.6 Derivatives of Trigonometric an Hyperbolic Functions Derivatives of the six trigonometric functions Derivatives of inverse trigonometric functions Hyperbolic functions, inverse hyperbolic functions, an their erivatives Derivatives of Trigonomteric Functions Because trigonometric functions have perioic oscillating behavior, an their slopes also have perioic oscillating behavior, it woul make sense if the erivatives of trigonometric functions were trigonometric. For example, the two graphs below show the function f(x) = sin x an its erivative f (x) =cosx. At each value of x, it turns out that the slope of the graph of f(x) = sin x is given by the height of the graph of f (x) =cosx. Check this for the values x = 5., x = π,anx =4: Slopes of f(x) =sinx at three points Heights of f (x) =cosx at three points 5. π 4 5. π 4 The six trigonometric functions have the following erivatives: Theorem.7 Derivatives of the Trigonometric Functions For all values of x at which the functions below are efine, we have: x (sin x)=cosx x (cos x) = sin x (c) () x (tan x)=sec x x (sec x) =secx tan x (e) (f) x (cot x) = csc x x (csc x)= csc x cot x It is important to note that these erivative formulas are only true if angles are measure in raians; see Exercise 5. Proof. We will prove the formulas for sin x an tan x from parts an (c) an leave the proofs of the remaining four formulas to Exercises 8 84. The proof of the first formula is nothing more than an annotate calculation using the efinition of erivative. To simplify the limit we obtain we will rewrite sin(x + h) with a trigonometric ientity. Our goal after that will be to rewrite the limit so that we can apply the
.6 Derivatives of Trigonometric an Hyperbolic Functions 3 two trigonometric limits from Theorem.34 in Section.6. sin(x + h) sin x x (sin x) = lim h 0 h (sin x cos h + sin h cosx) sin x = lim h 0 h sin x(cos h ) + sin h cosx = lim h 0 h = lim sin x cosh +cosx sin h ) h 0 ( ( = sin x lim h 0 h cos h h ) +cosx h ( lim h 0 sin h h ) efinition of erivative sum ientity for sine algebra algebra limit rules = (sin x)(0) + (cos x)() = cos x. trigonometric limits (c) We o not have to resort to the efinition of erivative in orer to prove the formula for ifferentiating tan x. Instea we can use the quotient rule, the fact that tan x = sin x cos x, an the formulas for ifferentiating sin x an cos x: ( x (tan x) = sin x ) x cos x = x (sin x) (cos x) (sin x) x (cos x) (cos x) (cos x)(cos x) (sin x)( sin x) = cos x = cos x + sin x cos = x cos x =sec x. quotient rule erivatives of sin x an cos x algebra an ientities Derivatives of Trigonometric Functions We can use the formulas for the erivatives of the trigonometric functions to prove formulas for the erivatives of the inverse trigonometric functions. Interestingly, although inverse trigonometric functions are transcenental, their erivatives are algebraic: Theorem.8 Derivatives of Inverse Trigonometric Functions For all values of x at which the functions below are efine, we have: x (sin x)= x x (tan x)= +x (c) x (sec x)= x x These erivative formulas are particularly useful for fining certain antierivatives, an in Chapter xxx they will be part of our arsenal of integration techniques. Of course, all of these rules can be use in combination with the sum, prouct, quotient, an chain rules. For example, x (sin (3x +))= (3x +) (6x)= 6x x. Proof. We will prove the rule for sin x an leave the remaining two rules to Exercises 85 an 86. We coul apply Theorem.3 here, but it is just as easy to o the implicit ifferentia-
.6 Derivatives of Trigonometric an Hyperbolic Functions 4 tion by han. Since sin(sin x)=x for all x in the omain of sin x,wehave: sin(sin x)=x x (sin(sin x)) = x (x) cos(sin x) x (sin x)= x (sin x)= x (sin x)= cos(sin x) sin (sin x) x (sin x)=. x sin x is the inverse of sin x ifferentiate both sies chain rule algebra since sin x +cos x = sin x is the inverse of sin x We coul also have use triangles an the unit circle to show that the composition cos(sin x) is equal to the algebraic expression x, as we i in Example 4 of Section 0.4. An interesting fact about the erivatives of inverse sine an inverse secant is that their omains are slightly smaller than the omains of the original functions. Below are the graphs of the inverse trigonometric functions an their omains. f(x) =sin x has omain [, ] " " g(x)=tan x has omain (, ) " " h(x) =sec x has omain (, ] [, ) " " -!"!" - -!"!" -!"!" If you look closely at the first an thir graphs above you shoul notice that at the ens of the omains the tangent lines will be vertical. Since a vertical line has unefine slope, the erivative oes not exist at these points. This means that the erivatives of sin x an sec x are not efine at x =or x = ; see the first an thir graphs below. f (x) = x has omain (, ) 3 g (x) = +x has omain (, ) 3 h (x) = x x has omain (, ) (, ) 3 - - - - - - Hyperbolic Functions an Their Derivatives* The trigonometric functions sine an cosine are circular functions in the sense that they are efine to be the coorinates of a parameterization of the unit circle. This means that the circle efine by x + y =is the path trace out by the coorinates (x, y)=(cost, sin t) as t varies; see the figure below left.
