Transcendental Functions

Transcription

1 78 Chapter 9 Transcenental Functions º½ 9 Transcenental Functions ÁÒÚ Ö ÙÒØ ÓÒ Informally, two functions f an g are inverses if each reverses, or unoes, the other More precisely: DEFINITION 9 Two functions f an g are inverses if for all in the omain of g, f(g()) =, an for all in the omain of f, g(f()) = EXAMPLE 9 f = an g = / are inverses, since ( ) / = an ( / ) = EXAMPLE 9 f = an g = / are not inverses While ( / ) =, it is not true that ( ) / = For eample, with =, (( ) ) / = / = The problem in the previous eample can be trace to the fact that there are two ifferent numbers with square equal to This turns out to be precisely escriptive of functions without inverses DEFINITION 9 Let A an B be sets an let f : A B be a function We say that f is injective or one-to-one if f() = f(y) implies that = y We say that f is surjective or onto if for every b B there is an a A such that f(a) = b If f is both injective an surjective, then f is bijective or one-to-one an onto We are intereste only in the case that A an B are sets of real numbers, an in this case there is a nice geometric interpretation of injectivity It is often easy to use this interpretation to ecie whether a function is or is not - THEOREM 95 Horizontal line test If f is a function efine on some subset of the real numbers, then f is injective if an only if every horizontal line intersects the graph of f at most once EXAMPLE 96 The function f = fails this test: horizontal lines y = k for k > intersect the graph of f twice (The horizontal line y = oes intersect it only once, an lines y = k, k <, o not intersect the graph at all) EXAMPLE 97 In each of these cases, we assume that f:a R, where A is the set of all real numbers for which f makes sense The function f() = is bijective The function f() = isneither injective nor surjective If we think of f as a function from R to the non-negative real numbers, then f is surjective; in other wors, if a function is not surjective this is not a major stumbling block The function f() = / is injective but not surjective since there is no value of such that f() = The function f() = ( )(+) is surjective but not injective; f() = for three ifferent values of On the other han lim f() = an lim f() = Since f is continuous on R, the intermeiate value theorem (56) guarantees that f takes all values between an The erivative furnishes us with a convenient criterion for injectivity without eplicitly looking for points where injectivity may fail THEOREM 98 If f is ifferentiable an f () > then f is injective Proof Suppose that f(a) = f(b) for some a < b By Rolle s theorem (65) there eists c (a,b) such that f (c) = (f(b) f(a))/(b a) =, which contraicts the hypothesis that f () > Hence, if f(a) = f(b) then a = b In the same way, we can see that if f () < then f is injective EXAMPLE 99 Let f() = 5 + Since f () = 5 + >, f is injective EXAMPLE 9 Let f() = + sin Then f () = + cos for every Hence, f is injective 77 9 Inverse functions 79 EXAMPLE 9 Let f() = This f is injective although the above theorem oes not apply, since f () = Therefore, the conitions in the theorem are sufficient but not necessary Our knowlege of erivatives can also lea us to conclue that a function is not injective THEOREM 9 If f is continuous an has a local maimum or minimum then f is not injective Proof Supposethatf hasalocalmaimumat = c Theninsomeinterval(c h,c+h), f() f(c) Let a (c h,c) If f(a) = f(c) then f is not injective; otherwise, f(a) < f(c) Let b (c,c + h) If f(b) = f(c) or f(b) = f(a) then f is not injective Otherwise, either f(b) < f(a) < f(c) or f(a) < f(b) < f(c) If f(b) < f(a) then by the intermeiate value theorem, there is a number (c,b) such that f() = f(a) an so f is not injective Likewise, if f(a) < f(b) then there is a number in (a,c) such that f() = f(b) an so f is not injective In every case, we see that f is not injective To return to our principal interest, inverse functions, we now connect