Chapter 1 Overview: Review of Derivatives

Transcription

1 Chapter Overview: Review of Derivatives The purpose of this chapter is to review the how of ifferentiation. We will review all the erivative rules learne last year in PreCalculus. In the net several chapters, we will be eploring the why of ifferentiation as well. As a quick reference, here are those rules: The Power Rule: un nun- The Prouct Rule: u v u uv u v The Quotient Rule: u v u v u v v The Chain Rule: f ( g( )) f '( g( )) g '( ) u sin u cos u u cos u sin u u tan u sec u e u e u u u a a ln a sin cos tan u u u u u u u u u u u u csc u csc u cot u u sec u sec u tan u u cot u csc u u ln u u u loga u u ln a csc sec cot u u u u u u u u u u u

2 Here is a quick review of from last year: Ientities While all will eventually be use somewhere in Calculus, the ones that occur most often early are the Reciprocals an Quotients, the Pythagoreans, an the Double Angle Ientities. sin tan ; cos cos cot ; sin sec cos ; csc sin ; tan sec ; cot csc sin cos sin sin cos ; cos cos sin Inverses Because of the quarants, taking an inverse yiels two answers, only one of which your calculator can show. How the secon answer is foun epens on the kin of inverse: cos - calculator n - calculator n sin calculator n calculator n tan - calculator n calculator n calculator n log a log a y log a ( y) log a log a y log a y n log n log a a

3 .: The Power Rule an the Eponential Rules In PreCalculus, we evelope the iea of the Derivative geometrically. That is, the erivative initially arose from our nee to fin the slope of the tangent line. In Chapter an, that meaning, its link to limits, an other conceptualizations of the Derivative will be Eplore. In this Chapter, we are primarily intereste in how to fin the Derivative an what it is use for. f ( h) f ( ) Derivative Def n: f '( ) lim h 0 h Means: The function that yiels the slope of the tangent line. f ( ) f ( a) Numerical Derivative Def n: f '( a) lim a a Means: The numerical value of the slope of the tangent line at = a. Symbols for the Derivative y = "-y--" f '() = "f prime of " y' = "y prime" = "--" D = " sub " OBJECTIVES Use the Power Rule an Eponential Rules to fin Derivatives. Fin the equations of tangent an normal lines. Use the equation of a tangent line to approimate a function value. Key Iea from PreCalc: The erivative yiels the slope of the tangent line. [But there is more to it than that.] Key Concept: The erivative of a function is the slope of the function. The efinitions of the erivative, above, both come from the epression for the slope of a line, combine with the iea of a limit. The numerical erivative of a function is the slope of that function at a point.

4 E. Fin the values of f ', f ', g ', an g ' y f() g() g 5 f ', f ', g ', ' Note that at = that the erivative of both f an g oes not eist because there is not a single tangent line at this point. We will iscuss this further in the chapter on limits an ifferentiability. E. Fin the values of where the function whose graph is below has a erivative value of 0. y

5 y A slope of 0 is a horizontal line, so this function has tangent lines with slope of 0 at =, = 0, an =. The first an most basic erivative rule is the Power Rule. Among the last rules we learne in PreCalculus were the Eponential Rules. They look similar to one another, therefore it woul be a goo iea to view them together. The Power Rule: n n- n The Eponential Rules: e e a a ln a The ifference between these is where the variable is. The Power Rule applies when the variable is in the base, while the Eponential Rules apply when the variable is in the Eponent. The ifference between the two Eponential rules is what the base is. e = , while a is any positive number other than. E Fin a) 5 an b) 5 5

6 The first is a case of the Power Rule while the secon is a case of the secon Eponential Rule. Therefore, a) 5 = 5 b) 5 = 5 ln5 There were a few other basic rules that we nee to remember. D [constant] is always 0 n n D c cn D f g D f D g These rules allow us to easily ifferentiate a polynomial--term by term. E y 5; fin y y [ 5] ( ) 5( ) ( 0) 65 E 5 f ( ) e; fin f '( ). f '( ) e 6

7 y E 6 y e; fin. y e y e y Note in E 6 that e is a constant, therefore, its erivative is 0. E 7 Fin the equations of the lines tangent an normal to f () at The slope of the tangent line will be f '( ) f '( ) f '( ) [Note that we coul have gotten this more easily with the nderiv function on our calculator.] f ( ), so the tangent line will be y ( ) or y The normal line is perpenicular to the tangent line an, therefore, has the negative reciprocal slope =. The normal line is y ( ) One of the uses of the tangent line is base on the iea of Local Linearity. This means that in small areas, algebraic curves act like lines namely their 7

