Authors Gregory Hartman, Ph.D. Brian Heinold, Ph.D. Troy Siemers, Ph.D. Dimplekumar Chalishajar, Ph.D. Editor Jennifer Bowen, Ph.D.

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1 Section 7 Derivatives of Inverse Functions AP E XC I Version 0 Authors Gregory Hartman, PhD Department of Applie Mathema cs Virginia Military Ins tute Brian Heinol, PhD Department of Mathema cs an Computer Science Mount Saint Mary s University Troy Siemers, PhD Department of Applie Mathema cs Virginia Military Ins tute Dimplekumar Chalishajar, PhD Department of Applie Mathema cs Virginia Military Ins tute Eitor Jennifer Bowen, PhD Department of Mathema cs an Computer Science The College of Wooster

2 Copyright 04 Gregory Hartman License to the public uner Crea ve Commons A ribu on-noncommercial 30 Unite States License

3 Contents Preface Table of Contents iii v Limits An Introuc on To Limits Epsilon-Delta Defini on of a Limit 9 3 Fining Limits Analy cally 6 4 One Sie Limits 7 5 Con nuity 34 6 Limits involving infinity 43 Deriva ves 55 Instantaneous Rates of Change: The Deriva ve 55 Interpreta ons of the Deriva ve 69 3 Basic Differen a on Rules 76 4 The Prouct an Quo ent Rules 83 5 The Chain Rule 93 6 Implicit Differen a on 03 7 Deriva ves of Inverse Func ons 4 3 The Graphical Behavior of Func ons 3 Etreme Values 3 The Mean Value Theorem 9 33 Increasing an Decreasing Func ons Concavity an the Secon Deriva ve 4 35 Curve Sketching 49 4 Applica ons of the Deriva ve 57 4 Newton s Metho 57 4 Relate Rates 64

4 43 Op miza on 7 44 Differen als 78 5 Integra on 85 5 An eriva ves an Inefinite Integra on 85 5 The Definite Integral Riemann Sums The Funamental Theorem of Calculus 55 Numerical Integra on 33 6 Techniques of An ifferen a on 47 6 Subs tu on 47 A Solu ons To Selecte Problems A Ine A

5 Chapter Deriva ves 7 Deriva ves of Inverse Func ons 05, 0375 y, , 05 5, Figure 30: A func on f along with its inverse f Note how it oes not ma er which func on we refer to as f; the other is f a, b y b, a Figure 3: Corresponing tangent lines rawn to f an f Recall that a func on y = f is sai to be one to one if it passes the horizontal line test; that is, for two ifferent values an, we o not have f = f In some cases the omain of f must be restricte so that it is one to one For instance, consier f = Clearly, f = f, so f is not one to one on its regular omain, but by restric ng f to 0,, f is one to one Now recall that one to one func ons have inverses That is, if f is one to one, it has an inverse func on, enote by f, such that if fa = b, then f b = a The omain of f is the range of f, an vice-versa For ease of nota on, we set g = f an treat g as a func on of Since fa = b implies gb = a, when we compose f an g we get a nice result: f gb = fa = b In general, f g = an g f = This gives us a convenient way to check if two func ons are inverses of each other: compose them an if the result is, then they are inverses on the appropriate omains When the point a, b lies on the graph of f, the point b, a lies on the graph of g This leas us to iscover that the graph of g is the reflec on of f across the line y = In Figure 30 we see a func on graphe along with its inverse See how the point, 5 lies on one graph, whereas 5, lies on the other Because of this rela onship, whatever we know about f can quickly be transferre into knowlege about g For eample, consier Figure 3 where the tangent line to f at the point a, b is rawn That line has slope f a Through reflec on across y =, we can see that the tangent line to g at the point b, a shoul have slope f a This then tells us that g b = f a Consier: Informa on about f 05, 0375 lies on f Slope of tangent line to f at = 05 is 3/4 Informa on about g = f 0375, 05 lies on g Slope of tangent line to g at = 0375 is 4/3 f 05 = 3/4 g 0375 = 4/3 We have iscovere a rela onship between f an g in a mostly graphical way We can realize this rela onship analy cally as well Let y = g, where again g = f We want to fin y Since y = g, we know that fy = Using the Chain Rule an Implicit Differen a on, take the eriva ve of both sies of Notes: 4