bijections an inverses THEOREM 9 Suppose f:a B is a bijection Then f has an inverse function g:b A Proof Suppose b B Since f is onto, there is an a A such that f(a) = b Since f is, a is the only element of A with this property We let g(b) = a Now it is easy to see that for all a A, g(f(a)) = a an for all b B, f(g(b)) = b We really on t have any choice about how to efine g in this proof; if f is a bijection, its inverse is completely etermine Thus, instea of using a new symbol g, we normally refer to the inverse of f as f Unfortunately, it is often ifficult to fin an eplicit formula for the inverse of a given function, f, even if it is known that f is bijective Generally, we attempt to fin an inverse in this way: Write y = f() Interchange an y Solve for y Replace y with f () 8 Chapter 9 Transcenental Functions Step is the har part; inee it is sometimes impossible to perform using algebraic operations EXAMPLE 9 Fin the inverse of f() = ( 6)/( + 7) First we write = (y 6)/(y+7) Now we solve for y: y 6 = y +7 (y +7) = y 6 y +7 = y = y y 7+6 = y( ) 7+6 = y Finally, we say f () = (7+6)/( ) EXAMPLE 95 Fin the inverse function of f() = +8 where What are the omain an range of the inverse function? First, y = + 8 becomes = y y + 8 Now we complete the square: = (y ) +anrearrangetoget = (y ) Sinceintheoriginalfunction, an we have switche an y, we know that y Thus taking the square root, we know y =, not y = Finally we write y = f () = + The omain of f is an the range is y While simple in principle, this metho is sometimes ifficult or impossible to apply For eample, consier f() = + Since f () = + > for every, f is injective (In fact it is bijective) To fin the inverse as above, we woul nee to solve = y + y for y; while possible, this is consierably more ifficult than solving the quaratic of the previous eample Some simple looking equations are impossible to solve using algebraic manipulation For eample, consier f() = a quintic polynomial (ie, a fifth egree polynomial) Since f () = >, f is injective (an inee f is bijective) If there were a quintic formula, analogous to the quaratic formula, we coul use that to compute f Unfortunately, no such formula eists fifth egree equations cannot in general be solve (There are eceptions; 5 = can be solve, for eample) Fortunately, it is often more important to know that a function has an inverse then to be able to come up with an eplicit formula Once an inverse is known to eist, numerical

2 9 Inverse functions 8 techniques can often be employe to obtain approimations of the inverse function Thus, theorem 98 an proposition 9 provie useful criteria for eciing whether a function is invertible We now turn to the calculus of inverse functions THEOREM 96 Let A be an open interval an let f : A R be injective an continuous Then f is continuous on f(a) Proof Since A is an open interval an f is injective an continuous it follows by proposition 9 that f has no local maima or minima Hence, f is either strictly increasing or strictly ecreasing Without loss of generality, f is strictly increasing Fi b f(a) Then there eists a unique a A such that f(a) = b Let ǫ > an we may assume that (a ǫ,a+ǫ) A Let δ = min{b f(a ǫ),f(a+ǫ) b} an note that δ > since f is increasing Then the interval (b δ,b + δ) is mappe by f into (a ǫ,a+ǫ) Since ǫ was arbitrary, it follows that f is continuous at b Our principal interest in inverses is the simple relationship between the erivative of a function an its inverse THEOREM 97 Inverse function theorem Let A be an open interval an let f : A R be injective an ifferentiable If f () for every A then f is ifferentiable on f(a) an (f ) () = /f (f ()) Proof Fi b f(a) Then there eists a unique a A such that f(a) = b For y b, let = f (y) Since f is ifferentiable, it follows that f an hence f are continuous Then f (b) f (y) a lim = lim y b b y a f(a) f() = f (a) In Leibniz notation, this can be written as y =, which is easy to remember y/ since it looks like orinary fractional