8 tangent lines. Therefore, one can get an approimate y-value for points near the point of tangency by plugging -values into the equation of the tangent line. We will eplore this more in a later section. As we have seen, when the variable is in the Eponent, we use the Eponential Rules. When the variable was in the base, we use the Power Rule. But what if the variable is in both places, such as? It is efinitely an Eponential problem, but the base is not a constant as the rules above have. The Change of Base Rule allows us to clarify the problem: e ln but we will nee the Prouct Rule for this erivative. Therefore, we will save this for section -. 8

9 . Homework Set A Differentiate.. f (). f () t t 8. y. y 5 e 5. v() r r 6. g() 7. y 8. u t t 9. z y A y Be y 0 0. y e 9

10 . Fin an equation of the tangent to the curve y e at the point ( 0, ).. Fin the points on the curve y where the tangent is horizontal. 5. Fin the equation of the tangent line to f ( ) 5 at. 0

11 5. Fin the tangent line equation to F( ) at. 5. Fin the equation of all tangent lines that have a slope of 5 for the function y 6. Fin the equation of the tangent line for f 5 6 that has a slope of.

12 7. Fin the equations of the lines tangent an normal to the function f 5 at the point (,5)

13 Answers:. Homework Set A. f (). f () t t 8 f '( ) f '( t) t. y. y 5 y 5 y e 5 5. v() r v'( r) e r 6. g() g '( ) r 7. y y 8. u t t u t t t A y 9. z y 0 Be 0. y y y y z' y 0 Ay Be e e. Fin an equation of the tangent to the curve y e at the point ( 0, ). y. Fin the points on the curve y where the tangent is horizontal. (, 6) (, ) 5. Fin the equation of the tangent line to f ( ) 5 at. y Fin the tangent line equation to F( ) at. y

14 5. Fin the equation of all tangent lines that have a slope of 5 for the function y y Fin the equation of the tangent line for f 5 6 that has a slope of. y Fin the equations of the lines tangent an normal to the function f 5 at the point (,5) y5 5

15 .: Composite Functions an the Chain Rule Composite Function--A function mae of two other functions, one within the other. For Eample, y 6, y sin, y cos, an y ( 5 ). The general symbols are f ( g( )) or [ f g]( ). E Given f cos, g, an (b) hg, an (c) (a) f h g., so g h cos, fin (a) f g f 0. f g, (b) (c) g 0, so h g h 0 0. g 0 h an h g 0 0, so cos f h g f 0 So how o we take the erivative of a composite function? There are two (or more) functions that must be ifferentiate, but, since one is insie the other, the erivatives cannot be taken at the same time. Just as a raical cannot be istribute over aition, a erivative cannot be istribute concentrically. The composite function is like a matryoshka (Russian oll) that has a oll insie a oll. The erivative is akin to opening them. They cannot both be opene at the same time an, when one is opene, there is an unopene one within. You en up with two open olls net to each other. The Chain Rule: f ( g( )) f '( g( )) g '( ) If you think of the insie function (the g ) as equaling u, we coul write The u Chain Rule like this: f ( u) f ' u. This is the way that most of the erivatives are written with The Chain Rule. 5

16 The Chain Rule is one of the cornerstones of Calculus. It can be embee within each of the other Rules, as it was in the introuction to this chapter. So the Power Rule an Eponential Rules in the last section really shoul have been state as: The Power Rule: n n- u u n u The Eponential Rules: e u e u u u a a ln a u u where u is a function of. OBJECTIVE Fin the Derivative of Composite Functions. E If y 6, fin y. y 6 6 y In this case, the is the f function an the polynomial erivative is foun by the Power Rule, but, as the erivative of the. 6 is the g. Each 6 is insie the, it is insie 6

17 E E e e e 8 8 e E 5 Given this table of values, fin at. f g an g f f g f ' g' f g f ' g g ' an g f g ' f f '. At, At, f g f ' g g ' g f g ' f f ' f ' 6 g '

18 E 6 Using the table from eample 5, fin f f. f f = 0 Of course, this kin of problem can eten to where we on t have a table of values, but where we still only have a function that we can erive symbolically. E 7 If g 5 an g ', fin ' g f if f e g 6 g Notice that while we o not actually know the function that g represents, we still can take its erivative, because we know the erivative of g is g. Of course, the Chain Rule is still essential in this process. g 6 g ' ' ' 6 ' f e g g f e g g g g g f ' e g ' g ' 6 g g ' 5 f ' e g ' 5 f ' 00 8 e e 5 5 8

