spits out 2 vectors, a point in 2 space.0466

F(x,y) it takes a vector, a point in 2-space, and it spits out a number.0474

So what I have really done here, the composite function actually ends up being a map from R to R. That is what is happening here.0487

That is what is important to see. You can jump around from space to space, that is what makes multi-variable calculus so unbelievably powerful. 0826 That you can actually jump around from space to space like this, with well defined functions.0499

This is going to be the x,y plane, this is R2. I will write you another copy of the real number line.0513

So, this is the real number line, this is R2, 2-space, and this is the real number line.0520

C(t) maps from here to here.0527

A point to a vector... a vector in 2-space.0535

F goes from here, takes a point in 2-space, and maps to here, so this is c(t) and this is f.0540

When we find the composite function, f(c(t)), what I have now is a map from R to R. That is what this example is.0549

I have this that goes to a point in 2-space. F takes a point in 2-space and spits out some number.0566

Noticed I ended up with some function of t, ln(t2)t5 a specific value of t. This is just a single number.0572

This is what is going on here. You are just forming a composite function with curves and functions of several variables.0583

Hopefully this is reasonably clear. This is what we want to understand. This is what is happening mathematically.0593

You are mapping from one space to another, and then you are moving from that space to another space. In this case the space you end up with happens to be the space you started off with, which is the real number line.0596

So now the chain rule allows us to differentiate something like this. So now, let us go ahead and explicitly write down what the chain rule is.0607

I want you to see this simply because I want you also to start becoming accustomed to the expression of theorems, formal things, but again, it has to be based on understanding. 0618

It is a little long, but there is nothing here that is strange, so let f be a function defined and differentiable on an open set u.0630

Let c be a differentiable curve, all that means is a nice smooth curve with no whacky bumps or corners... be a differentiable curve such that the values of c(t) lie in the open set u.0666

What we were doing in the beginning of the lesson, we have an open set, we have a curve that happens to pass through that open set, therefore the points along the curve can be used for our function.0713

Then, the composite function, f(c(t)) is differentiable.0725

It is differentiable itself... as a function of t and the derivative of f(c(t)) with respect to t is equal to the gradient of f evaluated at c(t). 1250 The dot product of that vector with the vector c'(t), now you remember the gradient is a vector.0738

If I have some function, like f(x,y), the gradient is df/dx, and the second component is df/dy, I just differentiate as many variables, and that is my gradient vector.0777

So let us stop and think about what this says. If I have some function that is defined and differentiable on some open set, and c happens to be a differentiable curve that passes through that open set, in other words take some values in that open set, then the composite function f(c(t)) is also differentiable. 0794

It is differentiable as a function of t, as a single variable t and the derivative of that composite function is equal to the gradient of f at c(t) · c'(t). 0812

This is very, very, very important.0825

Now, for computations, when we actually do specific problems, we of course are going to be working with components, which is always the case.0831

With vectors we can go ahead and write out the definitions and the theorems using a shorter, more elegant notation, but when we actually do the computations with vectors we have to work with components.0841

x,y,z, whatever it is that we happen to be working with. So, let us go ahead and just sort of write out the component form of this so you see what is happening.0852

Again, the dot product is the same dot product that you know. There is nothing new here, trust what you know.0860

This is a vector, this is a vector, when you take the dot product of 2 vectors you get a number. That is what this says. You are getting a derivative. 0868

A function of t, if you evaluated a specific point of t, it is actually just a number. You are still just doing a derivative. The same thing you have been doing for years.0877

Let us just see here. So, if c(t) equals, now c1, that is the... so c(t) is a curve... c2(t), its component functions are component functions of t, just like the first example, and we have f(x1,x2).0886

This time I did not write it as x and y, I wrote it as x1 and x2, these are variables.0920

The first variable, the second variable. Then the derivative with respect to t of the f(c(t)) = well, we said it equals the gradient of f evaluated at c(t) · c'(t).0924

Okay, the gradient -- I should probably write this out -- the gradient of this function is going to be... tell you what, let me go ahead and before I write that, let me write out the gradient because I know it has been a couple of lessons since we did that. 0949

So, let me write the gradf = df/dx1, df/dx2.0968

This is a vector, the first component of which is the derivative with respect to the first variable.0982

The second component is the derivative of the function with respect to the second variable.0988

Now, c'(t) is also a vector. It is the derivative of this, c1'(t) and it is the derivative of this, c2'(t).0991

That is it, these are just functions, so now what we have is the derivative with respect to t of f(c(t)), in other words this thing right here. 1006

We said it is the gradient of f · c', this is the gradient of f, this is c'.1018

So let us see what this looks like in component form. It is... oh, you know what, I have a capital F, don't I?1025

I keep forgetting that, that small f is just so ubiquitous in most scientific literature.1032

So, we have df/dx1 × dc1/dt, that is all this is, c' is just dc1/dt.1039

It is just notation. dc2/dt. The dot product is this × that + this × that + df/dx2 × dc2/dt.1062

So that is it, that is all we are doing here. We are just doing it in component form.1090

Now personally, I think that what I have just written here is actually a little bit more confusing than just the statement of the theorem.1096

If you look at it as just the statement of the theorem, the gradient of f dotted with c', and if you know what the gradient is, you know what c' is, you know how to take derivatives.1105

You just do the dot product. This is sort of the component representation of it.1114