algebra EXAMPLE 98 Let f() = Since f() = 7, f ( 7) = Since f () = 9 +5, f () = 5 an so (f ) ( 7) = /f () = /5 Eercises 9 Which of the following functions are injective? Which are surjective? Which are bijective? Sketch the graph of each function to illustrate your answers a f:r R, f() = 8 Chapter 9 Transcenental Functions b f:[, ) R, f() = c f:r [, ), f() = f:(,] [, ), f() = Which of the following functions are injective? Which are surjective? Which are bijective? Sketch the graph of each function to illustrate your answers a f:r R, f() = b f:[, ) R, f() = c f:r [,], f() = sin f:[,π] [,], f() = cos Define f() = { = Show that f is not injective on R Show that f () > for Why oes this not contraict theorem 98? Define f() = { < = + < Show that f is injective an has a local minimum Why oes this not contraict theorem 9? 5 If A = R sketch the graph of the ientity function on A 6 Fin the inverse function of f() = ( 6)/(7 +5) What are the omain an range of f? 7 Fin the inverse function of f() = /( 6) What are the omain an range of f? 8 Fin the inverse function of f() = a+b when a What are the omain an range of f? 9 Fin the inverse function of f() = /(c+) when c What are the omain an range of f? Suppose that a bc Fin the inverse function of f() = (a+b)/(c+) What are the omain an range of f? (The omain an range will epen on which if any of a,b,c, an are zero) Note: The conition a bc is a technical conition which ensures that both the omain of f will be all real numbers with perhaps one eception an that that the range of f will be all real numbers with perhaps one eception Fin the inverse function of f() = for What are the omain an range of f? Fin the inverse function of f() = 5 What are the omain an range of f? Fin the inverse function of f() = 5 What are the omain an range of f? Fin the inverse function of f() = 7 What are the omain an range of f? 5 Fin the inverse function of f() = +8 for What are the omain an range of f? 9 The natural logarithm 8 6 Fin the inverse function of f() = 9+ for What are the omain an range of f? 7 Fin the inverse function of f() = + b + c for b/ What are the omain an range of f? 8 Fin the inverse function of f() = + b + c for b/ What are the omain an range of f? 9 Fin the inverse function of f() = (+ )/( ) What are the omain an range of f? Show that f() = 7 + has an inverse function on R Show that f() = 9/9 + 5 has an inverse function on R Let A R an let f : A R be injective Then f eists on f(a) Note that the point P(a,f(a)) is on the graph of f an that Q(f(a),a) is the corresponing point on the graph of f a Show that if a f(a) then the slope of the line segment PQ is b Conclue that if a f(a) the line segment PQ is perpenicular to the graph, L, of the ientity function on R c Show that the mipoint of PQ is on L Conclue that the graph of f is the graph of f reflecte through L Let f() = + Sketch the graph of f an f on the same iagram Let f() = Sketch the graph of f an f on the same iagram 5 a Suppose that f is an increasing function on R What can you say about f? b Suppose that f is a concave up function on R What can you say about f? In both parts, use eercise to illustrate your claim 6 Let f() = +9+ Compute (f ) () 7 Let f() = + Show that f is increasing at = Thus, there is an interval I containing such that f is injective on I Compute (f ) () 8 Let f() = a+b with a Compute (f ) (b) Why o we nee the conition a? 9 Let f() = a + b + c with b Compute (f ) (c) Why o we nee the conition b? Let f() = an n +an n + +a+a = n k= akk with a Compute (f ) (a) Why o we nee the conition a? Suppose that f is injective on some interval containing If f() = an f () = 6 what is (f ) ()? º¾ Ì Ò ØÙÖ Ð ÐÓ Ö Ø Ñ The function f(t) = /t is continuous on (, ) By the funamental theorem of calculus, f has an antierivative on on the interval with en points an whenever > This observation allows us to make the following efinition 8 Chapter 9 Transcenental Functions DEFINITION 9 The natural logarithm ln() is an antierivative of /, given by ln = t t Figure 9 gives a geometric interpretation of ln Note that when <, ln is negative area is ln Figure 9 ln() is an area Some properties of this function ln are now easy to see THEOREM 9 Suppose that,y > an q Q a ln = b ln() = c ln(y) = ln+lny ln(/y) = ln lny e ln q = qln area is ln Proof Part (a) is simply the Funamental Theorem of Calculus (7) Part (b) follows irectly from the efinition, since ln() = t t Part (c) is a bit more involve; start with: y ln(y) = t t = t t+ y t = ln()+ t y t t