19 . Homework Set A. 7. y e, fin y. f ( ), fin f '( )... Given the following table of values, fin the inicate erivatives. f ( ) f ( ) a. g(), where g( ) f ( ) b. h'( ), where h( ) f ( ) 9

20 5. Given this graph an u f g, v g f, an w f f (a) u '.5 (b) v ' (c) w ', fin y f() g() 6. If f () 7, fin f '( ). 0

21 7. If g() an g(), fin f () if g( ) f ( ) e. 8. f ; fin f ' y ; fin y

22 . Homework Set B. Given the table of values below, fin g ' if g f f ' h h' ln -7 ln ln e f h. If g an g ', fin ' f if g f e g. If h 5 an h', fin f ' if h f h e

23 . Fin f g h 5. If g 0 5 an g ' 0, fin '0 f if f g g 6 g

24 . Homework Set A y e, fin y. y e f ( ), fin f '( ). f '( ).. Given the following table of values, fin the inicate erivatives. f ( ) f ( ) a. g(), where g( ) f ( ) g() b. h'( ), where h( ) f ( ) h '( ) 6 5. Given this graph an u f g, v g f, an w f f (a) u '.5 (b) v ' (c) w ' y, fin f() g() 5 5 (a) u'.5 (b) v' (c) w' 9

25 7 6. If f () 6 f '( ) 7, fin f '( ). 7. If g() an g(), fin f () if f '( ) e g( ) f ( ) e. 8. f ; fin f ' f ' y ; fin y 6 y 7 7. Homework Set B. Given the table of values below, fin g ' if g f f ' h h' ln -7 ln ln g ' 6. If g f an g ', fin ' ' e f if e f h g f e g 5

26 . If h 5 an h', fin f ' if 5 f ' 5 e. Fin f g h f ' g h g ' h h' 5. If g 0 5 an g ' 0, fin '0 f ' 0 h f h e f if f g g 6 g 6

27 .: Introuction to Implicit Differentiation, an Avance Use of the Chain Rule One of the most common applications of the Chain Rule in Calculus is calle implicit ifferentiation. Recall that ifferentiation just means to take the erivative. All of what we i in precalculus was eplicit ifferentiation that is, the functions were eplicitly solve for the epenent variable (usually y). But we can actually use the Chain Rule to fin the erivative of functions an relations where we o not have the epenent variable isolate. E. Fin the y for the function y 5 y 5 y 0 Now, this is an obvious an easy eample, but notice what we have on the left sie of the equation in the process of taking the erivative, the y became a y y. That is because the erivative of y is! OBJECTIVES Fin Derivatives of functions an relations that are not eplicitly solve for y. As you saw in eample, the erivative of y is y. This applies wherever the y is locate in a problem. An if the y is locate within another function like y which is actually just (y), then the Chain Rule applies, as it has in every other aspect of the erivative we have ever worke with. 7

28 E. Fin y for 6 y 6 y 0 y y y y y y Simply take the erivative, using the Chain Rule on the y Then just use basic algebra to solve for y y So the erivative of 6 y is. y Note that just like the original equation, there is still a y in the solution of the problem. This is a very common occurrence in implicit ifferentiation. In the case above, we coul ve solve it eplicitly for y by using algebra, an one the erivative that way. But in many cases, this is not a possibility. E. Fin y for y e y 7 y e y 7 y y y y e 0 y y y y y e y y e 8

29 y y ye E. Fin the equations of the lines tangent to the relation y e y 7 at y = 0. Note that we were given a y-value, so we can plug that in an fin an -value. We nee a point to use to fin the equation of the tangent line. 0 0 e 7 6 So we have two tangent lines, one passing through (,0) an the other passing through (,0). From eample, we know that the two points. y ye y, so we fin the values for y at y,0 0 e 8 y 0,0 0 e 0 8 Now we use the points an slopes an get the tangent lines: y 8 an y8 9

30 . Homework Set A Fin y. for each of the following relations. y 6. e y y 5. 5y y 6. y e e 0y 5. Fin the equations for the lines tangent an normal to the relation y y 0 passing through the point (5,). 6. Fin the equations for the lines tangent to the relation y 8y 0 when = 0

