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

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Implicit Differentiation and Linear Approximation

Instructor: Joel Lewis

View the complete course: http://ocw.mit.edu/18-01SCF10

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Afficher la révision 1 créée 03/13/2011 par audreycermak.

1. Hi! Welcome back to recitation.
2. In lecture, you've been talking about implicitly defined functions and implicit differentiation.
3. One of the reasons that these are important, or that implicit differentiation
4. is important, is that sometimes you have a function defined implicitly
5. and you can't solve for it. You don't have any algebraic method for computing
6. the function values as a formula.
7. For example, this function that I've written on the board
8. that I've called w of x is defined implicitly by the equation that
9. w of x plus one, quantity times e to the w of x, is equal to x for all x.
10. So this function, some of its values you can guess.
11. Like at x equals zero, the function value is going to be negative one.
12. The reason is that this can't ever be zero, so the only way to get this side to be zero
13. is if w is negative one, if this term is zero.
14. So, some of its values are easy to compute, but some of its values aren't.
15. So for example, if I asked you what w of three halves is,
16. it's very hard, there's no algebraic way you can manipulate this equation
17. that will let you figure that out. So in that situation you might still care
18. about what the function value is. So what can you do?
19. Well, you can try and find a numerical approximation.
20. So in this problem I'd like you to try and estimate the value w of three halves
21. by using a linear approximation of the function w of x in order to compute this value.
22. As a hint, I've given you that w of one is zero. Right? If you put in
23. x equals one and w of one equals zero on the left hand side, you do indeed get one, as you should.
24. So, good! So that will give you a hint about where you could base your linear approximation.
25. Why don't you pause the video, and take a few minutes to work this out.
26. Come back and we can work it out together.
27. All right, welcome back! So, hopefully, you've had a chance to work on this question a little bit.
28. In order to do this linear approximation that we want, we need to know a base point
29. and we need to know the derivative of the function at that base point.
30. Those are the two pieces of data you need in order to construct a linear approximation.
31. We have a good candidate for a base point here, which is the point one, zero.
32. So this curve, whatever it looks like, it passes through the point one, zero.
33. And that's the point we're going to use for our approximation.
34. So we're going to use the linear approximation w of x is approximately equal to
35. w prime of one, times x minus one, plus w of one when x is approximately equal to one.
36. So this is the linear approximation we're going to use. And we have that w of one here is zero.
37. So this is equal to w prime of one, times x minus one.
38. (So the w one of zero just goes away.)
39. In order to estimate w of x, in particular w of three halves, what we need to know is
40. we need to know the derivative of w. And to get the derivative of w
41. we need to use, well, we have only one piece of information about w,
42. that it's defined by this implicit equation. So in order to get the derivative of w
43. we have to use implicit differentiation. OK?
44. So let's do that. So if we implicitly differentiate this equation --
45. so let's start with -- the right hand side's going to be really easy --
46. we're going to differentiate with respect to x. The right hand side's going to be one.
47. On the left hand side, it's going to be a little more complicated.
48. We have a product, and then this piece, we're going to have a chain rule situation.
49. We have e to the w of x. So, [implicit diff 'n], we're going to take an implicit derivative.
50. On the left, so, the product rule first. We take the derivative of the first part.
51. So that's just w prime of x, times the second part. That's e to the w of x.
52. Plus the first part. That's w of x plus one, times the derivative of the second part.
53. 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.
54. That's the chain rule. So that's what happens when I differentiate the left hand side.
55. And on the right hand side, I take the derivative of x and I get one.
56. OK? Good. So now I've got this equation and I need to solve this equation for w prime.
57. Because if you look up here, that's what I want. I want a particular value of w prime.
58. And as always happens in implicit differentiation, from the point of view of this w prime,
59. it's only involved in the equation in a very simple way.
60. So there's it multiplied by functions of x and w of x. But not... it's just, you know,
61. it's just multipled by something that doesn't involve w prime at all.
62. And here it's multiplied by something that doesn't involve w prime at all.
63. So you can just collect your w primes and divide through. It's just like solving a linear equation.
64. So here if we collect our w primes, this is w prime of x, times, it looks like, w of x, plus two
65. times e to the w of x. Did I get that right?
66. Looks good. OK, so that's still equal to one.
67. That means w prime of x is just, well not just, it's equal to
68. one over w of x plus two, times e to the w of x.
69. OK, so this is true for every x. But I don't need this equation for every x.
70. I just need the particular value of w prime at one.
71. So I'm going to take this equation then and I'm just going to put in
72. x equals one. So I put in x equals one. I'll do it over here.
73. So I get w prime of one. And everywhere I had an x, I put in a one.
74. So actually, in this equation, the only place x appears is in the argument of w.
75. So this is w of one, plus two, times e to the w of one.
76. OK, so in order to get w prime of one, I need to know what w of one is.
77. But I had that. I had it -- it was right back here. There was the - that was my hint to you.
78. 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.
79. So we take that value. So w of one at zero, so this is just one over two.
80. OK? So I take that back upstairs to this equation that I had here.
81. And I have that w of x is approximately equal to w prime of one, times x minus one.
82. So w of x is approximately equal to (w prime of one we saw is one half)
83. times x minus one. And that approximation was good near our base point,
84. so that's good when x is near one.
85. All right. So this is the linear approximation. I asked for the linear approximation
86. -- its value at the particular point x equals three halves.
87. So w of three halves is approximately one half times (well, three halves
88. minus one is also a half) so this is a quarter.
89. OK. So this is our estimate for w of three halves.
90. W of three halves is approximately one fourth.
91. If you wanted a better estimate, you could try iterating this process,
92. or choosing some base point even closer if you could figure out
93. the value of w and x near this point that you're interested in -- three halves.
94. Just to sum up what we did was we had this implicitly defined function w.
95. We wanted to estimate its value at a point where we couldn't compute it explicitly.
96. So what we did, was we did our normal linear approximation method.
97. We wrote down our normal linear approximation formula.
98. The only thing that was slightly unusual is that we had to use implicit differentiation.
99. In order to compute the derivative that appears in the linear approximation, we implicitly differentiated.
100. OK? So that happened just like normal. And then at the end, we plugged in the values
101. that we were interested in to actually compute the particular value of that approximation.
102. So I'll end there.