3 In the remaining integral, use the substitution u = t/ to get y y t t = y u u = u = ln(y) u Parts () an (e) are left as eercises 9 The natural logarithm 85 Part (e) is in fact true for any real number q (not just rationals) but one of the points of our approach here is to give a rigorous efinition of real powers which so far we have not one We now turn to the task of sketching the graph of ln THEOREM 9 ln is increasing an its graph is concave own everywhere Proof Since ln = / ispositivefor >, the Mean Value Theorem (65)implies that ln is increasing The secon erivative of ln is then / which is negative, so the graph is concave own Notice that this theorem implies that ln is injective THEOREM 9 lim ln = Proof Note that ln > an for n N, ln n = nln Since ln is increasing, when > n, ln() > nln Since lim nln =, also lim ln = n COROLLARY 95 lim + ln = Proof If < <, then (/) > an lim +(/) = Let y = /; then lim + ln = lim y ln(/y) = lim y ln() ln(y) = lim y ln(y) = Thus, the omain of ln is (, ) an the range is R; ln() is shown in figure 9 86 Chapter 9 Transcenental Functions By the intermeiate value theorem (56) there is a number e such that lne = The number e is also known as Napier s constant It turns out that e is not rational In fact, e is not the root of a polynomial with rational coefficients which means that e is a transcenental number We will not prove these assertions here The value of e is approimately 78 EXAMPLE 96 Let f() = ln( 5 +7+) Compute f () Using the chain rule: f () = (5+7) EXAMPLE 97 Let f() = ln( ) for < Compute f () f () = ( ) = So the erivatives of ln() an ln( ) are the same Thus, you will often see ln +C as the general antierivative of / EXAMPLE 98 Compute tan Use u = cos: sin tan = cos = u = ln u +C = ln cos +C u Using one of the properties of the logarithm, we coul go further: ln cos +C = ln (cos) +C = ln sec +C = EXAMPLE 99 Let f() = 8 (+) 6/7 ( + +) 6 Compute f () Computing the erivative irectly is straightforwar but irritating We therefore take an inirect approach Note that f() > for every Let g() = lnf() Then g () = f ()/f() an so f () = f()g () Now ( 8 (+) 6/7 ) g() = ln ( + +) 6 e 5 6 Figure 9 The graph of ln() Hence, Therefore, = 8ln+ 6 7 ln(+) 6ln( + +) g () = (+) 6(9 +8) + + f () = 8 (+) 6/7 ( ) 8 ( + +) (+) 6(9 +8) + + Eercises 9 Prove parts () an (e) of theorem 9 9 The eponential function 87 In subsequent eercises, it is unerstoo that the arguments in any logarithms are positive unless otherwise state Epan ln((+5) 7 ( )) Epan ln 5+(7/) Sketch the graph of y = ln( 7) + 5 Sketch the graph of y = ln for 6 Write ln+7ln( ) ln( ++) as a single logarithm 7 Differentiate f() = ln 8 Differentiate f() = ln(ln()) 9 Sketch the graph of ln( ) Differentiate f() = +ln( ) +ln() Differentiate f() = ln sec + tan Fin the secon erivative of f() = ln( ) Fin the equation of the tangent line to f() = ln at = a Differentiate f() = 8 ( ) / 7 6 ( 6) 8 5 If f() = ln( +) compute f (e / ) e 6 Compute ln 7 Compute the erivative with respect to of lntt (Assume that > ) π/6 8 Compute tan() ln 9 Compute sin() Compute +cos Fin the volume of the soli obtaine by rotating the region uner y = / from to e about the -ais º Ì ÜÔÓÒ ÒØ Ð ÙÒØ ÓÒ In this section, we efine what is arguably the single most important function in all of mathematics We have alreay note that the function ln is injective, an therefore it has an inverse 88 Chapter 9 Transcenental Functions DEFINITION 9 The inverse function