.6 Derivatives of Trigonometric an Hyperbolic Functions 5 Points on the circle x + y = Points on the hyperbola x y = y y (x,y) = (cos t, sin t) (x,y) = (cosh t, sinh t) x - - x - - - - - - Now let s consier the path trace out by the hyperbola x y =as shown above right. One parameterization of the right half of this hyperbola is trace out by the hyperbolic functions (cosh t, sinh t) that we will spen the rest of this section investigating. The hyperbolic functions are nothing more than simple combinations of the exponential functions e x an e x : Definition.9 Hypberbolic Sine an Hyperbolic Cosine For any real number x, the hyperbolic sine function an the hyperbolic cosine function are efine as the following combinations of exponential functions: sinh x = ex e x cosh x = ex + e x The hyperbolic sine function is pronounce sinch an the hyperbolic cosine function is pronounce cosh. The h is for hyperbolic. As we will soon see, the properties an interrelationships of the hyperbolic functions are similar to the properties an interrelationships of the trigonometric functions. These properties will be particularly useful in Chapter?? when we are attempting to solve certain forms of integrals. It is a simple matter to use the efinition above to verify that for any value of t, the point (x, y)=(cosht, sinh t) lies on the hyperbola x y =; see Exercise 87. We will usually think of this fact with the variable x, as this ientity: cosh x sinh x =. Here we are using the familiar convention that, for example, sinh x is shorthan for (sinh x). Note the similarity between the hyperbolic ientity cosh t sinh t =an the Pythagorean ientity for sine an cosine. Hyperbolic functions also satisfy many other algebraic ientities that are reminiscent of those that hol for trigonometric functions, as you will see in Exercises 88 90. Just as we can efine four aitional trigonometric functions from sine an cosine,we can efine four aitional hyperbolic functions from hyperbolic sine an hyperbolic cosine. We will be primarily intereste in the hyperbolic tangent function: tanh x = sinh x cosh x = ex e x e x + e x. We can also efine csch x, sechx, an coth x as the reciprocals of sinh x, cosh x, antanh x, respectively. The graphs of sinh x, cosh x, antanh x are shown below. In Exercises 3 6 you will investigate various properties of these graphs.
.6 Derivatives of Trigonometric an Hyperbolic Functions 6 y =sinhx y =coshx y =tanhx 8 4 8 6-3 - - 3-4 -8 4-3 - - 3-3 - - 3 - In Chapter 4 we will see that the mile graph of y =coshx is an example of a catenary curve, which is the shape forme by a hanging chain or cable. As with any functions that we stuy, we are intereste in fining formulas for the erivatives of sinh x, coshx, an tanh x. The similarity between hyperbolic functions an trigonometric functions continues here. These erivatives follow a very familiar pattern, iffering from the pattern for trigonometric functions only by a sign change. Theorem.0 Derivatives of Hyperbolic Functions For all real numbers x, wehave: x (sinh x)=coshx x (cosh x) = sinh x (c) x (tanh x) =sech x If you prefer to stay away from the hyperbolic secant functionsechx, you can write the thir erivative above as cosh x. Proof. The proofs of these ifferentiation formulas follow immeiately from the efinitions of the hyperbolic functions as simple combinations of exponential functions. For example, x (sinh x) = The remaining proofs are left to Exercises 9 9. x ( (ex e x )) = (ex + e x ) = coshx. Although hyperbolic functions may seem somewhat exotic, they work with the other ifferentiation rules just like any other functions. For example, with the prouct an chain rules we can calculate: x (5x sinh3 x ) = 5 sinh 3 x +5x(3 sinh x )(cosh x )(x). The erivatives of the remaining three hyperbolic functions are also very similar to those of their trigonometric cousins, but at the moment we will be focusing only on hyperbolic sine, cosine, an tangent. Inverse Hyperbolic Functions an Their Derivatives* For a function to have an inverse, it must be one-to-one. Looking back at the graphs of sinh x, cosh x, antanh x, we see that only cosh x fails to be one-to-one. Just as when we efine the trigonometric inverses, we will restrict the omain of cosh x to a smaller omain on which it is one-to-one. We will choose the restricte omain of cosh x to be x 0. The notation we will use for the inverses of these three functions is what you woul expect: sinh x, cosh x an tanh x. Since the hyperbolic functions are efine as combinations of exponential functions, it woul seem reasonable to expect that their inverses coul be expresse in terms of logarithmic functions. This is in fact the case, as you will see in Exercises 95 97. However, our main
