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Slide 1Slide 2Slide 3Slide 4Slide 5Slide 6Slide 7Slide 8Slide 9Slide 10Slide 11Slide 12Slide 13Slide 14Slide 15Slide 16Slide 17Slide 18Slide 19Slide 20Slide 21Derivitive RulesEkstrom Math 115b Mathematics for Business Decisions, part IIDifferentiationMath 115bEkstrom Math 115b Differentiation, Part IWhat comes to mind when you think of “rate” RateEkstrom Math 115b Differentiation, Part IDescribe the graph:Where is the function…increasing?decreasing?decreasing the fastest? Properties of Graphs$0$5$10$15$20$250 1,000 2,000 3,000 4,000Ekstrom Math 115b Differentiation, Part IDescribe f(x).Where is f:positive?negative?zero?increasing?decreasing? Properties, cont.-4 -3 -2 -1 0 1 2-25-20-15-10-5051015FUNCTIONxf(x)Ekstrom Math 115b Differentiation, Part IRate of change of a linear function is called “slope”Denoted as m in y = mx + bHow is it defined?What if the function is not linear? Rate of ChangeEkstrom Math 115b Differentiation, Part IConsider the function from earlier:Can we define a “slope” of this line? Rate of Change, cont.-4 -3 -2 -1 0 1 2-25-20-15-10-5051015FUNCTIONxf(x)Ekstrom Math 115b Differentiation, Part I•Consider the following set of data points (Tucson temperatures before, during, and after a thunderstorm): Example DataTimeTemp (F)12:00 91.913:00 93.914:00 9515:00 9516:00 9316:18 87.816:29 78.816:43 7716:45 78.816:48 80.617:00 8218:00 8619:00 84Ekstrom Math 115b Differentiation, Part IPerhaps plotting the data will give us a better description:What is the rate of change of the temperature at 4:29 (16:29)? Example, cont.11:00 12:12 13:24 14:36 15:48 17:00 18:12 19:247580859095100Tucson TemperatureTimeTemp. (F)Ekstrom Math 115b Differentiation, Part ISo what do we want to do?To evaluate the rate of change (slope) of f (x) at x, we should find the slope between the points before and after the point in question:for some h. Finding the “slope” at a pointhhxfhxfhxhxhxfhxfxym2)()()()()()( Ekstrom Math 115b Differentiation, Part IAs h gets smaller and smaller, the approximation of the slope gets better and better.The derivative of a function is slope of a tangent line at a point on any curve, and can be calculated by:It is usually denoted as or Slope at a pointhhxfhxfh2)()(lim0)(' xfdxdfEkstrom Math 115b Differentiation, Part IWhat does f (x + h) mean?Ex. Soln:It means you evaluate the function at the quantity, x + h. Do NOT simply add h to f(x)! This will ultimately lead to a slope of 1. Algebra Review 114 384 324)2( ).2( find ,34)( IfxxxxfxfxxfEkstrom Math 115b Differentiation, Part IExample: Calculate the derivative of the function f (x) = 5x + 2 using the difference quotient.Solution:Surprised? Algebra Review, cont.       52102255255225252)()(hhhhxhxhhxhxhhxfhxfmEkstrom Math 115b Differentiation, Part ICalculate the derivatives of the following functions: Example calculations2at 231)(14)(23)(12)(2xxxfxxfxxfxfEkstrom Math 115b Differentiation, Part IThe derivative of a function is the slope of the line tangent to any point on the curve, f (x).It is calculated by finding the limit:This gives an instantaneous rate of change of the function, f (x). Difference Quotienthhxfhxfxfh2)()(lim)('0Ekstrom Math 115b Differentiation, Part IWhat do we mean by instantaneous?If h was one unit, and we calculated the difference quotient, then we would be finding the average rate of change between the points before and after the point in question.We want h to be smaller and smaller (closer and closer to 0) so that the length 2h is approximately 0 so our quotient will stabilize. Instantaneous Rate of ChangeEkstrom Math 115b Differentiation, Part ITo visualize the tangent line, think of a bird’s eye view of a curvy road at night. The headlights of a car traveling along this road will not follow the curves of the pavement. The path of the headlights represents the tangent line to the curvy road. Tangent LineEkstrom Math 115b Differentiation, Part IThe equation of the tangent line should be y = mx + bSlope of tangent line is equal to the derivative at every point xm = f ’(x), where m is the slope of the tangent lineSince we know the slope and a point on the line, we can find the equation of the tangent lineIf the derivative at the point exists, then the tangent line to the graph of f at the point (a, f (a)) has the equation Tangent Line)()()(' afaxafy Ekstrom Math 115b Differentiation, Part IFind the slope of the line tangent to the graph of at the point (3, f (3)). Find an equation for the tangent line at that point. First Example44)(2 xxxfEkstrom Math 115b Differentiation, Part ILet f (x) = x3 + 6Find the equation of the line tangent to f (x) at the point (-1, f (-1)).Luckily, you don’t have to do this by hand every timeDifferentiating.xls Second ExampleEkstrom Math 115b


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UA MATH 115B - Differentiation

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