of ln() is y = ep(), calle the natural eponential function The omain of ep() is all real numbers an the range is (, ) Note that because ep() is the inverse of ln(), ep(ln) = for >, an ln(ep) = for all Also, our knowlege of ln() tells us immeiately that ep() = e, ep() =, lim ep =, an lim ep = THEOREM 9 ep() = ep() Proof By the Inverse Function Theorem (97), ep() has a erivative everywhere The theorem also tells us what the erivative is Alternately, we may compute the erivative using implicit ifferentiation: Let y = ep, so lny = Differentiating with respect to we get y y = Hence, y = y = ep COROLLARY 9 Since ep >, ep is an increasing function whose graph is concave up The graph of the natural eponential function is inicate in figure 9 Compare this to the graph of ln, figure e Figure 9 The graph of ep() COROLLARY 9 The general antierivative of ep is ep+c

4 9 The eponential function 89 Of course, the wor eponential alreay has a mathematical meaning, an this meaning etens in a natural way to the eponential function ep() LEMMA 95 For any rational number q, ep(q) = e q Proof Let y = e q Then lny = ln(e q ) = qlne = q, an so y = ep(q) In view of this lemma, we usually write ep() as e for any real number Conveniently, it turns out that the usual laws of eponents apply to e THEOREM 96 For every,y R an q Q: (a) e +y = e e y (b) e y = e /e y (c) (e ) q = e q Proof Parts(b)an(c)areleftaseercises Forpart(a),ln(e e y ) = lne +lne y = +y, so e e y = e +y EXAMPLE 97 Solve e +5 = for If e +5 = then +5 = ln an so = ln 5 EXAMPLE 98 Fin the erivative of f() = e sin() By the prouct an chain rules, f () = e sin()+e cos() EXAMPLE 99 Evaluate e Let u =, so u = Then Eercises 9 Prove parts (b) an (c) of theorem 96 Solve ln(+ ) = 6 for Solve e = 8 for Solve ln(ln()) = for 5 Sketch the graph of f() = e Sketch the graph of f() = e +6 e = e u u = eu = e +C 7 Fin the equation of the tangent line to f() = e at = a 8 Compute the erivative of f() = e Chapter 9 Transcenental Functions 9 Compute the erivative of f() = e (++ + n + +! n! Prove that e > for Then prove that e > + for Using the previous two eercises, prove (using mathematical inuction) that e > + + n + n + +! n! = k for k! k= Use the preceing eercise to show that e > 7 Differentiate ek +e k with respect to e +e Compute lim e e 5 Integrate 5 e 5 with respect to π/ 6 Compute cos()e sin e / 7 Compute 8 Let F() = e e t t Compute F () 9 If f() = e k what is f (9) ()? º ÇØ Ö Notice that if q Q an a > then a q = e ln(aq) = e qlna This equation motivates the following efinition DEFINITION 9 For a > an R, we efine a = e lna The function f() = a is the eponential function with base a Separately, we efine = for > Notice also that for R an a >, lna = ln(e lna ) = lna Hence, the power rule for the natural logarithm works even when the power is irrational We now show that the familiar rules for eponents are vali THEOREM 9 For,y R an a,b > : Proof a a +y = a a y b a y = a /a y c (a ) y = a y (ab) = a b ) (a) We compute: a +y = e (+y)lna = e lna+ylna = e lna e ylna = a a y The proof of (b) is similar an left as an eercise (c) We compute: (a ) y = e yln(a) = e ylna = a y () We compute: (ab) = e ln(ab) = e lna+lnb = e lna e lnb = a b THEOREM 9 If f() = a (with a > ) then f () = a lna Proof f () = (elna ) = e lna lna = a lna COROLLARY 9 For a > an a, a = a lna +C We are now in a position to prove the general power rule 9 Other bases 9 THEOREM 95 Power Rule If f() = n, >, an n is any real number, then f () = n n Proof f () = n = enln = e nlnn = nn = nn The restriction that > is necessary since we have not efine eponential epressions with negative bases an arbitrary real powers We now turn to logarithms base a Note that if a > an a then a lna for every Hence, the function f() = a is injective DEFINITION 96 If a > an a, the inverse of a is calle the logarithmic function base a In symbols, we write this function as log a We eclue a = because = is not injective on any omain containing more than