6 7 Deriva ves of Inverse Func ons this last equality fy = f y y = y = f y y = f g This leas us to the following theorem Theorem Deriva ves of Inverse Func ons Let f be ifferen able an one to one on an open interval I, where f 0 for all in I, let J be the range of f on I, let g be the inverse func on of f, an let fa = b for some a in I Then g is a ifferen able func on on J, an in par cular, f b = g b = f an f = g = a f g The results of Theorem are not trivial; the nota on may seem confusing at first Careful consiera on, along with eamples, shoul earn unerstaning In the net eample we apply Theorem to the arcsine func on Eample 73 Fining the eriva ve of an inverse trigonometric func on Let y = arcsin = sin Fin y using Theorem S Aop ng our previously efine nota on, let g = arcsin an f = sin Thus f = cos Applying the theorem, we have g = f g = cosarcsin This last epression is not immeiately illumina ng Drawing a figure will help, as shown in Figure 33 Recall that the sine func on can be viewe as taking in an angle an returning a ra o of sies of a right triangle, specifically, the ra o opposite over hypotenuse This means that the arcsine func on takes as input a ra o of sies an returns an angle The equa on y = arcsin can be rewri en as y = arcsin/; that is, consier a right triangle where the Notes: 5

7 Chapter Deriva ves hypotenuse has length an the sie opposite of the angle with measure y has length This means the final sie has length, using the Pythagorean Theorem y Therefore cossin = cos y = / =, resul ng in Figure 33: A right triangle efine by y = sin / with the length of the thir leg foun using the Pythagorean Theorem arcsin = g = π π 4 y = sin π π 4 y π 3, 3 π 4 3, π 3 π Remember that the input of the arcsine func on is a ra o of a sie of a right triangle to its hypotenuse; the absolute value of this ra o will never be greater than Therefore the insie of the square root will never be nega ve In orer to make y = sin one to one, we restrict its omain to [ π/, π/]; on this omain, the range is [, ] Therefore the omain of y = arcsin is [, ] an the range is [ π/, π/] When = ±, note how the eriva ve of the arcsine func on is unefine; this correspons to the fact that as ±, the tangent lines to arcsine approach ver cal lines with unefine slopes In Figure 34 we see f = sin an f = sin graphe on their respec ve omains The line tangent to sin at the point π/3, 3/ has slope cos π/3 = / The slope of the corresponing point on sin, the point 3/, π/3, is y = sin π 4 = = = 3/ 3/4 /4 / =, π Figure 34: Graphs of sin an sin along with corresponing tangent lines verifying yet again that at corresponing points, a func on an its inverse have reciprocal slopes Using similar techniques, we can fin the eriva ves of all the inverse trigonometric func ons In Figure 3 we show the restric ons of the omains of the stanar trigonometric func ons that allow them to be inver ble Notes: 6

8 7 Deriva ves of Inverse Func ons Func on Domain Range Inverse Func on Domain Range sin [ π/, π/] [, ] sin [, ] [ π/, π/] cos [0, π] [, ] cos [, ] [0, π] tan π/, π/, tan, π/, π/ csc [ π/, 0 0, π/], ] [, csc, ] [, [ π/, 0 0, π/] sec [0, π/ π/, π], ] [, sec, ] [, [0, π/ π/, π] cot 0, π, cot, 0, π Figure 3: Domains an ranges of the trigonometric an inverse trigonometric func ons Theorem 3 Deriva ves of Inverse Trigonometric Func ons The inverse trigonometric func ons are ifferen able on their omains as liste in Figure 3 an their eriva ves are as follows: sin = sec = 3 tan = + 4 cos = 5 csc = 6 cot = + Note how the last three eriva ves are merely the opposites of the first three, respec vely Because of this, the first three are use almost eclusively throughout this tet In Sec on 3, we state without proof or eplana on that ln = We can jus fy that now using Theorem, as shown in the eample Eample 74 Fining the eriva ve of y = ln Use Theorem to compute ln S View y = ln as the inverse of y = e Therefore, using our stanar nota on, let f = e an g = ln We wish to fin g Theorem Notes: 7

9 Chapter Deriva ves gives: g = f g = e ln = In this chapter we have efine the eriva ve, given rules to facilitate its computa on, an given the eriva ves of a number of stanar func ons We restate the most important of these in the following theorem, intene to be a reference for further work Theorem 4 Glossary of Deriva ves of Elementary Func ons Let u an v be ifferen able func ons, an let a, c an n be a real numbers, a > 0, n 0 cu = cu 3 u v = uv + u v uv = u vv 5 7 = e = e 9 ln = 3 sin = cos 5 csc = csc cot 7 tan = sec sin = 9 csc = 3 tan = + u ± v = u ± v = u v uv 4 u v v c = 0 n = n n a = ln a a loga = ln a 4 cos = sin 6 sec = sec tan 8 cot = csc cos = 0 4 sec = cot = + Notes: 8

