Smith R., Minton R. Calculus

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6-29 SECTION 6.4

The Inverse Trigonometric Functions 403

hypotenuse 5, by the Pythagorean Theorem. Using the picture as a guide, we can read

off the desired value.

sinθ = sin



tan

−1



While we’re at it, we should also observe that we have found that

cosθ = cos



tan

−1





BEYOND FORMULAS

There is a subtle device we have used throughout this section. Since the function sin x

is not one-to-one, it clearly does not have an inverse. However, inverse functions are

so useful that in cases such as this, mathematicians don’t like to take this as the ﬁnal

answer and so, restrict the domain of the function, effectively deﬁning a new function

that does have an inverse. In the case of sin x, we restricted the domain to the interval

[−π/2,π/2], thus deﬁning a new function that equals sin x on this interval and yet,

does have an inverse. The general idea of manipulating various features of a function

to produce a desired result is common in mathematics. The alternative would mean

meekly accepting the existing limitations of a given function, which is not part of the

culture of mathematics.

EXERCISES 6.4

WRITING EXERCISES

1. Discuss how to compute sec

−1

x, csc

−1

x and cot

−1

x on a cal-

culator that only has built-in functions for sin

−1

x, cos

−1

x and

tan

−1

2. Give a different range for sec

−1

x than that given in the text.

For which x’s would the value of sec

−1

x change? Using the

calculator discussion in exercise 1, give one reason why we

might have chosen the range that we did.

3. Inverse functions are necessary for solving equations. The re-

strictedrangewehadtousetodeﬁneinversesofthetrigonomet-

ric functions also restricts their usefulness in equation solving.

Explain how to use sin

−1

x to ﬁnd all solutions of the equation

sinu = x.

4. The idea of restricting ranges can be used to deﬁne inverses

for a variety of functions. Explain how to deﬁne an inverse for

f (x) = x

In exercises 1–6, evaluate the inverse function by sketching a

unit circle and locating the correct angle on the circle.

1. (a) sin

−1

(0) (b) sin

−1

(−

) (c) sin

−1

(−1)

2. (a) cos

−1

(0) (b) cos

−1

(1) (c) cos

−1

(

)

3. (a) tan

−1

(1) (b) tan

−1

(0) (c) tan

−1

(−1)

4. (a) cot

−1

(0) (b) cot

−1

(1) (c) cot

−1

(

√

5. (a) sec

−1

(1) (b) sec

−1

(2) (c) sec

−1

(

√

6. (a) csc

−1

(1) (b) csc

−1

(−1) (c) csc

−1

(−2)

............................................................

In exercises 7–18, use a triangle to simplify each expression.

Where applicable, state the range of x’s for which the simpliﬁ-

cation holds.

7. cos(sin

−1

x) 8. cos(tan

−1

9. tan(sec

−1

x) 10. cot(cos

−1

11. sin



cos

−1



12. cos



sin

−1



13. tan



cos

−1



14. csc



sin

−1



15. cos

−1

(cos(−π/8)) 16. sin

−1

(sin3π/4)

17. sin(2sin

−1

x/3) 18. cos(2sin

−1

2x)

............................................................

In exercises 19–24, sketch a graph of the function.

19. cos

−1

(2x) 20. cos

−1

)

21. sin

−1

(3x) 22. sin

−1

(x/4)

23. tan

−1



− 1



24. sin

−1



1 −



............................................................

In exercises 25–32, ﬁnd all solutions.

25. cos(2x) = 0 26. cos(3x) = 1

27. sin(3x) = 0 28. sin(x/4) = 1

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404 CHAPTER 6

Exponentials, Logarithms and Other Transcendental Functions 6-30

29. tan(5x) = 1 30. sin(x/4) =−

31. sec(2x) = 2 32. csc(2x) =−2

............................................................

33. Give precise deﬁnitions of csc

−1

x and cot

−1

APPLICATIONS

34. In baseball, outﬁelders are able to easily track down and catch

ﬂy balls that have very long and high trajectories. Physicists

have argued for years about how this is done. A recent expla-

nation involves the following geometry.

Outfielde

Ball

Home

plate

The player can catch the ball by running to keep the angle

ψ constant (this makes it appear that the ball is moving in

a straight line). Assuming that all triangles shown are right

triangles, show that tanψ =

tanα

tanβ

and then solve for ψ.