.6 Derivatives of Trigonometric an Hyperbolic Functions 7 concern here is to fin formulas for the erivatives of the inverse hyperbolic functions, which we can o irectly from ientities an properties of inverses. Theorem. Derivatives of Inverse Hyperbolic Functions For all x at which the following are efine, we have: x (sinh x)= x + x (cosh x)= x (c) x (tanh x)= x Similar formulas can be evelope for the remaining three inverse hyperbolic functions. Notice the strong similarities between these erivatives an the erivatives of the inverse trigonometric functions. Proof. We will prove the rule for the erivative of sinh x an leave the remaining two rules to Exercises 93 an 94. Starting from the fact that sinh(sinh x) for all x, wecanapply implicit ifferentiation: sinh(sinh x)=x x (sinh(sinh x)) = x (x) cosh(sinh x) x (sinh x)= x (sinh x)= x (sinh x)= cosh(sinh x). x (sinh x)=. +x + sinh (sinh x) sinh x is the inverse of sinh x ifferentiate both sies chain rule, erivative of sinh x algebra since cosh x sinh x = sinh x is the inverse of sinh x Compare this proof with our proof earlier in this section for the erivative of sin x; the two are very similar. Examples an Explorations Example Differentiating combinations of trigonometric functions Fin the erivatives of each of the following functions. f(x)= tan x x 3 f(x) =x sin (3x+) (c) f(x) =sec e x Solution. By the quotient rule an the rule for ifferentiating tangent, we have: ( tan x ) x x 3 = x (tan x) (x3 ) (tan x) x (x3 ) (x 3 ) = (sec x)(x 3 ) (tan x)(3x ) (x 3 ). This is a prouct of functions, an thus we begin with the prouct rule. We will also nee the chain rule to ifferentiate the composition sin (3x +): f (x) = () sin (3x +)+x (3) = 3x (3x +) sin (3x +)+. (3x +)
.6 Derivatives of Trigonometric an Hyperbolic Functions 8 (c) This is a composition of three functions, an thus we nee to apply the chain rule twice: x (sec e x )= x ((sec(ex )) ) =(sece x ) x (sec ex ) =(sece x )(sec e x )(tan e x ) x (ex ) =(sece x )(sec e x )(tan e x )e x rewrite so compositions are clear first application of chain rule secon application of chain rule erivative of e x Perhaps the trickiest part of this calculation is that the erivative of sec x has two instances of the inepenent variable: x (sec x)=secxtan x. This means that in the calculation above, we neee to put the insie function e x into both of these variable slots. Example Differentiating combinations of hyperbolic functions Fin the erivatives of each of the following functions. f(x) = ln(tanh (x 3 +x +)) f(x) = cosh (e 3x ) Solution. This is a neste chain rule problem, since f(x) is a composition of multiple functions. We will work from the outsie to the insie, one step at a time: f (x)= tanh (x 3 x +x +) (tanh (x 3 +x +)) = tanh (x 3 +x +) ( tanh(x3 +x +)) x (tanh(x3 +x +)) = tanh (x 3 +x +) ( tanh(x3 +x +))(sech (x 3 +x +))(3x +). Once again we have a neste chain rule situation. Notice in particular how the e 3x works with the erivative of inverse hyperbolic cosine: f (x) = (cosh (e 3x )) x (cosh (e 3x )) ( ) = (cosh (e 3x )) (3e 3x ). (e 3x ) Example 3 Fining antierivatives that involve inverse trigonometric functions Fin a function f whose erivative is f (x) = +4x. Solution. Since the erivative of tan x is +x, we might suspect that the function f we are looking for is relate to inverse tangent. We will use an intelligent guess-an-check metho to fin f. Clearlyf(x) =tan x isn t exactly right, since its erivative is missing the 4. A goo guess might be f(x)=tan (4x); let s try that: x (tan (4x)) = +(4x) (4) = 4 +6x. Obviously that wasn t quite right either; but by examining the results we can make a new guess. We might try tan (x), since the x will be square in the erivative an become
.6 Derivatives of Trigonometric an Hyperbolic Functions 9 the 4x we are looking for in the enominator: x (tan (x)) = +(x) () = +4x. Now we are getting somewhere; this iffers by a multiplicative constant from the erivative f (x) we are looking for, an that is easy to fix. We nee only ivie our guess by that constant. Try the function f(x)= tan (x): x ( tan (x)) = ( ) +(x) () = +4x. We now know that f(x) = tan (x) is a function whose erivative is f (x) = +4x. Of course, we coul also a any constant to f(x) an not change its erivative; for example, f(x)= tan (x)+5woul also work. In fact, any function of the form f(x)= tan (x)+ C will have f (x)= +4x. Example 4 Fining antierivatives that involve hyperbolic functions Fin a function f whose erivative is f (x) = e x e x. Solution. Until we learn more specific anti-ifferentiation techniques in Chapter??, aproblem like this is best one by an intelligent guess-an-check proceure. Given that we have the inverse hyperbolic functions in min, the best match of the three is the erivative of cosh x. Since the expression for f (x) also involves an e x, let s revise that guess right away to cosh e x. Now we check by ifferentiating with the chain rule: x (cosh e x )= (ex ) ex = e x e x. We guesse it on the first try! We have just shown that f(x) =cosh e x has the esire erivative.? Questions. Test your unerstaning of the reaing by answering these questions: What trigonometric limits were use to fin the erivative of sin x? How can we obtain the erivative of sec x from the erivative of cos x? What is the graphical reason that the omains of the erivatives of sin x an sec x are slightly smaller than the omains of the functions themselves? How are hyperbolic functions similar to trigonometric functions? How are they ifferent? How can we obtain the erivative of sinh x from the erivative of sinh x?
