Implicit Differentiation and Linear Approximation  MIT 18.01SC Single Variable Calculus, Fall 2010

0:07  0:09Hi! Welcome back to recitation.

0:09  0:15In lecture, you've been talking about implicitly defined functions and implicit differentiation.

0:15  0:20One of the reasons that these are important, or that implicit differentiation

0:20  0:23is important, is that sometimes you have a function defined implicitly

0:23  0:28and you can't solve for it. You don't have any algebraic method for computing

0:28  0:31the function values as a formula.

0:31  0:36For example, this function that I've written on the board

0:36  0:40that I've called w of x is defined implicitly by the equation that

0:40  0:46w of x plus one, quantity times e to the w of x, is equal to x for all x.

0:46  0:51So this function, some of its values you can guess.

0:51  0:57Like at x equals zero, the function value is going to be negative one.

0:57  1:01The reason is that this can't ever be zero, so the only way to get this side to be zero

1:01  1:04is if w is negative one, if this term is zero.

1:04  1:08So, some of its values are easy to compute, but some of its values aren't.

1:08  1:13So for example, if I asked you what w of three halves is,

1:13  1:18it's very hard, there's no algebraic way you can manipulate this equation

1:18  1:22that will let you figure that out. So in that situation you might still care

1:22  1:26about what the function value is. So what can you do?

1:26  1:29Well, you can try and find a numerical approximation.

1:29  1:33So in this problem I'd like you to try and estimate the value w of three halves

1:33  1:48by using a linear approximation of the function w of x in order to compute this value.

1:48  1:59As a hint, I've given you that w of one is zero. Right? If you put in

1:59  2:09x equals one and w of one equals zero on the left hand side, you do indeed get one, as you should.

2:09  2:19So, good! So that will give you a hint about where you could base your linear approximation.

2:19  2:23Why don't you pause the video, and take a few minutes to work this out.

2:23  2:32Come back and we can work it out together.

2:32  2:37All right, welcome back! So, hopefully, you've had a chance to work on this question a little bit.

2:37  2:43In order to do this linear approximation that we want, we need to know a base point

2:43  2:46and we need to know the derivative of the function at that base point.

2:46  2:51Those are the two pieces of data you need in order to construct a linear approximation.

2:51  2:56We have a good candidate for a base point here, which is the point one, zero.

2:56  3:00So this curve, whatever it looks like, it passes through the point one, zero.

3:00  3:03And that's the point we're going to use for our approximation.

3:03  3:16So we're going to use the linear approximation w of x is approximately equal to

3:16  3:33w prime of one, times x minus one, plus w of one when x is approximately equal to one.

3:33  3:38So this is the linear approximation we're going to use. And we have that w of one here is zero.

3:38  3:48So this is equal to w prime of one, times x minus one.

3:48  3:51(So the w one of zero just goes away.)

3:51  3:58In order to estimate w of x, in particular w of three halves, what we need to know is

3:58  4:03we need to know the derivative of w. And to get the derivative of w

4:03  4:08we need to use, well, we have only one piece of information about w,

4:08  4:16that it's defined by this implicit equation. So in order to get the derivative of w

4:16  4:19we have to use implicit differentiation. OK?

4:19  4:26So let's do that. So if we implicitly differentiate this equation 

4:26  4:28so let's start with  the right hand side's going to be really easy 

4:28  4:32we're going to differentiate with respect to x. The right hand side's going to be one.

4:32  4:35On the left hand side, it's going to be a little more complicated.

4:35  4:39We have a product, and then this piece, we're going to have a chain rule situation.

4:39  4:54We have e to the w of x. So, [implicit diff 'n], we're going to take an implicit derivative.

4:54  4:59On the left, so, the product rule first. We take the derivative of the first part.

4:59  5:06So that's just w prime of x, times the second part. That's e to the w of x.

5:06  5:15Plus the first part. That's w of x plus one, times the derivative of the second part.

5:15  5:24So the second part is e to the w of x. So that gives me an e to the w of x, times w prime of x.