31 . Homework Set A Fin y. y for each of the following relations. y 6. e y y 5 y y e y. 5y y 6. y y or y6 0y y 6 0y y 6 y e e 0y y e 0 ye y 5. Fin the equations for the lines tangent an normal to the relation y y 0 passing through the point (5,). Tangent: y 5 Normal: y 5 6. Fin the equations for the lines tangent to the relation y 8y 0 when = Tangent to (,0): y0 Tangent to (, ): y

32 .: Trig an Log Rules Trigonometric--Defn: "A function (sin, cos, tan, sec, csc, or cot) whose inepenent variable represents an angle measure." Means: an equation with sine, cosine, tangent, secant, cosecant, or cotangent in it. Logarithmic--Defn: "The inverse of an Eponential function." Means: there is a Log or Ln in the equation. u sin u cos u u cos u sin u u tan u sec u u csc u csc u cot u u sec u sec u tan u u cot u csc u u lnu u u loga u u ln a Note that all these Rules are epresse in terms of the Chain Rule. OBJECTIVES Fin Derivatives involving Trig an Logarithmic Functions.

33 E sin sin sin cos E sin sin cos cos E ln ln We coul have also simplifie algebraically before taking the erivative: ln ln ln Oftentimes it is significantly easier to take a erivative if we simplify the function first in the above case, it eliminate the nee for The Chain Rule.

34 Of course, composites can involve more than two functions. The Chain Rule has as many erivatives in the chain as there are functions. E sec 5 sec 5 5sec sec tan 5 60 sec tan E 5 ln cos ln cos sin cos tan tan

35 . Homework Set A For #-, fin the erivatives of the given functions. 5. y sin. y sec y a cos. y cot (sin ). 5. f ( t) tant 6. f ( ) ln cos 7. y cosa 8. y tan 5

36 9. f ( ) cos ln 5 0. f ( ) ln. f ( ) log0 sin. f ( ) log. Fin the equation of the tangent line to y cos at the point 0,.. Fin the equation of the tangent line to y sec cos. at the point, 6

37 5. Fin all points on the graph of f ( ) sin sin at which the tangent line is horizontal. 6. Fin the equations of the lines tangent an normal to the function f e ln 5 at the point (0,+ln5) when 7. Fin the equation of the line tangent to y cos. 7

38 8. Fin the equation of the line tangent to y sec cot point, 8. Use eact values in your answers. through the. Homework Set B. Given the table of values below, fin g ' if g f h f f ' h h' -7-8

39 . e csc ln cot sec. If g, g ', an f ' f g g cos e, fin sin g. If z ln cot secln, fin z 9

40 ln sec 5ln 7 5. z ln cost sec e 7, fin z t t 6. If z ln tant sin e 7, fin z t t 7. If 8. If z ln cos sinln, fin z 0

41 9. Fin f ' 6 when f cos 0. Fin ' g when g ln. ln 5 t. 5 sin ln 7t. csc ln 7. t ln e t6

42 sec 5 cot e 0ln 5. For problems 6 to 8, fin the first erivative for the following functions z y A y f r A Be 8. r y sinr 6. Be 7. 0 y ln tan 5e 7.. cos ln 5 sin ln 5

43 . Homework Set A. y sin. 5 y sec y cos 5 5 y' 0 sec tan y a cos. y cot (sin ) y cos sin y cos cot(sin )csc (sin ). 5. f ( t) tant 6. f ( ) ln cos sec t f '( t) tant 7. y cosa sin f '( ) tan 8. y tan 6tan sec y a 9. f ( ) cos ln sin ln y 5 0. f ( ) ln f '( ) f '( ) 5ln 5. f ( ) log0 sin. f ( ) log cos f '( ) ln0 f '( ) sin ln. Fin the equation of the tangent line to y cos y 0 0,. at the point. Fin the equation of the tangent line to y sec cos y. at the point,

44 5. Fin all points on the graph of f ( ) sin sin at which the tangent line is horizontal. n,, n, 6. Fin the equations of the lines tangent an normal to the function f e ln 5 at the point (0,+ln5) 5 y ln 5 0 when 7. Fin the equation of the line tangent to y cos y. 8. Fin the equation of the line tangent to y sec cot point, 8. Use eact values in your answers. y 8. Homework Set B through the. Given the table of values below, fin g ' if g f h f f ' h h' -7 g ' -