Exercises.6 Thinking Back Trigonometric an Inverse Trigonometric Values: Fin the exact values of each of the quantities below by han, without using a calculator. sin( π 3 ) tan( π 4 ) sec( 5π 6 ) sin tan ( 3) sec ( ) Compositions: For each function k below, fin functions f, g,anh so that k = f g h. There may be more then one possible answer. k(x) = sin(x 3 ) k(x) =sin (cos x) k(x) =tan (3x +) k(x) =sec(x 3 )tan(x 3 ) Writing trigonometric compositions algebraically: Prove each of the following equalities, which rewrite compositions of trigonometric an inverse trigonometric functions as algebraic functions. cos(sin x)= x sin(cos x)= x q sec (tan x)=+x tan(sec x)= x x Concepts 0. Problem Zero: Rea the section an make your own summary of the material.. True/False: Determine whether each of the following statements is true or false. If a statement is true, explain why. If a statement is false, provie a counterexample. True or False: To fin the erivative of sin x we ha to use the efinition of erivative. True or False: To fin the erivative of tan x we have to use the efinition of erivative. 4x3 (c) True or False: The erivative of is sin x cos x. () True or False: If a function is algebraic, then so is its erivative. (e) True or False: If a function is transcenental, then so is its erivative. (f) True or False: If f is a trigonometric function, then f is also a trigonometric function. (g) (h) True or False: If f is an inverse trigonometric function, then f is also an inverse trigonometric function. True or False: If f is a hyperbolic function, then f is also a hyperbolic function.. Examples: Give examples of each of the following. Try to fin examples that are ifferent than any in the reaing. (c) x 4 A function that is its own fourth erivative. A function whose omain is larger than the omain of its erivative. Three non-logarithmic functions that are transcenental, but whose erivatives are algebraic. 3. What limit facts an trigonometric ientities are use in the proof that (sin x) =cosx? x 4. Sketchgraphsof sin x an cos x on [ π, π]. (c) Use the graph of sin x to etermine where sin x is increasing an ecreasing. Use the graph of cos x to etermine where cos x is positive an negative. Explain why your answers to parts an suggest that cos x is the erivative of sin x. 5. The ifferentiation formula (sin x) =cosx is vali x only if x is measure in raians. In this problem you will explore why this erivative relationship oes not hol if x is measure in egrees. Set your calculator to egree moe an sketch agraphofsin x that shows at least two perios. If the erivative of sine is cosine, then the slope of your graph at x =0shoul be equal to cos 0 =. Use your graph to explain why this is not the case when using egrees. (Hint: Think about your graphing winow scale.) Now set your calculator back to raians moe! 6. Suppose you wish to ifferentiate g(x) =sin (x)+ cos (x). Whatisthefastestwaytoothis,anwhy? 7. The following erivatives of the function f(x) = cos(3x ) are incorrect. What misconception occurs in each case? Incorrect: f (x)=( sin x)(3x )+(cosx)(6x) Incorrect: f (x)= sin(6x) 8. The following erivatives of the function f(x) = cos(3x ) are incorrect. What misconception occurs in each case? Incorrect: f (x)= sin(3x ) Incorrect: f (x)= sin(3x )(6x)(6)
.6 Derivatives of Trigonometric an Hyperbolic Functions 3 9. In the proof that x (sin x)=, we use x It is also true that the fact that sin(sin x)=x. sin (sin x) =x; coul we have starte with that inequality instea? Why or why not? 0. Both of the following equations are true: tan(tan x)=x an tan (tan x) =x. We can fin the erivative of tan x by ifferentiating both sies of one of these equations an solving for x (tan x). Whichone,anwhy?. The following erivatives of the function f(x) = cos(3x ) are incorrect. What misconception occurs in each case? Incorrect: f (x) =( sin x)(3x )+(cosx)(6x) Incorrect: f (x) = sin(6x). Suppose you wish to ifferentiate g(x) =sin (x)+ cos (x). Whatisthefastestwaytoothis,anwhy? 