one point Remark If a = we usually write log instea of log, an of course log e = ln In more avance tets, log refers to the natural logarithm THEOREM 97 The following hol for a,,y >, a, an q R: a log a (y) = log a +log a y b log a y = log a log a y c log a q = qlog a Proof (a) Let u = log a an v = log a y Then a u = an a v = y, an y = a u a v = a u+v, so log a (y) = u+v = log a +log a y 9 Chapter 9 Transcenental Functions The other parts are left as eercises When computing ecimal approimations to logs of arbitrary bases with a calculator or a computer algebra system the following result comes in hany LEMMA 98 If a,b >, a,b, an >, log a = log b log b a Proof Let y = log a, so a y = Then log b = log b (a y ) = ylog b a = log a log b a Typically this is useful when b = e an b =, since calculators can typically compute logarithms to those bases THEOREM 99 log a = lna Proof By the preceing lemma, f() = ln/lna, an the erivative is then easy Finally, we epress e as a limit When = we get a limit epression for e which is sometimes taken as the efinition of e ( THEOREM 9 If, e = lim + n n n) Proof If = both epressions are If > we begin by rewriting the right sie as we have before: Now because e is continuous, ( + ) n ( ) n = e ln(+/n) = e nln(+/n) n lim n enln(+/n) = e limn nln(+/n) So really we nee to compute lim nln(+/n), for which we use L Hôpitals rule: n ln(+/n) +/n n lim nln(+/n) = lim = lim n n /n n /n = lim n +/n = This same simple fact, a = e lna, is useful in many similar situations EXAMPLE 9 Let f() =, > Compute f () an lim +f()

5 Start with f() = = e ln Then For the limit, we again notice that f () = e ln ( +ln ) = (+ln) lim = lim e ln = e lim + ln + + Then we compute the limit by L Hôpital s rule again: Thus lim + = e = ln lim +ln = lim + / = lim / + / = lim +( ) = 9 Other bases 9 π/ EXAMPLE 9 Compute cos sin π/6 Let u = cos, so u = sin Changing the limits, when = π/6, u = /, an when = π/, u = / Then Eercises 9 π/ / cos sin = u u = u / π/6 / ln = / + / / ln Prove part (b) of theorem 9 Sketch the graph of y = a in the three cases a >, a =, an < a < What happens to the graph as a +? What happens to the graph as a? Sketch the graph of y = log a in the two cases a > an < a < What happens to the graph as a +? What happens to the graph as a? (Use the previous eercise together with eercise in section 9) Prove parts (b) an (c) of theorem 97 5 Sketch the graph of y = Sketch the graph of y = (/) 7 Sketch the graph of y = log (+6) 8 Compute the secon erivative of f() = 9 Compute f (π/) when f() = 5 sin +log 7 Compute the erivative of f() = +sin() π e 9 Chapter 9 Transcenental Functions Compute Compute sin( ) Fin the area of the region given by {(,y), y } Fin the average of the function f() = 5 on the interval [,9] 5 Fin the volume of the soli obtaine by rotating the region {(,y),(log )/ y } about the line y = 6 Show that log a = log /a for any a >,a Interpret this result geometrically; that is, sketch the graph of y = log a an y = log /a on the same iagram an point out how the graphs are relate to each other º ÁÒÚ Ö ÌÖ ÓÒÓÑ ØÖ ÙÒØ ÓÒ The trigonometric functions frequently arise in problems, an often it is necessary to invert the functions, for eample, to fin an angle with a specifie sine Of course, there are many angles with the same sine, so the sine function oesn t actually have an inverse that reliably unoes the sine function If you know that sin = 5, you can t reverse this to iscover, that is, you can t solve for, as there are infinitely many angles with sine 5 Nevertheless, it is useful to have something like an inverse to the sine, however imperfect The usual approach is to pick out some collection of angles that prouce all possible values of the sine eactly once If we iscar all other angles, the resulting function oes have a proper inverse The sine takes on all values between an eactly once on the interval [ π/,π/] If we truncate the sine, keeping only the interval [ π/,π/], as shown in figure 95, then