5:24  5:29That's the chain rule. So that's what happens when I differentiate the left hand side.

5:29  5:33And on the right hand side, I take the derivative of x and I get one.

5:33  5:40OK? Good. So now I've got this equation and I need to solve this equation for w prime.

5:40  5:45Because if you look up here, that's what I want. I want a particular value of w prime.

5:45  5:51And as always happens in implicit differentiation, from the point of view of this w prime,

5:51  5:54it's only involved in the equation in a very simple way.

5:54  6:01So there's it multiplied by functions of x and w of x. But not... it's just, you know,

6:01  6:06it's just multipled by something that doesn't involve w prime at all.

6:06  6:09And here it's multiplied by something that doesn't involve w prime at all.

6:09  6:14So you can just collect your w primes and divide through. It's just like solving a linear equation.

6:14  6:29So here if we collect our w primes, this is w prime of x, times, it looks like, w of x, plus two

6:29  6:36times e to the w of x. Did I get that right?

6:36  6:41Looks good. OK, so that's still equal to one.

6:41  6:48That means w prime of x is just, well not just, it's equal to

6:48  6:58one over w of x plus two, times e to the w of x.

6:58  7:04OK, so this is true for every x. But I don't need this equation for every x.

7:04  7:08I just need the particular value of w prime at one.

7:08  7:12So I'm going to take this equation then and I'm just going to put in

7:12  7:17x equals one. So I put in x equals one. I'll do it over here.

7:17  7:22So I get w prime of one. And everywhere I had an x, I put in a one.

7:22  7:27So actually, in this equation, the only place x appears is in the argument of w.

7:27  7:34So this is w of one, plus two, times e to the w of one.

7:34  7:39OK, so in order to get w prime of one, I need to know what w of one is.

7:39  7:44But I had that. I had it  it was right back here. There was the  that was my hint to you.

7:44  7:51This is why we are using this point as a base point. Which is, we know the value of w for this value of x.

7:51  8:04So we take that value. So w of one at zero, so this is just one over two.

8:10  8:14OK? So I take that back upstairs to this equation that I had here.

8:14  8:21And I have that w of x is approximately equal to w prime of one, times x minus one.

8:21  8:31So w of x is approximately equal to (w prime of one we saw is one half)

8:31  8:41times x minus one. And that approximation was good near our base point,

8:41  8:47so that's good when x is near one.

8:47  8:51All right. So this is the linear approximation. I asked for the linear approximation

8:51  8:55 its value at the particular point x equals three halves.

8:55  9:04So w of three halves is approximately one half times (well, three halves

9:04  9:13minus one is also a half) so this is a quarter.

9:13  9:16OK. So this is our estimate for w of three halves.

9:16  9:19W of three halves is approximately one fourth.

9:19  9:22If you wanted a better estimate, you could try iterating this process,

9:22  9:28or choosing some base point even closer if you could figure out

9:28  9:35the value of w and x near this point that you're interested in  three halves.

9:35  9:41Just to sum up what we did was we had this implicitly defined function w.

9:41  9:45We wanted to estimate its value at a point where we couldn't compute it explicitly.

9:45  9:48So what we did, was we did our normal linear approximation method.

9:48  9:54We wrote down our normal linear approximation formula.

9:54  9:59The only thing that was slightly unusual is that we had to use implicit differentiation.

9:59  10:04In order to compute the derivative that appears in the linear approximation, we implicitly differentiated.

10:04  10:08OK? So that happened just like normal. And then at the end, we plugged in the values

10:08  10:15that we were interested in to actually compute the particular value of that approximation.

10:15 So I'll end there.
 Titre:
 Implicit Differentiation and Linear Approximation  MIT 18.01SC Single Variable Calculus, Fall 2010
 Description:

Implicit Differentiation and Linear Approximation
Instructor: Joel Lewis
View the complete course: http://ocw.mit.edu/1801SCF10
License: Creative Commons BYNCSA
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More courses at http://ocw.mit.edu  Langue de la vidéo:
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 Durée:
 10:17