45 . e csc ln cot sec e csc csc csc cot sec tan cot. If g, g ' f ', an. If z ln cot secln z csc seclntan ln, fin z cot ln sec 5ln g f g g cos e 5ln 7 tan 5ln 7 t 6. If z ln cost sec e 7, fin z t z tant e t sece t tan e t t z ln tant sin e 7, fin z t z sec t t t e cos e t tant t 7. If 8. If z ln cos sinln z cosln, fin z tan 9. Fin f ' 6 f ' 0 6 when f cos sin, fin f ' 5

46 0. Fin ' g ' g when g ln. ln 5 5. csc ln 7 t t t. 5 sin ln 7t. csc ln 7 cot ln 7 7 sec 5 cot e 0ln 5sec5 tan 5 csc 0 e e 5. z y A y f r 5sin ln 7 cos ln 7 t 6t y sinr 6. Be 7. 0 ln 7t e A Be 8. r sin f r Ar Bcos r e y z' y 0Ay Be t6 y y r

47 ln tan 5e 7.. e e tan 5e sec 5 7 csc cos ln 5 sin ln 5 ln 5 7

48 .5: Prouct an Quotient Rules Remember: The Prouct Rule: v u uv u v u v u u v The Quotient Rule: v v OBJECTIVE Fin the Derivative of a prouct or quotient of two functions. E sin sin cos sin cos sin E u U, so v V, so u v V U f ( ) V ( ) ( ) ( ) ( ) 6 8 ( ) 85 ( ) 8

49 E 5 Notice that this problem becomes much easier if we simplify before applying the Quotient Rule. ( )( ) 5 ( )( ) ( )( ) ( )( ) ( ) ( ) E cot csc cot cot csc csc cot E 5 cos 5 5 cos 5 sin cos 5 ln5 5 ln5 cos sin 9

50 Remember that, in section., we sai we woul nee the Prouct Rule to eal with the erivative of a function where the variable is in both the base an the Eponent. We can now aress that situation. E 6 cos lncos cos e lncos e sin ln cos cos cos ln cos tan 50

51 .5 Homework Set A Fin the erivative of the following functions y t cost g( ). y tan sec. y sin 6. y y e 5

52 cos 7. y e 8. y r r 9. y sin 0. 5 y e cos. f ( ) ln. y ln e e. Fin the equation of the line tangent to y e at the point, e. 5

53 . If f(), fin f '( e ). 5. ht () ln 7, fin h'( t ) 6. y 5, fin y' 7. f ( ) 7 sin tan, fin f '. 5 5

54 8. Fin the equation of the lines tangent an normal to y sin ln when e 9. Fin the equation of the line tangent to sin ye when 0 0. Fin the equation of the lines tangent an normal to. y sin when 5

55 .5 Homework Set B t. Fin the first erivative for the following function: t e sin t 5t 5t. Fin the first erivative for the following function: t e tan t. Fin the first erivative for the following function: y 5 t. Fin the first erivative for the following function: t e t 5t 55

56 5. e sin sin y e ln cot e y y 7. sin ln 8. If h 5 an h', fin f ' if ln f h h 9. Fin g' z if gz 8 5z e 0. cos e ln z 56

57 .. cos 5 e ln g t if. 5 ln 5. Fin ' gt t t 5 5. e cos 6. tan ln 57

58 7. sinttant t 8. ln csc e 9. 5 ln sin 0. p 5psin p e ln p p p p. tan e 5. 5 ln 7 58

59 . 90. sin 5. 5 c c 9c c c g f h sin h 6. Given the table of values below, fin g ' if f f ' h h' -7-59

60 7. If g, g ', an f ' g f g g cos e, fin sin 60

61 .5 Homework Set A y t cost g( ) 7 g '( ) 7 9 y' t cost tsint. 5. y y' tan sec tan sec. y y' y e sin 6. y cos sin y' e y' cos 7. y e 8. cos ' y e cos sin y y' r r r 9. y sin 0. 5 y e cos y y 5 cos sin e 5cos sin. f ( ) ln. y ln e e ln f '( ) ln. Fin the equation of the line tangent to y e y e e y at the point, e. 6

62 . If f(), fin f '( e ). 5. ht () ln f '( e) 0 6. y y'() 5, fin y' 7 7. f ( ) 7 sin tan, fin f ' t t h'( t) t, fin h'( t) f '( ) 5 sin tan sin cos 8 tan sec 8. Fin the equation of the lines tangent an normal to y sin ln when e y e e y e e Tangent: Normal: Fin the equation of the line tangent to y sin ye when 0 0. Fin the equation of the lines tangent an normal to. Tan: Normal: y 8 y y sin when.5 Homework Set B t. Fin the first erivative for the following function: t e sin t 5t t ' t te 5t cost 5t sin t 5t 6