35. A picture hanging in an art gallery has a frame 20 inches high

and the bottom of the frame is 6 feet above the ﬂoor. A person

whose eye is 6 feet above the ﬂoor stands x feet from the wall.

Let A be the angle formed by the ray from the person’s eye

to the bottom of the frame and the ray from the person’s eye

to the top of the frame. Write A as a function of x and graph

y = A(x).

20"

36. In golf, the goal is to hit a ball into a hole of diameter

4.5 inches. Suppose a golfer stands x feet from the hole trying

to putt the ball into the hole. A ﬁrst approximation of the mar-

gin of error in a putt is to measure the angle A formed by the

ray from the ball to the right edge of the hole and the ray from

the ball to the left edge of the hole. Find A as a function of x.

EXPLORATORY EXERCISES

1. An oil tank with circular cross sections lies on its side. A stick

is inserted in a hole at the top and used to measure the depth

d of oil in the tank. Based on this measurement, the goal is to

compute the percentage of oil left in the tank.

To simplify calculations, suppose the circle is a unit circle with

center at (0, 0). Sketch radii extending from the origin to the

top of the oil. The area of oil at the bottom equals the area of

the portion of the circle bounded by the radii minus the area of

the triangle formed above.

Start with the triangle, which has area one-half base times

height. Explain why the height is 1 − d. Find a right triangle

in the ﬁgure (there are two of them) with hypotenuse 1 (the

radius of the circle) and one vertical side of length 1 − d. The

horizontal side has length equal to one-half the base of

the larger triangle. Show that this equals



1 − (1 − d)

. The

area of the portion of the circle equals πθ/2π = θ/2, where θ

is the angle at the top of the triangle. Find this angle as a func-

tion of d. (Hint: Go back to the right triangle used previously

with upper angle θ/2.) Then ﬁnd the area ﬁlled with oil and

divide by π to get the portion of the tank ﬁlled with oil.

2. In this exercise, you will design a movie theater with all seats

havingan equal viewof the screen. Suppose the screen extends

vertically from 10 feet to 30 feet above the ﬂoor. The ﬁrst row

of seats is 15 feet from the screen. Your task is to determine a

function h(x) such thatif seats x feet from thescreen are raised

h(x) feet above ﬂoor level, then the angle from the bottom of

the screen to the viewer to the top of the screen will be the

same as for a viewer sitting in the ﬁrst row. You will be able to

accomplish this only for a limited range of x-values. Beyond

the maximum such x, ﬁnd the height that maximizes the view-

ing angle.



Hint: Write the angle as a difference of inverse

tangents and use the formula tan(a − b) =

tana − tan b

1 + tan a tanb



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6-31 SECTION 6.5

The Calculus of the Inverse Trigonometric Functions 405

6.5 THE CALCULUS OF THE INVERSE

TRIGONOMETRIC FUNCTIONS

Nowthatwehavedeﬁnedtheinversetrigonometricfunctionsinsection6.4,wewillexamine

the calculus of these functions brieﬂy in the present section. First, to ﬁnd the derivative of

sin

−1

x, we recall the deﬁnition of sin

−1

x given in (4.1):

y = sin

−1

x if and only if sin y = x and y ∈

−

Differentiating the equation sin y = x implicitly, we have

sin y =

and so, cos y

= 1.

Solving this for

, we ﬁnd (for cos y = 0) that

cos y

This is not entirely satisfactory, though, since this gives us the derivative in terms of y.

Notice that for y ∈



−



, cos y ≥ 0 and hence,

cos y =

1 − sin

y =



1 − x

This leaves us with

cos y

√

1 − x

for −1 < x < 1. That is,

sin

−1

x =

√

1 − x

, for −1 < x < 1.

Alternatively, we can derive this same derivative formula using Theorem 2.3 in section 6.2.

We leave it as an exercise to show that

cos

−1

x =

−1

√

1 − x

, for −1 < x < 1.

To ﬁnd

tan

−1

x, we rely on its deﬁnition in (4.4). Recall that we have

y = tan

−1

x if and only if tan y = x and y ∈



−



Using implicit differentiation, we then have

tan y =

and so, (sec

= 1.

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406 CHAPTER 6

Exponentials, Logarithms and Other Transcendental Functions 6-32

We solve this for

, to obtain

sec

1 + tan

1 + x

That is,

tan

−1

x =

1 + x

You can likewise show that

sec

−1

x =

|x|

√

− 1

, for |x| > 1.