this truncate sine has an inverse function We call this the inverse sine or the arcsine, an write y = arcsin() π π/ π/ π π/ π/ π/ π π/ π π/ π/ Figure 95 The sine, the truncate sine, the inverse sine 95 Inverse Trigonometric Functions Chapter 9 Transcenental Functions Recall that a function an its inverse uno each other in either orer, for eample, ( ) = an = This oes not work with the sine an the inverse sine because the inverse sine is the inverse of the truncate sine function, not the real sine function It is true that sin(arcsin()) =, that is, the sine unoes the arcsine It is not true that the arcsine unoes the sine, for eample, sin(5π/6) = / an arcsin(/) = π/6, so oing first the sine then the arcsine oes not get us back where we starte This is because 5π/6 is not in the omain of the truncate sine If we start with an angle between π/ an π/ then the arcsine oes reverse the sine: sin(π/6) = / an arcsin(/) = π/6 What is the erivative of the arcsine? Since this is an inverse function, we can iscover the erivative by using implicit ifferentiation Suppose y = arcsin() Then π π/ π/ π Figure 95 The truncate cosine, the inverse cosine sin(y) = sin(arcsin()) = π/ Now taking the erivative of both sies, we get y cosy = y = cosy π/ π/ π/ π/ π/ As we epect when using implicit ifferentiation, y appears on the right han sie here We woul certainly prefer to have y written in terms of, an as in the case of ln we can actually o that here Since sin y + cos y =, cos y = sin y = So cosy = ±, but which is it plus or minus? It coul in general be either, but this isn t in general : since y = arcsin() we know that π/ y π/, an the cosine of an angle in this interval is always positive Thus cosy = an arcsin() = Note that this agrees with figure 95: the graph of the arcsine has positive slope everywhere We can o something similar for the cosine As with the sine, we must first truncate the cosine so that it can be inverte, as shown in figure 95 Then we use implicit ifferentiation to fin that arccos() = Note that the truncate cosine uses a ifferent interval than the truncate sine, so that if y = arccos() we know that y π The computation of the erivative of the arccosine is left as an eercise Figure 95 The tangent, the truncate tangent, the inverse tangent Finally we look at the tangent; the other trigonometric functions also have partial inverses but the sine, cosine an tangent are enough for most purposes The tangent, truncate tangent an inverse tangent are shown in figure 95; the erivative of the arctangent is left as an eercise Eercises 95 Show that the erivative of arccos is Show that the erivative of arctan is + The inverse of cot is usually efine so that the range of arccot is (,π) Sketch the graph of y = arccot In the process you will make it clear what the omain of arccot is Fin the erivative of the arccotangent Show that arccot+arctan = π/ 5 Fin the erivative of arcsin( ) 6 Fin the erivative of arctan(e ) 7 Fin the erivative of arccos(sin ) 8 Fin the erivative of ln((arcsin) ) 9 Fin the erivative of arccose

6 Fin the erivative of arcsin+arccos Fin the erivative of log 5 (arctan( )) arcsec Compute ln(arcsin) Compute arcsin Compute 5 Compute ( + +e + ) +9 6 Fin the equation of the tangent line to f() = arccsc at = π/6 7 Let º { A = (,y) }, y ( ) / 96 Hyperbolic Functions 97 Sketch the region A Let S be the soli obtaine from rotating A about the -ais Compute the volume of S ÀÝÔ Ö ÓÐ ÙÒØ ÓÒ The hyperbolic functions appear with some frequency in applications, an are quite similar in many respects to the trigonometric functions This is a bit surprising given our initial efinitions DEFINITION 96 The hyperbolic cosine is the function an the hyperbolic sine is the function cosh = e +e, sinh = e e Notice that cosh is even (that is, cosh( ) = cosh()) while sinh is o (sinh( ) = sinh()), an cosh +sinh = e Also, for all, cosh >, while sinh = if an only if e e =, which is true precisely when = LEMMA 96 The range of cosh is [, ) 