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5 Pages

### 0928

Course: M 0928, Fall 2009
School: University of Texas
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Word Count: 445

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3 Contents 2 4 5 Lecture topics 09 - 28 - 98 Whats it mean when the derivative does not exist? Part 1 Whats it mean when the derivative does not exist? Part 2 Whats it mean when the derivative does not exist? Part 3 Lecture topics 09 - 28 - 98 2) Used chain rule to dierentiate (x2 + 1)3 , sin x, sin (cos(x)) Stated chain rule for sin ( ). 3) Used chain rule to dierentiate y = |x2 1|, then to locate on the...

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3 Contents 2 4 5 Lecture topics 09 - 28 - 98 Whats it mean when the derivative does not exist? Part 1 Whats it mean when the derivative does not exist? Part 2 Whats it mean when the derivative does not exist? Part 3 Lecture topics 09 - 28 - 98 2) Used chain rule to dierentiate (x2 + 1)3 , sin x, sin (cos(x)) Stated chain rule for sin ( ). 3) Used chain rule to dierentiate y = |x2 1|, then to locate on the graph where y = 0, y dne. 4) Talked about meaning of chain rule for [sin(2x)] versus [sin(x)] . 1) Stated chain rule for [f (g (x))] , [f ( )] and dy dx . Whats it mean when the derivative does not exist? Part 1 In class I showed that for y = |x2 1| then you can use the chain rule to get x2 1 (2x) y= |x2 1| Then you check that y dne when x = 1. If you look at the graph, you this is two places on the curve where theres pointy places. And that cuz theres pointy places, the function cant have a tangent there. Check the pic: Whats it mean when the derivative does not exist? Part 2 x This isnt really all there is because the piece of the derivative |x2 1| 1 actually has jumps at x = 1. The point I want to make is the that limits at the jump mean something. Check it out: 2 x1+ lim x2 1 |x2 1| x2 1 |x2 1| (2x) = x1+ lim x2 1 x2 1 (x2 1) x2 1 (2x) = x1+ lim (2x) = 2 x1 lim (2x) = lim x1 (2x) = lim (2x) = 2 x1 The derivative means something geometrically: its the slope of a tangent line. The function y = |x2 1| has two tangent lines at x = +1; the one at the left has slope slope = 2; the one on the right has slope slope = +2. Check the right pic below. right tangent; slope = +2 left tangent; slope = -2 Whats it mean when the derivative does not exist? Part 3 Ive made a point about how, when a derivative does not exist, that can tell you something about the ...

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University of Texas - M - 0930
Contents2 Lecture topics 09 - 30 - 98 3 Another example of implicit differentiationLecture topics 09 - 30 - 981) Did basic dierentiation rules using the d d d 2 3 3 dx (x sin x); dx (x + 1) ; dx (sin(x ). 2) Found 3) Found 4) Found 5) Found 7) F
University of Texas - M - 1005
Contents2 3 4 5 6 Lecture topics 10 - 05 - 98 Worked Word Problem Page 1 Worked Word Problem Page 2 Worked Word Problem Page 3 Worked Word Problem Page 4Lecture topics 10 - 05 - 981) Did variant word problem: Mt. St. Helens erupts, sending a clo
University of Texas - M - 1007
Contents2 Lecture topics 10 - 07 - 98 3 Worked worked out derivative from classLecture topics 10 - 07 - 981) Warned that Oct 21 was the last day for an easy drop. 2) Stated Rolles Theorem, then the Mean Value theorem. 3) Remarked that the MVT al
University of Texas - M - 1009
Contents2 3 4 5 6 7 Lecture topics 10 - 09 - 98 The Second Derivative Test Why I Dont Like The Second Derivative Test Why I Still Dont Like . . . Let The Computer Do The Work Maybe the Second Derivative Test Isnt That BadLecture topics 10 - 09 - 9
University of Texas - M - 1028
Contents2 3 4 5 Lecture topics 10 - 28 - 98 Even Functions Integrating Even Functions Odd FunctionsLecture topics 10 - 28 - 981) Used formula for true area:b|f (x)| dx =a{f &gt;0}f (x) dx f (x) dx{f &lt;0}and computed 0| sin x| dx2)
University of Texas - M - 1104
Contents2 3 4 5 Lecture topics 11 - 04 - 98 Slices of the hemisphere I Return of Slices of the hemisphere II Revenge of Slices of the hemisphere IIILecture topics 11 - 04 - 981) I worked the problem: Find the area between the curves x = y 2 , x
University of Texas - M - 1106
Contents2 3 4 5 Lecture topics 11 - 06 - 98 Rotating a curve about the x - axis, I Rotating a curve about the x - axis, II Rotating a curve about the x - axis, IIILecture topics 11 - 06 - 981) Computed the volume of a half-sphere, by slicing par
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626.37 9 . 7 274 c lerkn O37@gma I l.cO meDuCationCalifornia polyteChniC state university, san luis oBispo major: Graphic Communication concentration: Design Reproduction Technology degree: Bachelor of Science expected graduation date: December 2
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Cal Poly - PROJ - 339
Education:California Polytechnic State University, San Luis Obispo 2002-2006Bachelor of Science, Graphic Communication Concentrations: Design Reproduction Technology and Printing and Imaging ManagementWork Experience:Dolphin Shirt Company,
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Math 212 Exam 2 Spring 00 KerckhoveName Pledge1. Two farmers, husband and wife, wish to purchase a eld in which to grow yams. The eld is rectangular in shape and lies in a valley between a riverbed and a hillside. The Farmers estimate that land
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Math 212 | Final Exam KerckhoveName: Pledge:1. Consider the system of di erential equations given bydx dt dy dt= ,2x = 2x , 2yx0 = 10 y 0 = 3a. Find the function xt. b. Use your answer to part a and an integrating factor to nd the function
Richmond - M - 212
Math 212 | Final Exam Spring '95 | Kerckhove 1. Evaluate the following integrals.Name PledgeaZt cost dt dZ0 1bZt cost dt2cZ0sin x dx21 dx 1+x2eZ1 y , 2y , 3 dy2. Here's a plot of sinx on the interval 0 x :5.2
Richmond - M - 212
Math 212 Homework through Quiz 1Two types of homework will be given for each class period. One set of problems will be done in preparation for the day's lecture (these will not be turned in) and the other set will pertain to topics covered in the l
Richmond - M - 212
Math 212 Fall 04 Kerckhove Exam 2 (Math 212 visits Age of Iron)Name: Pledge:1. Consider a group of people who have received treatment for a disease such as breast cancer. Let T be the measured number of years that a particular patient might live
Richmond - M - 212
Richmond - M - 212
Richmond - M - 212
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hw_solns_11_03.nb1M212 HW 3 November Pharmacology Problem1 1 a) halflife = 6.3 hours implies Q0 = QH6.3L = Q0 * e -k*6.3 implies k = * lnH2L = 0.110 2 6.3 -0.110*24 = Q * a with a = 0.071 to three decimal places. Then QH24L = Q0 * e 0Answer:
Richmond - M - 212
m212_commentary_on_review.nb1Commentary on M212 Ch 5 Review The Computational Problems (10,15,16,19,23,24)d I used #10 to talk about the fundamental theorem, T 0 f HxL x = f HT L and about the relationship between the graph of d F HT L and th
Richmond - M - 212
Richmond - M - 212
Richmond - M - 212
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Richmond - M - 212
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Richmond - M - 212
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Richmond - M - 212
Richmond - M - 212
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Richmond - M - 212
Richmond - M - 212
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