I personally do not like seeing all of these things because again it is notationally intensive.1121

The idea is to understand what this is, and then you can do the rest.1125

So, personally, my favorite, I still think it is great to learn it this way. Gradient · c'. Gradient of f · c'. Just keep telling yourself that about 5 or 6 times, and you will know what to do.1130

So, let us go ahead and just do an example, that is the best way to make sense of this.1145

So, let me go back to my black ink here. Actually you know what, let me go ahead and go to blue.1151

So example 2. Now, we will let our curve t be t, e(t) and t2, and we will let our function x, y, z, so we are definitely talking about a curve in 3-space, and a function of 3 variables, equals xy2z.1157

First of all, let us talk about what is going on here. We are going to form the composite function. We are going to be forming f(c(t)).1193

That is what we are going to be doing. Well, f(c(t)), x,y,z, in this case x is this thing, t, y is this thing e(t), and z is this t2.1201

We want to write everything out. Now the gradient of f, that is it, we are just going to build this step by step by step, that is all we are doing here.1220

The gradient of f is equal to, well it is the first component is the first partial, the second component is the second partial, the third component is the third partial.1231

If you like the other notation it is going to be df/dx, df/dy, and it is going to be df/dz.1244

Now, let us go ahead and actually compute that. The first partial, the derivative of this function with respect to x is y2z.1254

The derivative with respect to y is 2xyz.1264

And, the derivative with respect to z is going to be xy2, so this is my gradient of f.1271

Now, my gradient of f, evaluated at c(t), so now we will take the next step, now we will do the grad of f evaluated at c(t) which is the actual expression that is in the definition for the chain rule.1284

All that says is that take my gradient f, this thing, and I just put in the values c(t) in here.1300

Well, x is t, y is e(t), and z is t2.1308

So when I put these things into here, here is what I get.1312

y2 is just, so it is going to end up being t2e(2t), right?1323

y2 is just e(2t), z is t2, so that is t2e(2t).1329

2 × x, which is t, y which is e(t), z which is t2, I end up with 2t3e(t).1336

xy2 is t × e(t) × t2, I get t × e of... wait, e(2t), yes, there we go...1346

Okay, so that takes care of this one. That is the grad of f evaluated at c(t), now I just have to find c'.1362

That is really, really simple. C'(t). Well here is my c right here, I will just take the derivative of each one, that is it.1369

The derivative of t is 1, the derivative of e(t) is et, and the derivative of t2 is 2t.1378

Well, now I just form my dot product, so the gradient of f evaluated at c(t) dotted with c'(t), it equals this vector dotted with this vector.1390

Well the dot product is this × that, so it is t2, the dot product is not a vector, this × that so it is t2e(2t) + this × that, 2t3e(2t) + this × that + 2t2e(2t), and that is it. 1411

Let us see if there is anything that I can combine here, t2e(2t), 2t2e(2t), yes, there is.1446

So it is going to be 3t2e(2t) + 2 × t3e(2t), and that is my final answer. That is it. Let me go back.1453

I was given some, you know, a curve, and I was given a function, and I just hammered it out.1469

I took the gradient as a function, as a vector in x, y, z, I evaluated it at c(t), in other words I put these values in for x, y, z, and I got this.1480

Now it is the gradient vector expressed in t. I took the derivative of c which is c', that is easiest enough to do.1490

Then I just took the dot product of those vectors. That is it, that is all that is going on here.1498

So this happens to be the derivative of f(c(t)). That is what this is equal to. This is equal to f(c(t))... no, we want to definitely, f'(c(t)), that is it. That is all that is going on here.1501

It is just a way of differentiating a composite function. 1523

Now, you are probably asking yourself, Okay, well if I start with a function of t, t goes from R to R3, and then I take the function R3 to R so that it is a function R to R, what I actually have is a function of t, right?1526

Yes, it is just a function of t that you are differentiating. You are saying, well wait a minute, if I found f(t) up here, couldn't I just do this directly? Do I have to use the chain rule?1542

The answer is no, you do not have to use the chain rule, you can do it directly. Which one is better?1552

Well, actually it depends on the situation. It depends on the function, it depends on what you are doing, that is all it is. 1556

So, let us go ahead and actually do it directly just to confirm that you can do it directly.1562

I think it will shed a little bit more light on this relationship between the curve and the function.1566

Let us do this in blue again... so we said that f(x,y,z) is equal to xy2z.1574

We said that c(t), let us just rewrite them over again.1589

Let us see, what did we say c(t) was... t, e(t), and t2, so now let us just form f(c(t)).1595

So f(c(t) = well, xy2z, x is t, so that is t, y2 is e(2t), and z is going to be t2, I am just plugging those in, and I end up with t3e(2t).1606

Well, not if I just take... this is just a function of t, so if I just take df with respect to t, I end up getting, so it is going to be this × the derivative of that, which is going to be 2t3e(2t) + that × the derivative of that, 3t2e(2t).1633

What do you know, you end up with the same exact answer.1656

Which is better? Again it just depends on your particular situation.1660

Sometimes you want to do it directly if it makes more sense, sometimes you want to use the chain rule if it makes more sense.1667

The problem at hand we will actually decide which one is better. 1672

Ok, so that is the chain rule, thank you very much for joining us here at educator.com. We will see you next time. Take care, bye-bye.1676