3. The graph below left shows y =sinhx, y =coshx, an y = ex.foreachfactbelow, explain graphically why the fact is true. Then prove the fact algebraically using the efinitions of the hyperbolic functions. sinh x ex cosh x for all x lim x sinh x ex =an lim x Graph for Exercise 3 4 3 - - - - -3-4 cosh x ex = Graph for Exercise 4 4 3 - - - - -3-4 4. The graph above right shows y =coshx, y = ex, an y = e x.foreachfactbelow, explain graphically why the fact is true. Then prove the fact algebraically using the efinitions of the hyperbolic functions. cosh x = ex + e x lim x cosh x e x = 5. The graphbelow left shows y =sinhx, y = ex,an y = e x. For each fact below, explain graphically why the fact is true. Then prove the fact algebraically using the efinitions of the hyperbolic functions. sinh x = ex e x lim x sinh x = e x Graph for Exercise 5 4 3 - - - - -3-4 Graph for Exercise 6-3 - - 3 6. The graph above right shows y =tanhx, y =,an y =. Foreachfactbelow, explain graphically why the fact is true. Then prove the fact algebraically using the efinitions of the hyperbolic functions. tanh x lim x tanh x =an lim x - tanh x = Skills Fin the erivatives of each of the functions in Exercises 7 50. In some cases it may be convenient to o some preliminary algebra. 7. f(x) = x + cos x 8. f(x)=cos(x 3 ) 9. f(x) =cotx csc x 0. f(x)=tan (3x +). f(x)=4sin x+4cos x. f(x)=sec x 3. f(x) =3secx tan x 4. f(x)=3 x sec x +7 5. f(x)=sin(cos(sec(x))) 6. f(x)=csc (e x ) 7. f(x) =e csc x 8. f(x)=e x csc x 9. f(x) = x 5x sin x 30. f(x)= 3. f(x) =x sin x cos x 3. f(x)= log 3 (3 x ) sin x +cos x sin x csc x cot x cos x 33. f(x)= 3x ln x tan x 34. f(x) = ln(3x ) tan x 35. f(x)= sin(lnx) 36. f(x) = ln(x sin x) 37. f(x)=sin (3x ) 38. f(x) =3(sin x) 39. f(x)=x arctan x 40. f(x) =tan (ln x) 4. f(x)=sec x 4. f(x) =sin(sin x) 43. f(x)=sin (sec x) 44. f(x) =sin (sec x) 45. f(x)= sin x tan x 46. f(x) = sin x cos x 47. f(x)=ln(arcsec(sin x)) 48. f(x) =x e 4x sin x 49. f(x)=sec(+tan x) 50. f(x) = sin(arcsin x) arctan x
.6 Derivatives of Trigonometric an Hyperbolic Functions 3 Fin the erivatives of each of the functions in Exercises 5 6. In some cases it may be convenient to o some preliminary algebra. These problems involve hyperbolic functions an their inverses. 5. f(x) =x sinh x 3 5. f(x)=x sinh 3 x 53. f(x) =cosh(ln(x +)) 54. f(x)=3tanh e x 55. f(x) = p cosh x + 56. f(x)= tanh x sinh x 57. f(x) =sinh (x 3 ) 58. f(x)=tanh (tan x ) 59. f(x) = sinh x cosh x 6. f(x) =sin(e sinh x ) 60. f(x)=x tanh x 6. f(x)=cosh (cosh x) Use logarithmic ifferentiation to fin the erivatives of each of the functions in Exercises 63 65. 63. (sin x) x 64. (sec x) x 65. (sin x) cos x In Exercises 66??, fin a function f that has the given erivative f. In each case you can fin the answer with an eucate guess-an-check process. 66. f (x) = 68. f (x) = x 4x 67. f (x)= 4x 69. f (x)= 3x +9x +9x 70. f (x) = 7. f 3 (x)= 9+x 4 9x In Exercises 7??, fin a function f that has the given erivative f. In each case you can fin the answer with an eucate guess-an-check process. Some of these exercises involve hyperbolic functions. 7. f (x) = 74. f (x) = x +4x 73. f (x)= +4x 75. f (x)= 3x 9x 9x 76. f (x) = 77. f 3 (x)= 9 x 4+9x Applications 78. In Problem 83 from Section.6 we saw that the oscillating position of a mass hanging from the en of aspring,neglectingairresistance,isgivenbyfollowing equation, where A, B, k,anm are constants: s(t) =A sin `q k m t +B cos `q k m t, Show that this function s(t) has the property that s (t)+ k s(t) =0. This is the ifferential m equation for the spring motion, which means it is an equation involving erivatives that escribes the motion of the bob on the en of the spring. Suppose the spring is release from an initial position of s 0,anwithaninitialvelocityofv p 0. Show that A = v m 0 an B = s0. k 80. Suppose your frien Max rops a penny from the top floor of the Empire State Builing, 50 feet from the groun. You are staning about a block away, 50 feet from the base of the builing. Fin a formula for the angle of elevation α(t) from the groun at your feet to the height of the penny t secons after Max rops it. Multiply by an appropriate constant so that α(t) is measure in egrees. Fin a formula for the rate at which the angle of elevation α(t) is changing at time t, an use it to etermine the rate of change of the angle of elevation at the time the penny hits the groun. 79. InProblem84fromSection.6welearnethattheoscillating position of a mass hanging from the en of a spring, taking air resistance into account, is given by the following equation, where A, B, k, f, anm are constants: s(t)=e f t`a m sin` 4km f t +B cos` 4km f t, m m Show that this function s(t) has the property that s (t)+ a m s (t)+ k s(t) =0for some constant a. This is the ifferential equation for m spring motion, taking air resistance into account. (Hint: Fin the first an secon erivatives of s(t) first an then show that s(t), s (t),ans (t) have the given relationship.) Suppose the spring is release from an initial position of s 0,anwithaninitialvelocityofv 0. Show that A = mv 0+fs 0 an B = s0. 4km f