This is left as an exercise. The derivatives of the two remaining inverse trigonometric

functions are not important and are not discussed here.

EXAMPLE 5.1 Finding the Derivative of an Inverse

Trigonometric Function

Compute the derivative of (a) cos

−1

(3x

) and (b) (sec

−1

Solution From the chain rule, we have

(a)

cos

−1

(3x

) =

−1



1 − (3x

)

(3x

)

−6x

√

1 − 9x

and

(b)

(sec

−1

= 2(sec

−1

(sec

−1

= 2(sec

−1

|x|

√

− 1



EXAMPLE 5.2 Modeling the Rate of Change of a Ballplayer’s Gaze

One of the guiding principles of most sports is to “keep your eye on the ball.” In

baseball, a batter stands 2 feet from home plate as a pitch is thrown with a velocity of

130 ft/s (about 90 mph). Assuming that the ball only moves horizontally, at what rate

does the batter’s angle of gaze need to change when the ball crosses home plate?

Solution First, look at the triangle shown in Figure 6.29. We denote the distance from

the ball to home plate by d and the angle of gaze by θ. Since the distance is changing

with time, we write d = d(t). The velocity of 130 ft/s means that d



(t) =−130. [Why

would d



(t) be negative?] From Figure 6.29, notice that

θ(t) = tan

−1



d(t)



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6-33 SECTION 6.5

The Calculus of the Inverse Trigonometric Functions 407

Overhead view

FIGURE 6.29

A ballplayer’s gaze

The rate of change of the angle is then



(t) =

1 +



d(t)





(t)



(t)

4 + [d(t)]

radians/second.

When d(t) = 0 (i.e., when the ball is crossing home plate), the rate of change is then



(t) =

2(−130)

=−65 radians/second.

One problem with this is that most humans can accurately track objects only at the rate

of about 3 radians/second. Keeping your eye on the ball in this case is thus physically

impossible. How do those ballplayers do it? Research indicates that ballplayers must

anticipate where the ball is going, instead of continuing to track the ball visually. (See

Watts and Bahill, Keep Your Eye on the Ball.)



Integrals Involving the Inverse Trigonometric Functions

You may have already guessed the next step. Each of our newdifferentiationformulas gives

rise to a new integration formula. First, since

sin

−1

x =

√

1 − x

we also have that



√

1 − x

dx = sin

−1

x +c. (5.1)

Likewise, since

cos

−1

x =

−1

√

1 − x

we also have that



√

1 − x

dx =−cos

−1

x +c. (5.2)

However, since the integral in (5.2) is the same integral as in (5.1), we will ignore (5.2).

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408 CHAPTER 6

Exponentials, Logarithms and Other Transcendental Functions 6-34

In the same way, we obtain



1 + x

dx = tan

−1

x +c

and



|x|

√

− 1

dx = sec

−1

x +c.

EXAMPLE 5.3 An Integral Related to tan

−1

Evaluate



9 + x

dx.

Solution Notice that the integrand is nearly the derivative of tan

−1

x. However, the

constant in the denominator is 9, instead of the 1 we need. With this in mind, we rewrite

the integral as



9 + x

dx =



1 +





dx.

If we let u =

, then du =

dx and hence,



9 + x

dx =



1 +







1 +





  

1+u







1 + u

tan

−1

u + c

tan

−1





+ c.



We leave it as an exercise to prove the more general formula:



+ x

dx =

tan

−1





+ c.

EXAMPLE 5.4 An Integral Requiring a Simple Substitution

Evaluate



1 + e

dx.

Solution Think about how you might approach this. You probably won’t recognize an

antiderivative immediately. Remember that it often helps to look for terms that are

derivatives of other terms. You should also recognize that e

= (e

)

. With this in

mind, we let u = e

, so that du = e

dx. We then have



1 + e

dx =



1 + (e

)

  

1+u





1 + u

= tan

−1

u + c

= tan

−1

) + c.



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6-35 SECTION 6.5

The Calculus of the Inverse Trigonometric Functions 409

EXAMPLE 5.5 Another Integral Requiring a Simple Substitution

Evaluate



√

1 − x

dx.