63 5t. Fin the first erivative for the following function: t e tan t 5 t ' t e t sec t 5tan t 5. Fin the first erivative for the following function: y y t. Fin the first erivative for the following function: t e t 5t ' t tet 5t 0t t 5. e sin 7 5 e sin e 7 5 cos sin 6. sin y e ln cot e y y sin y y e y e csc e e cos y e cot y sin y y ln cot e 7. sin ln cos sin ln ln 8. If h 5 an h', fin f ' if 750 f ' ln Fin g' z if gz g' z 8 ln f h h 5z e 0. cos e ln z 590z 8e 5z 5zln z sin cos 9 z ln z e 6

64 cos 5 e 5cos sin 5 e ln g t if. 5 ln 5. Fin ' ln 5 ln 5 5 ln 5 g' t gt 50tt t 6 t t 5 5. e cos e sin cos 6. tan ln sec ln tan ln 7. sinttant t 8. ln csc cot cot ln sintsec t costtan t csc e 5 e cot 0 ln sin 9. 5 ln sin 0. p 5psin p e ln p p p p p p p 6p p 5p e cos p e 5sin p e p p 6

65 . tan e 5. 5 ln 7 tan e 5 sec e 5 7ln ln 7 5. sin 5 cos c c 9c c c 8 c c g f sin f f ' h h' -7 - g ' 6. Given the table of values below, fin g ' if 7. If g, g ', an f ' 8 f ' g f g g cos e, fin sin 65

66 .6: Higher Orer Derivatives What we have been calling the Derivative is actually the First Derivative. There can be successive uses of the erivative rules, an they have meanings other than the slope of the tangent line. In this section, we will eplore the process of fining the higher orer erivatives. Secon Derivative--Defn: The erivative of the erivative. Just as with the First Derivative, there are several symbols for the n Derivative: Higher Orer Derivative Symbols Liebnitz: Function: Combination: y y = square y, square; y ; n n IV n f " = f ouble prime of ; f ''' ; f ; f y'' = y ouble prime OBJECTIVE Fin higher orer erivatives. E

67 E Fin y if y sin y sin y cos cos y sin sin 9 y cos cos 9 7 More complicate functions, in particular Composite Functions, have a complicate process. When the Chain Rule is applie, the answer often becomes a prouct or quotient. Therefore, the n Derivative will require the Prouct or Quotient Rules as well as, possibly, the Chain Rule again. E y e, fin y''. y e 6 6e y 6 e 6 e 6 6 e 6e 6e 6 E y sin, fin y'' y' sin cos y'' sin sin cos 6sin cos sin cos sin 67

68 E 5 f ln f., fin f f E 6 g, fin g. g 8 g

69 .6 Homework Set A In #-5, fin the first an secon erivatives of the given function. f ( ) 6 7. h( ). 5. H( t) tant. y t g() t t e 6. If 5. 5 y e, fin y ''. 69

70 7. If y sin, fin y ''. 8. If f ( t) tcost, fin f '''(0)..6 Homework Set B. Fin f ', f '', an f '" if f ln sec.. Fin y an y for y 0cot. For the function y tan, show that y 8sec tan. 70

71 . Fin " f for f ln 5. Fin the first, secon, an thir erivative for f 56 e 6. Fin the first, secon, an thir erivative for f tan 7

72 7. A fourth ifferentiable function is efine for all real numbers an satisfies each of the following: g 5, g ', an g If the function f is given by " a. Fin f, f ', f '' b. Show that the fourth erivative of f is k f e g, where k is a constant. k k e 6 g '''' 7

73 .6 Homework Set A 5. f ( ) 6 7. h( ) f '( ) 5 7 h'( ) f "( ) 0 h"( ). H( t) tant. y y y H t '( ) sec H "( t) 8sec t tan t t t g() t t e 6. If 5. 5 y e, fin y ''. 5t y'' 6e 6 g '( t) t e 5t g t te t t 5 "( ) t If y sin, fin y ''. 8. If f ( t) tcost, fin f '''(0). y'' sin cos sin f '''(0).6 Homework Set B. Fin f ', f '', an f '" if f ln sec tan sec f ' f '' f ''' sec tan. 7