.6 Derivatives of Trigonometric an Hyperbolic Functions 33 Proofs 8. Use the efinition of erivative, a trigonometric ientity, an known trigonometric limits to prove that (cos x) = sin x. x 8. Use the quotient rule an the erivative of cosine to prove that (sec x)=secxtanx. x 83. Use the quotient rule an the erivative of sine to prove that (csc x)= csc x cot x. x 84. Use the quotient rule an the erivatives of sine an cosine to prove that (cot x)= x csc x. 85. Use implicit ifferentiation an the fact that tan(tan x)=xfor all x in the omain of tan x to prove that x (tan x)=. +x 86. Use implicit ifferentiation an the fact that sec(sec x)=x for all x in the omain of sec x to prove that x (sec x)= x.youwillhaveto consier the cases x> an x< separately. 87. Prove that for any value of t, the point (x, y) = (cosh t, sinh t) lies on the hyperbola x y =. Bonus question: In fact, these points will always lie on the right han sie of the hyperbola; why? Use the efinitions of the hyperbolic functions to prove that each of the ientities in Exercises 88 90 hol for all values of x an y. Note how similar these ientities are to those that hol for trigonometric functions. x 88. sinh( x) = sinh x,an cosh( x)= cosh x 89. sinh(x + y)=sinhx cosh y +coshx sinh y 90. cosh(x + y) =coshx cosh y +sinhx sinh y Prove each of the ifferentiation formulas in Exercises 9 94. 9. 9. 93. 94. (cosh x) =sinhx x (tanh x x)=sech x x (cosh x)= x x (tanh x) x Prove that the inverse hyperbolic functions can be written in terms of logarithms as in Exercises 95 97. (Hint for the first problem: Solve sinh y = x for y, by using algebra to get an expression that is quaratic in e y, that is, of the form ae y + be y + c, an then applying the quaratic formula.) 95. sinh x =ln(x + x +),foranyx. 96. cosh x =ln(x + x ),forx. 97. tanh x = ln ` +x,for x <x<. Thinking Forwar Local extrema an inflection points: In the problems below you will investigate how erivatives can help us fin the locations of the maxima an minima of a function. Suppose f has a maximum or minimum value at x = c. If f is ifferentiable at x = c, what must be true of f (c),anwhy? If f is a ifferentiable function, then the values x = c at which the sign of the erivative f (x) changes are the locations of the local extrema of f. Use this information to fin the local extrema of the function f(x) =sinx. Illustrate your answer on a graph of y =sinx. If f is a ifferentiable function, then the values x = c at which the sign of the secon erivative f (x) changes are the locations of the inflection points of f. Use this information to fin the inflection points of the function f(x) =sinx. Illustrate your answer on a graph of y =sinx.
Appenix A Answers To O Problems
3 79. If Lina sells magazines at a rate of m = t magazine subscriptions per week, an she makes D =4ollars per magazine, then m by the chain rule, the amount of money she makes each week is D = D m =4()= t m t 48 ollars per week. A 8. =πr. No; Yes. (c) A = r t t (π(r(t)) )=πr(t)r (t)=πr r.()yes; t Yes. (e) A r=4 =π(4)() = 96π. t 83. Proof: f(x) = x g(x) x (f(x)(g(x)) ) = (f(x)) x (g(x)) + f(x) x ((g(x)) )= f (x) (g(x)) + f(x) ( g(x)) g (x) = f (x) f(x)g (x) = f (x)g(x) f(x)g (x). g(x) (g(x)) (g(x)) 85. Mimic the proof in Example 5. 87. Let y = x k,where k is a negative integer. Then x k y =, so by implicit ifferentiation an the prouct rule, we have kx k y + x k y =0,anthereforey = kxk x k x k = kx k. Section.5. T, T, F, F, T, T, F, T. 3. No, since the base is a variable. No, since the exponent is a variable. 5. When k =we have x (ex )= x (ex )= e x = e x. When b = e we have x (bx )= x (ex )=(lne)e x =()e x = e x. 7. ln x is of the form log b x with b = e. 9. The graph passing through (0, ) is ( x ); the graph passing through (0, ) an (, 4) is 4 x ; the graph passing through (, 4) is x.. f(3) = 9 =5an g(3) = 8 =64. The solutions to (x) =( x ) are x =0an x =. 3. Consier that logarithmic functions are the inverses of exponential functions, an use the efinition of one-to-one. 5. Logarithmic ifferentiation is the process of applying ln x to both sies of an equation y = f(x) an then ifferentiating both sies in orer to solve for f (x). It is useful for fining erivatives of functions that involve multiple proucts or quotients, an functions that have variables in both a base an an exponent. 7. f (x) = ( e 5x ) ( 5e 5x ) 9. f (x) =6xe 4x x e 4x. f (x)= ( )ex ( x)e x e x 3. e x (x +3x ) + e x (x +3) 5. f (x)=3x,withx>0 7. f (x)=e (ex) e x 9. f (x)=e ex+ 3. f (x)=0 33. f (x)= 3x 4 e x + x 3 (e x ). 