98 Chapter 9 Transcenental Functions Proof Let y = cosh We solve for : y = e +e y = e +e ye = e + = e ye + e = y ± y e = y ± y From the last equation, we see y, an since y, it follows that y Now suppose y, so y± y > Then = ln(y± y ) is a real number, an y = cosh, so y is in the range of cosh() DEFINITION 96 The other hyperbolic functions are tanh = sinh cosh coth = cosh sinh sech = cosh csch = sinh The omain of coth an csch is while the omain of the other hyperbolic functions is all real numbers Graphs are shown in figure 96 Figure 96 The hyperbolic functions: cosh, sinh, tanh, sech, csch, coth 96 Hyperbolic Functions 99 Certainly the hyperbolic functions o not closely resemble the trigonometric functions graphically But they o have analogous properties, beginning with the following ientity THEOREM 96 For all in R, cosh sinh = Proof The proof is a straightforwar computation: cosh sinh = (e +e ) (e e ) This immeiately gives two aitional ientities: = e ++e e + e tanh = sech an coth = csch = = The ientity of the theorem also helps to provie a geometric motivation Recall that the graph of y = is a hyperbola with asymptotes = ±y whose -intercepts are ± If (,y) is a point on the right half of the hyperbola, an if we let = cosht, then y = ± = ± cosh t = ±sinht So for some suitable t, cosht an sinht are the coorinates of a typical point on the hyperbola In fact, it turns out that t is twice the area shown in the first graph of figure 96 Even this is analogous to trigonometry; cost an sint are the coorinates of a typical point on the unit circle, an t is twice the area shown in the secon graph of figure 96 (cosht, sinht) (cost, sint) Figure 96 Geometric efinitions of sin, cos, sinh, cosh: t is twice the shae area in each figure Given the efinitions of the hyperbolic functions, fining their erivatives is straightforwar Here again we see similarities to the trigonometric functions THEOREM 965 cosh = sinh an sinh = cosh Chapter 9 Transcenental Functions Proof cosh = e +e = e e = sinh, an sinh = e e = e +e = cosh Of course, this immeiately gives us two anti-erivatives as well Since cosh >, sinh is increasing an hence injective, so sinh has an inverse, arcsinh Also, sinh > when >, so cosh is injective on [, ) an has a (partial) inverse, arccosh The other hyperbolic functions have inverses as well, though arcsech is only a partial inverse We may compute the erivatives of these functions as we have other inverse functions THEOREM 966 arcsinh = + Proof Let y = arcsinh, so sinhy = Then sinhy = cosh(y) y =, an so y = coshy = +sinh y = + The other erivatives are left to the eercises Eercises 96 Show that the range of sinh is all real numbers (Hint: show that if y = sinh then = ln(y + y +)) Compute the following limits: a lim cosh b lim sinh c lim tanh lim (cosh sinh) Show that the range of tanh is (,) What are the ranges of coth, sech, an csch? (Use the fact that they are reciprocal functions) Prove that for every,y R, sinh( + y) = sinhcoshy + coshsinhy Obtain a similar ientity for sinh( y) 5 Prove that for every,y R, cosh( + y) = coshcoshy + sinhsinhy Obtain a similar ientity for cosh( y) 6 Use eercises an 5 to show that sinh() = sinhcosh an cosh() = cosh +sinh for every Conclue also that (cosh() )/ = sinh 7 Show that (tanh) = sech Compute the erivatives of the remaining hyperbolic functions as well 8 What are the omains of the si inverse hyperbolic functions?

7 96 Hyperbolic Functions 9 Sketch the graphs of all si inverse hyperbolic functions The following four eercises epan on the geometric interpretation of the hyperbolic functions Refer to figure 96 Use eercises an 5 to show that sinh() = sinhcosh an cosh() = cosh +sinh for every Conclue that (cosh() )/ = sinh Compute (Hint: make the substitution u = arccosh an then use the preceing eercise) Fi t > Sketch the region R in the right half plane boune by the curves y = tanht, y = tanht, an y = Note well: t is fie, the plane is the -y plane Prove that the area of R is t