74 . Fin y an y y 0cot y 0csc y 80csc cot for. For the function y tan, show that y y y sec secsec tan. Fin " f '' 8sec tan f for f ln 6 y 8sec tan. 5. Fin the first, secon, an thir erivative for f 56 e f ' 5 e f '' e f ''' e 6. Fin the first, secon, an thir erivative for f tan f ' sec f '' 8sec tan f ''' 5sec sec tan 7

75 7. A fourth ifferentiable function is efine for all real numbers an satisfies each of the following: 5 ' g" g, g, an k a. Fin f, f ', f '' If the function f is given by f e g, where k is a constant. b. Show that the fourth erivative of f is a. f 6, f b. ' k k f ' ke g ' f '' k, k f '' k e g '' k f ''' k e 8 g ''' k f '''' k e 6 g '''' k k e 6 g '''' 75

76 .7: Derivatives of Inverse Functions We alreay know something about inverse functions. Eponential an Logarithmic functions are inverses of each other, as are Raicals an Powers. Inverse Functions Defn: Two functions wherein the omain of one serves as the range of the other an vice versa. --Means: Two functions which cancel (or uno) each other. The efinition gives us a way to fin the inverse for any function, the symbol for which is f. You just switch the an y variables an isolate y. E Fin f if f f y So, for f, y y f This is little more interesting with a Rational Function. E Fin f if f f y So, for f, y y y y y y y y y y f 76

77 Note that. For Eample, 0 - f f f an f - 0. General inverses are not all that interesting. We are more intereste in particular inverse functions, like the natural logarithm (which is the inverse of the eponential function). Another particular kin of inverse function that bears more stuy is the Trig Inverse Function. Interestingly, as with the Logarithmic Functions, the erivatives of these Transcenental Functions become Algebraic Functions. Inverse Trig Derivative Rules sin cos tan u u u Du u Du u Du u csc - u u u sec - u u u cot - u D u D u u D u OBJECTIVE Fin the erivatives of inverse trig functions. Proof that sin. y sin sin y D sin y y cos y y cos y Note the use of implicit ifferentiation when we took the erivative of sin y 77

78 But this erivative is not in terms of, so we are not one. Consier the right triangle that woul yiel this SOHCAHTOA relationship: y - Note that for sin y to equal, must be the opposite leg an the hypotenuse is (SOH). The Pythagorean Theorem gives us the ajacent leg. By CAH, cos y Therefore, sin cos y Note that in the proof above, when we wante to take the erivative of sin y, we ha to use The Chain Rule (since the y is a function other than, an the erivative of y is y. E tan - tan

79 E 5 sec - sec - 79

80 .7 Homework Set A Fin the erivative of the function. Simplify where possible.. y sin e. y tan. ysin. H ( ) tan ( ) 5. y cos 6. f ( ) e arctan 8. ycsc 7. y sin 80

81 0. y cos 9. y cot tan. sec y for 0. y ln tan.7 Homework Set B. ytan z. y cos e 8

82 y sin. y cos. y sec csc 6. f ( t) csin t c t c f ( ) ln tan 5 7. f ( ) arccos 9. g w sin 5w cos 5w 0. f ( t) sec 9 t 8

83 . e y tan e. y ln u ucot u 8

84 .7 Homework Set A y sin e. y tan. y e e y. ysin. y H ( ) tan ( ) H '( ) tan ( ) 5. y cos 6. f ( ) e arctan y cos f '( ) arctan e 7. y sin y 8. ycsc y 0. y cos 9. y cot tan y 0 y. sec y for 0. y ln tan y sec y tan.7 Homework Set B z y cos e. ytan. y e z y 6z e 8

85 y sin. cos y y y. y sec csc 6. f ( t) csin t c t c y 0 c t f '( t) c t f ( ) ln tan 5 7. f ( ) arccos f '( ) arccos f '( ) 5 5 tan 5 9. g w sin 5w cos 5w 0. f ( t) sec 9 t t g' w 0 f '( t) 9t 8t. e y tan e y u cot y e u u u e. y ln u ucot u 85

86 .8: Local Linearity an Approimations Before calculators, one of the most valuable uses of the erivative was to fin approimate function values from a tangent line. Since the tangent line only shares one point on the function, y-values on the line are very close to y-values on the function. This iea is calle local linearity near the point of tangency, the function curve appears to be a line. This can be easily emonstrate with the graphing calculator by zooming in on the point of tangency. Consier the graphs of y. 5 an its tangent line at, y. 75. y The closer you zoom in, the more the line an the curve become one. The y-values on the line are goo approimations of the y-values on the curve. For a goo animation of this concept, see Since it is easier to fin the y-value of a line arithmetically than for other functions especially transcenental functions the tangent line approimation is useful if you have no calculator. OBJECTIVES Use the equation of a tangent line to approimate function values. 86

