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

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Hi! Welcome back to recitation.
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In lecture, you've been talking about implicitly defined functions and implicit differentiation.
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One of the reasons that these are important, or that implicit differentiation
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is important, is that sometimes you have a function defined implicitly
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and you can't solve for it. You don't have any algebraic method for computing
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the function values as a formula.
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For example, this function that I've written on the board
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that I've called w of x is defined implicitly by the equation that
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w of x plus one, quantity times e to the w of x, is equal to x for all x.
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So this function, some of its values you can guess.
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Like at x equals zero, the function value is going to be negative one.
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The reason is that this can't ever be zero, so the only way to get this side to be zero
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is if w is negative one, if this term is zero.
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So, some of its values are easy to compute, but some of its values aren't.
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So for example, if I asked you what w of three halves is,
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it's very hard, there's no algebraic way you can manipulate this equation
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that will let you figure that out. So in that situation you might still care
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about what the function value is. So what can you do?
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Well, you can try and find a numerical approximation.
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So in this problem I'd like you to try and estimate the value w of three halves
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by using a linear approximation of the function w of x in order to compute this value.
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As a hint, I've given you that w of one is zero. Right? If you put in
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x equals one and w of one equals zero on the left hand side, you do indeed get one, as you should.
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So, good! So that will give you a hint about where you could base your linear approximation.
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Why don't you pause the video, and take a few minutes to work this out.
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Come back and we can work it out together.
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All right, welcome back! So, hopefully, you've had a chance to work on this question a little bit.
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In order to do this linear approximation that we want, we need to know a base point
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and we need to know the derivative of the function at that base point.
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Those are the two pieces of data you need in order to construct a linear approximation.
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We have a good candidate for a base point here, which is the point one, zero.
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So this curve, whatever it looks like, it passes through the point one, zero.
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And that's the point we're going to use for our approximation.
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So we're going to use the linear approximation w of x is approximately equal to
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w prime of one, times x minus one, plus w of one when x is approximately equal to one.
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So this is the linear approximation we're going to use. And we have that w of one here is zero.
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So this is equal to w prime of one, times x minus one.
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(So the w one of zero just goes away.)
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In order to estimate w of x, in particular w of three halves, what we need to know is
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we need to know the derivative of w. And to get the derivative of w
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we need to use, well, we have only one piece of information about w,
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that it's defined by this implicit equation. So in order to get the derivative of w
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we have to use implicit differentiation. OK?
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So let's do that. So if we implicitly differentiate this equation --
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so let's start with -- the right hand side's going to be really easy --
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we're going to differentiate with respect to x. The right hand side's going to be one.
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On the left hand side, it's going to be a little more complicated.
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We have a product, and then this piece, we're going to have a chain rule situation.
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We have e to the w of x. So, [implicit diff 'n], we're going to take an implicit derivative.
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On the left, so, the product rule first. We take the derivative of the first part.
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So that's just w prime of x, times the second part. That's e to the w of x.
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Plus the first part. That's w of x plus one, times the derivative of the second part.
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So 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.
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That's the chain rule. So that's what happens when I differentiate the left hand side.
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And on the right hand side, I take the derivative of x and I get one.
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OK? Good. So now I've got this equation and I need to solve this equation for w prime.
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Because if you look up here, that's what I want. I want a particular value of w prime.
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And as always happens in implicit differentiation, from the point of view of this w prime,
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it's only involved in the equation in a very simple way.
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So there's it multiplied by functions of x and w of x. But not... it's just, you know,
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it's just multipled by something that doesn't involve w prime at all.
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And here it's multiplied by something that doesn't involve w prime at all.
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So you can just collect your w primes and divide through. It's just like solving a linear equation.
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So here if we collect our w primes, this is w prime of x, times, it looks like, w of x, plus two
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times e to the w of x. Did I get that right?
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Looks good. OK, so that's still equal to one.
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That means w prime of x is just, well not just, it's equal to
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one over w of x plus two, times e to the w of x.
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OK, so this is true for every x. But I don't need this equation for every x.
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I just need the particular value of w prime at one.
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So I'm going to take this equation then and I'm just going to put in
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x equals one. So I put in x equals one. I'll do it over here.
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So I get w prime of one. And everywhere I had an x, I put in a one.
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So actually, in this equation, the only place x appears is in the argument of w.
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So this is w of one, plus two, times e to the w of one.
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OK, so in order to get w prime of one, I need to know what w of one is.
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But I had that. I had it -- it was right back here. There was the - that was my hint to you.
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This is why we are using this point as a base point. Which is, we know the value of w for this value of x.
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So we take that value. So w of one at zero, so this is just one over two.
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OK? So I take that back upstairs to this equation that I had here.
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And I have that w of x is approximately equal to w prime of one, times x minus one.
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So w of x is approximately equal to (w prime of one we saw is one half)
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times x minus one. And that approximation was good near our base point,
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so that's good when x is near one.
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All right. So this is the linear approximation. I asked for the linear approximation
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-- its value at the particular point x equals three halves.
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So w of three halves is approximately one half times (well, three halves
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minus one is also a half) so this is a quarter.
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OK. So this is our estimate for w of three halves.
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W of three halves is approximately one fourth.
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If you wanted a better estimate, you could try iterating this process,
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or choosing some base point even closer if you could figure out
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the value of w and x near this point that you're interested in -- three halves.
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Just to sum up what we did was we had this implicitly defined function w.
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We wanted to estimate its value at a point where we couldn't compute it explicitly.
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So what we did, was we did our normal linear approximation method.
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We wrote down our normal linear approximation formula.
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The only thing that was slightly unusual is that we had to use implicit differentiation.
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In order to compute the derivative that appears in the linear approximation, we implicitly differentiated.
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OK? So that happened just like normal. And then at the end, we plugged in the values
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that we were interested in to actually compute the particular value of that approximation.
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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/18-01SCF10

More courses at http://ocw.mit.edu

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Langue de la vidéo:
English
Équipe: MIT OpenCourseware
Durée:
10:17

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