Solution Look carefully at the integrand and recognize that you don’t know an anti-

derivative. However, you might observe that



√

1 − x

dx =





1 − (x

)

dx.

Once you’ve recognized this one small step, you can complete the problem with a

simple substitution. Letting u = x

,wehavedu = 2xdxand



√

1 − x

dx =





1 − (x

)



√

1 − u

sin

−1

u + c

sin

−1

) + c.



You will explore integrals involving the inverse trigonometric functions further in the

exercises. Whenever dealing with these functions, it is easiest if you keep the basic deﬁni-

tions in mind (including the domains and ranges). You will only need to knowthe derivative

formulas for sin

−1

x, tan

−1

x and sec

−1

x. With these basic formulas, you can quickly de-

velop anything else that you need, using elementary techniques (such as substitution).

EXERCISES 6.5

WRITING EXERCISES

1. Explain why, as is indicated in the text, antiderivatives cor-

responding to three of the six inverse trigonometric functions

“can be ignored.”

2. From equations (5.1) and (5.2), explain why it follows that

sin

−1

x =−cos

−1

x + c. From the graphs of y = sin x and

y = cos x, explain why this is plausible and identify the con-

stant c for 0 < x <π/2.

In exercises 1–10, ﬁnd the derivative of the function.

1. sin

−1

(3x

) 2. cos

−1

+ 1)

3. sec

−1

) 4. csc

−1

(

√

5. x cos

−1

(2x) 6. sin x sin

−1

(2x)

7. cos

−1

(sin x) 8. tan

−1

(cos x)

9. tan

−1

(sec x) 10. sec

−1

(tan x)

............................................................

In exercises 11–26, evaluate the given integral.

11.



1 + x

dx 12.



9 + x

13.



1 + x

dx 14.



1 + x

15.



√

1 − x

dx 16.



√

1 − x

17.



√

− 1

dx 18.



|x|

√

− 1

19.



4 + x

dx 20.



4 + x

21.



√

1 − e

dx 22.



cos x

4 + sin

23.



4 + x

dx 24.



√

2 + 2x

25.



1/4

√

1 − 4x

dx 26.



√

4 − x

............................................................

27. Derive the formula

cos

−1

x =

−1

√

1 − x

28. Derive the formula

sec

−1

x =

|x|

√

− 1

29. Derive the formula



+ x

dx =

tan

−1





+ c for

any positive constant a.

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410 CHAPTER 6

Exponentials, Logarithms and Other Transcendental Functions 6-36

30. In this exercise, we give a different derivation from the text

for the derivative of sin

−1

x. Use Theorem 2.3 of section 6.2

to show that

sin

−1

x =

cos(sin

−1

. Then, evaluate

cos(sin

−1

x).

31. In example 5.2, it was shown that by the time the baseball

reached home plate, the rate of rotation of the player’s gaze

(θ



) was too fast for humans to track. Given a maximum ro-

tational rate of θ



=−3 radians per second, ﬁnd d such that



=−3. That is, ﬁnd how close to the plate a player can track

the ball. In a major league setting, the player must start swing-

ing by the time the pitch is halfway (30



) to home plate. How

does this correspond to the distance at which the player loses

track of the ball?

32. Suppose the pitching speed x



in example 5.2 is different. Then



will be different and the value of x for which θ



=−3 will

change. Find x as a function of x



for x



ranging from 30 ft/s

(slowpitch softball) to 140 ft/s (major league fastball) and

sketch the graph.

33. Use Theorem 2.3 of section 6.2 with f (x) = cos x to derive

the formula for

cos

−1

34. Use Theorem 2.3 of section 6.2 with f (x) = tan x to derive

the formula for

tan

−1

35. Show that



−1

√

1 − x

dx = cos

−1

x + c and



−1

√

1 − x

dx =−sin

−1

x + c. Explain why this does not

imply that cos

−1

x =−sin

−1

x. Find an equation relating

cos

−1

x and sin

−1

36. Evaluate



|x|

√

− 1

dx by rewriting the integrand as



1 − 1/x

and then making the substitution u = 1/x. Use

your answer to derive an identity involving sin

−1

(1/x) and

sec

−1

37. Show that both





1 − x

dx and



1 + x

dx equal

UseSimpson’sRuleon each integralwithn = 4 andn = 8 and

compare to the exact value. Which integral provides a better

algorithm for estimating π?