10 Eercises 7 Terms an Concepts T/F: Every func on has an inverse In your own wors eplain what it means for a func on to be one to one 3 If, 0 lies on the graph of y = f, what can be sai about the graph of y = f? 4 If, 0 lies on the graph of y = f an f = 5, what can be sai about y = f? Problems In Eercises 5 8, verify that the given func ons are inverses 5 f = + 6 an g = 3 6 f = + 6 +, 3 an g = 3, 7 f = 3 5, 5 an g = 3 + 5, 0 8 f = +, an g = f In Eercises 9 4, an inver ble func on f is given along with a point that lies on its graph Using Theorem, evaluate f at the inicate value 9 f = Point=, 0 Evaluate f 0 0 f = + 4, Point= 3, 7 Evaluate f 7 f = sin, π/4 π/4 Point= π/6, 3/ Evaluate f 3/ f = Point=, 8 Evaluate f 8 3 f = +, 0 Point=, / Evaluate f / 4 f = 6e 3 Point= 0, 6 Evaluate f 6 In Eercises 5 4, compute the eriva ve of the given func- on 5 ht = sin t 6 ft = sec t 7 g = tan 8 f = sin 9 gt = sin t cos t 0 ft = ln te t h = sin cos g = tan 3 f = sec / 4 f = sinsin In Eercises 5 7, compute the eriva ve of the given func- on in two ways: a By simplifying first, then taking the eriva ve, an b by using the Chain Rule first then simplifying Verify that the two answers are the same 5 f = sinsin 6 f = tan tan 7 f = sincos In Eercises 8 9, fin the equa on of the line tangent to the graph of f at the inicate value 8 f = sin at = 9 f = cos at = 3 4 Review 30 Fin y, where y y = 3 Fin the equa on of the line tangent to the graph of + y + y = 7 at the point, 3 Let f = 3 + f + s f Evaluate lim s 0 s

12 Solutions to O Eercises 3 y y = y siny cosy y+ cos+cosy+sin+y sinysin+y++cosy y 0 +y y+ 3 a y = 0 b y = a y = 4 b y = Sec on 5 T 3 F 5 T 7 f t = 53t 4 9 h t = 6t + e 3t +t f = 3 sin3 3 h t = 8 sin 3 t cost 5 f = tan 7 f = / 9 g t = ln 5 5 cos t sin t m w = ln3/3/ w 3 f = ln ln 5 g t = 5 cost +3t cos5t 7 t+3 sint +3t sin5t 7 7 Tangent line: y = 0 Normal line: = 0 9 Tangent line: y = 3θ π/ + Normal line: y = /3θ π/ + 3 In both cases the eriva ve is the same: / 33 a F/mph b The sign woul be nega ve; when the win is blowing at 0 mph, any increase in win spee will make it feel coler, ie, a lower number on the Fahrenheit scale Sec on 6 Answers will vary 3 T 5 f = 3 /3 = g t = t cos t + sin t t 9 y = 43 y+ y = sin secy 7 a y = b y = y = 3 y 3/ /5 5 y 6/5 y = 0 33 y = ln + Tangent line: y = + 4 ln + 35 y = sin+ sin + cos ln + Tangent line: y = 3π /4 π/ + π/ 3 37 y = Tangent line: y = /7 + /6 Sec on 7 F 3 The point 0, lies on the graph of y = f assuming f is inver ble 5 Compose fg an gf to confirm that each equals 7 Compose fg an gf to confirm that each equals 9 f 0 = f = /5 f 3/ = f π/6 = 3 f / = f = 5 h t = 4t 7 g = +4 9 g t = cos t cost sint t h = sin +cos cos 3 f = 5 a f =, so f = b f = cossin = 7 a f =, so f = b f = coscos 9 y = 4 3/4 + π/6 3 y = 4/5 + =

Section 6.2 Integration by Parts

Section 6.2 Integration by Parts Section 6 Integration by Parts AP E XC II Version 0 Authors Gregory Hartman, PhD Department of Applied Mathema cs Virginia Military Ins tute Brian Heinold, PhD Department of Mathema cs and Computer Science