35. f (x)=x log x + x ln +3x 37. f (x)= x +e (x + x ex ( x )) 39. f (x) = (log (3x 5)) ( ((ln ln 3 x 5 3)3x ) 4. f (x)=xln(ln x)+x ln x x 43. f (x)=(ln) x (x) 8 < (ln ) x, if x< 45. f (x)= DNE, if x = :, if x> x 8 3 < x, if x< 47. f (x)= DNE, if x = :, if x x 49. f (x) = (x ln x + ) (ln x + + (ln )x x x + ) 5. f (x) = x x 3 x(x ) (ln + x ) 53. f (x)=x ln x ()(ln x)( x ) 3x (x 3 ) x 55. f (x)=( x x )x (ln x ln(x ) + 57. f (x)=(lnx) ln x ln(ln x) ( + ) x x 59. f(x) = e4x 3x 5 6. f(x) = ln x +3 + C 63. f(x) =ln(+e x )+C x x ) 65. A(t) = 000e 0.077t ; A(30) $0,074.4; (c)t 8 years. 66. 70 egrees; 350 egrees; (c) minutes. () Fin T (t) an show that it is always positive; this means that the temperature of the yam increases over time. Fin T (t) an show that it is always negative; this means that the rate of change of the temperature of the yam ecreases over time. 69. x (ekx )=e kx x (kx)=ekx k = ke kx. 7. If f(x) =Ae kx is exponential, then f (x) = Ake kx = k(ae kx )=kf(x), sof (x) is proportional to f(x). 73. For x>0 we have x = x an therefore (ln x )= (ln x)=.forx<0we have x x x x = x an therefore by the chain rule we have (ln x ) = (ln( x)) = ( ) = x x x x. Section.6. T, F, F, T, F, T, F, T.
3. See the proof in the reaing. 5. If x is in egrees, then the slope of the graph of sin x at x =0is very small, an in particular not equal to cos 0 =. To convince yourself of this, graph sin x (in egrees) together with the line y = x (which has slope at x =0). 7. cos(3x ) is a composition, not a prouct, but the prouct rule was applie; the chain rule was applie incorrectly, with the erivative of 3x written on the insie instea of the outsie. 9. No, because to ifferentiate sin (sin x)=x we woul first nee to know how to ifferentiate sin x,whichisexactlywhatwe woul be trying to prove.. cos(3x ) is a composition, not a prouct, but the prouct rule was applie; the chain rule was applie incorrectly, with the erivative of 3x written on the insie instea of the outsie. 3. Use the fact that e x 0 for all x, an split the expressions in the efinitions of cosh x an sinh x into sums. Calculate the two limits by iviing top an bottom by e x an show they are both equal to. 5. Think about the graph of the sum of ex an e x. Calculate lim x (ex e x ) e x by iviing top an bottom by e x, an Show that this limit is equal to. 7. f (x) = x cos x+(x +) sin x cos x 9. f (x) = csc x +cscx cot x. f (x) =0 3. f (x) =3secx sec x +3secx tan x 5. f (x)=cos(cos(secx))( sin(sec x))(sec x tan x) 7. f (x) =e csc x ( csc x)( csc x cot x) 9. f (x) = (ln )x (5x sin x)+ x (5 sin x+5x cos x) 5x sin x 3. f (x) = sin x cos x + x(sin x cos x) (cos x sin x) 33. f (x) = (6x ln x+3x) tan x 3x ln x sec x tan x 35. f (x) =cos(lnx)( ) x 37. f (x) = 6x 9x 4 39. f (x) =x arctan x + x ( x 4. f (x) = 43. f (x) = 45. x x tan x x sin x +x 47. f (x)= (tan x) x 4 +x 4 ) ( sec x)(sec x tanx) sec 4 x arcsec(sin x) sin x sin 4 x ( sin x cos x) 49. f (x)=sec(+tan x)tan(+tan x) ( ) +x 5. f (x)=sinhx 3 +3x 3 cosh x 3 53. f (x)=sinh(ln(x +))( )(x) x + 55. f (x)= (cosh x +) ( cosh x sinh x) 57. f (x)= 3x x 6 + 59. f (x)= (cosh x) sinh x( ) x + x (cosh x) 6. f (x)=cos(e sinh x )e sinh x x + 63. f (x)=(sinx) x (ln(sin x)+ x cos x sin x ) 65. f (x)=(sinx) cos x ( sinx ln(sin x)+ cos x sin x ) 67. f(x) =sin x 69. f(x) = 6 ln( + 9x ) 7. f(x) =sin ( 3x ) 73. f(x) =sinh x 75. f(x) = 6 ln( 9x ) 77. f(x) =sinh ( 3x ) 4km f 79. To simplify things, let C = ; m since k, m, anf are all constants, so is C. Then follow the given hint. Set s(0) = s 0 an s (0) = v 0 an solve for A an B. 8. Mimic the proof in the reaing for (sin x), x except using a sum ientity for cosine in the secon step. ` 83. (csc x)= x x sin x = (0)(sin x) ()(cos x) = (sin x) cos x = ` `cos x sin x sin x sin x = csc x cot x. 85. Differentiating both sies of tan(tan x)= x gives us sec (tan x) x (tan x)=, an therefore x (tan x)= By one of the Thinking Back problems in this section, we have sec (tan x)=+x, sec (tan x). an therefore x (tan x)=. +x 87. x y = cosh t sinh t = ( et e t ) ( et +e t ) = e t e t e t +e t e t e t e t +e t =. For the bonus question, consier whether cosh t can ever be negative. 89. Expan the right han sie using the efinitions an simplify to get the left han sie. 9. Mimic the proof in the reaing that was given for the erivative of sinh x. 93. Mimic the proof in the reaing that was given for the erivative of sinh x. 95. e y e y = x = e y =x = ey = e y e y x = e y xe y =0. This is quaratic in e y, an the quaratic formula gives e y = x± 4x +4, an thus y =ln(x ± x +). Since logarithms are only efine for positive numbers we have y =ln(x + x +).