87 E Fin the tangent line equation to f () at = an use it to approimate value of f ( 09. ). The slope of the tangent line will be f '( ) f '( ) f '( ) [Note that we coul have gotten this more easily with the nderiv function on our calculator.] f ( ), so the tangent line will be y ( ) or y While we can fin the eact value of f ( 09. ) with a calculator, we can get a quick approimation from the tangent line. If = 0.9 on the tangent line, then f y This last eample is somewhat trite in that we coul have just plugge 0.9 into f () an figure out the eact value even without a calculator. It woul have been a pain, but a person coul actually o it by han because of the operations involve (there is only basic arithmetic involve). Consier the net eample, though. 87

88 f E Fin the tangent line equation to approimate value of e 0.. e at 0 an use it to Without a calculator, we coul not fin the eact value of e 0.. In fact, even the calculator only gives an approimate value. f ' e an f ' 0 e 0 f 0 e So the tangent line equation is y In orer to fin e 0., we are looking for. = 0. 0 or y 0. e 0.. f 0. Therefore, we plug in Note that the value that you get from a calculator for. e 0 is.0 Our approimation of. seems very reasonable. 88

89 .8 Homework Set A 5. Fin the equation of the tangent line to f ( ) 5 at an use it to get an approimate value of f Fin the tangent line equation to F( ) at an use it to get an approimate value of F... Fin all points on the graph of y sin sin where the tangent line is horizontal. 89

90 . Fin the equation of the tangent line at = for g ln to approimate g... Use this 5. Fin the equation of the line tangent to g 5 tan when. Use this tangent line to fin an approimation for g.. 6. Use a tangent line to fin the approimate the value of f.0 if sin f e. Then use your calculator to fin an actual value for.0 f. 90

91 7. Fin the equation of the tangent line for f ln sin for =. Plug in =. into both the original function an the tangent line. Eplain why the values are so similar. 8. Given the relation, y y, which passes through the point (,), fin each of the following: a. y b. y, c. The equation of the line tangent to the relation, passing through (,). An approimate value for y when =.0 9

92 .8 Homework Set A 5. Fin the equation of the tangent line to f ( ) 5 at an use it to get an approimate value of f.9. y 75 f Fin the tangent line equation to F( ) at an use it to get an approimate value of F.. y F..95. Fin all points on the graph of y sin sin where the tangent line is horizontal. n,, n,. Fin the equation of the tangent line at = for g ln to approimate g.. y g... Use this 5. Fin the equation of the line tangent to g 5 tan when. Use this tangent line to fin an approimation for g.. y 7 g Use a tangent line to fin the approimate the value of f.0 if sin f e. Then use your calculator to fin an actual value for.0 f.0.0; f f. 9

93 7. Fin the equation of the tangent line for f ln sin for =. Plug in =. into both the original function an the tangent line. Eplain why the values are so similar. y f..; f..090 The results are so similar because of local linearity. The tangent line is etremely close to the curve near the point of tangency, so values of y for the tangent line serve as goo approimations for values of f near the point of tangency. 8. Given the relation, y y, which passes through the point (,), fin each of the following: a. y y b. y, c. The equation of the line tangent to the relation, passing through (,) y. An approimate value for y when =.0 y 9

94 Chapter Test y tan cot ; fin y'. 7. D

95 . Fin the equation of the line tangent to f estimate f (0.98). at = an use it to g z 5. ln z z ; Fin g'(w) an g'(). 6. Fin the equation of the line tangent to y ln 5 at. 95

96 7. Fin f " f t 7 t. 6 if sin e cos 9. Fin y an the equation for the tangent line for the relation 5 y 6y 7 passing through the point (,). 96

97 Chapter Test ln y tan cot y D ; fin y' Fin the equation of the line tangent to f estimate f (0.98). y f g z 5. ln g' w z ; Fin g'(w) an g'(). z ww w w g ' Fin the equation of the line tangent to y ln y80 7. Fin f " f " 6 6 f t 7 t. 6 if sin at = an use it to 5 at. 97

98 e cos e cos sin 9. Fin y an the equation for the tangent line for the relation 5 y 6y 7 passing through the point (,). y 5 y 6 Tangent Line: y 98