97. Start with y =tanhx = sinh x an use the cosh x problems above. Chapter 3 Section 3.. T, T, T, F, F, T, F, F. 3. f () is either 0 or oes not exist. If f is ifferentiable at x =then f () must equal 0. 5. x =, x =0, x =4, x =5. 7. f has at least two zeros in the interval [ 4, ]. 9. There is some c (, 4) with f (c) = 3.. If f is continuous on [a, b], ifferentiable on (a, b), aniff =f =0,thenthereexists at least one value c (a, b) for which the tangent line to f at x = c is horizontal. 3. The graph of f(x) =(x )(x 6) is one example. 5. Your graph shoul have roots at x = an x =an horizontal tangent lines at three places between these roots. 7. One example is an upsie-own V with roots at x = an x =where the top point of the V occurs at x =. 9. One example is the function f that is equal to x +3for 3 x< an equal to 0 for x =.. Draw a graph that happens to have a slight cusp just at the place where its tangent line woul have been equal to the average rate of change. 3. f (x) =0at x = 3 an f (x) oes not exist at x =3. f has a local maximum at x = 3 an a local minimum at x =3. 5. f (x) =0at x 0.5, x, anx 3.5. f has a local minimum at x 0.5 an a local minimum at x =3.5. There is neither a maximum nor a minimum at x. 7. One critical point at x = 0.65, a local minimum. 9. Critical points x = 3, x =0, x =,alocal minimum, maximum, an minimum, respectively. 3. One critical point at x =ln( 3 ), a local maximum. 33. Unefine at x =0.Onecriticalpointatx = e, a local maximum. 35. Critical points at points of the form x = πk where k is an integer; local minima at the o multiples x = π(k +), local maxima at the even multiples x = π(k). 37. From the graph, f appears to be continuous on [ 3, ] an ifferentiable on ( 3, ), an moreover f( 3) = f() = 0, sorolle s Theorem applies. Therefore there is some c ( 3, ) such that f (c)=0. In this example there are three such values of c, namely c.3, c =, anc =0.3. 39. From the graph, f appears to be continuous on [0, 4] an ifferentiable on (0, 4), an moreover f(0) = f(4) = 0, sorolle s Theorem applies. Therefore there is some c (0, 4) such that f (c) =0. In this example there are three such values of c, namely c 0.5, c, anc 3.5. 4. f is continuous an ifferentiable everywhere, an f(0) = f(3) = 0, sorolle sthe- orem applies; c = (4 7) an c = (4 + 3 3 7). 43. f is continuous an ifferentiable everywhere, an f( ) = f() = 0, sorolle s Theorem applies; c.779, c 0, an c.779. 45. f is continuous an ifferentiable everywhere, an cos( π )=cos(3π )=0, so Rolle s Theorem applies; c =0,c= π. 47. f is continuous an ifferentiable everywhere, an f(0) = f() = 0, sorolle sthe- orem applies; c =. 49. f appears continuous on [0, ] an ifferentiable on (0, ); there is one value x = c that satisfies the conclusion of the Mean Value Theorem, rougly at x.. 5. f appears continuous on [ 3, 0] an ifferentiable on ( 3, 0); there are two values x = c that satisfy the conclusion of the Mean Value Theorem, at c.8 an c 0.9. 53. f is not continuous or ifferentiable on [ 3, ], so the Mean Value Theorem oes not apply. 55. f is continuous an ifferentiable on [, 3]; c 0.575 an c.575. 57. f is continuous on [0, ] an ifferentiable on (0, ), so the Mean Value Theorem applies; c 0.408. 59. f is continuous an ifferentiable everywhere, so the Mean Value Theorem applies; c = cos ( ) 0.88. π 6. C(h) is a ifferentiable function, an C (4) = 0.6 0,soC(h) cannot have a local minimum at h =4.