38. Find and simplifythe derivative of sin

−1



√

+ 1



. Use the

result to write out an equation relating sin

−1



√

+ 1



and

tan

−1

39. Use the Mean Value Theorem to show that |tan

−1

a| < |a| for

all a = 0 and use this inequality to ﬁnd all solutions of the

equation tan

−1

x = x.

40. Prove that |x| < |sin

−1

x| for 0 < |x| < 1.

APPLICATIONS

41. Inthediagram,ahockeyplayerisD feetfromthenetonthecen-

tral axis of the rink. The goalie blocks off a segment of width

w and stands d feet from the net. The shooting angle to the

left of the goalie is given by φ = tan

−1



3(1 − d/D) − w/2

D − d



Use a linear approximation of tan

−1

x at x = 0 to show that if

d = 0, then φ ≈

3−w/2

. Based on this, describe howφ changes

if there is an increase in (a) w or (b) D.

Exercise 41

42. The shooter in exercise 41 is assumed to be in the center of the

ice. Suppose that the line from the shooter to the center of the

goal makes an angle of θ with the center line. For the goalie to

completely block the goal, he must stand d feet away from the

net where d = D(1 −w/6cosθ). Show that for small angles,

d ≈ D(1 −w/6).

43. For a college football ﬁeld with the dimensions shown,

the angle θ for kicking a ﬁeld goal from a (horizon-

tal) distance of x feet from the goal post is given by

θ(x) = tan

−1

(29.25/x) − tan

−1

(10.75/x). Show that

f (t) =

+ t

is increasing for a > t and use this fact to

show that θ(x) is a decreasing function for x ≥ 30. Announc-

ers often say that for a short ﬁeld goal (50 ≤ x ≤ 60), a team

can improve the angle by backing up 5 yards with a penalty.

Is this true?

18.5

40'

␪

44. To start skating, you must angle your foot and push off the ice.

Alain Hach´e’s The Physics of Hockey derives the relationship

between the skate angle θ, the sideways stride distance s, the

stroke period T and the forward speed v of the skater, with

θ = tan

−1

(

). For T = 1 second, s = 60 cm and an acceler-

ation of 1 m/s

, ﬁnd the rate of change of the angle θ when the

skater reaches (a) 1 m/s and (b) 2 m/s. Interpret the sign and

size of θ



in terms of skating technique.

push

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6-37 SECTION 6.6

The Hyperbolic Functions 411

45. Suppose a painting hangs on a wall. The frame extends from

6 feet to 8 feet above the ﬂoor. A person whose eye is 5 feet

above the ﬂoor stands x feet from the wall and views the paint-

ing,withaviewingangle A formedby the ray fromtheperson’s

eye to the top of the frame and the ray from the person’s eye

to the bottom of the frame. Find the value of x that maximizes

the viewing angle A.

46. What changes in exercise 45 if the person’s eye is 6 feet above

the ﬂoor?

EXPLORATORY EXERCISES

1. The idea of a homeomorphism helps to identify when super-

ﬁcially different sets are essentially the same. For example,

the intervals (0, 1), (0,π) and (−π/2,π/2) are all ﬁnite open

intervals. They are homeomorphic. By deﬁnition, intervals I

and I

are homeomorphic if there is a function f from I

onto

that has an inverse with f and f

−1

both being continuous.

A homeomorphism from (0, 1) to (0,π)is f (x) = π x. Show

that this is a homeomorphism byﬁnding its inverse and verify-

ing that both are continuous. Find a homeomorphism from

(0,π)to(−π/2,π/2). [Hint: Sketch a picture, and decide

how you could move the interval (0,π) to produce the inter-

val (−π/2,π/2).] Find a homeomorphism for any two ﬁnite

open intervals (a, b) and (c, d). It remains to decide whether

the interval (−∞, ∞) is different because it is inﬁnite or the

same because it is open. In fact, (−∞, ∞) is homeomorphic to

(−π/2,π/2) and hence to all other open intervals. Show that

tan

−1

x is a homeomorphism from (−∞, ∞)to(−π/2,π/2).

2. Explore the graphs of e

−x

, xe

−x

, x

−x

and x

−x

. Find all lo-

calextremaandgraphicallydeterminethebehavioras x →∞.

You can think of the graph of x

−x

as showing the results of

a tug-of-war: x

→∞as x →∞but e

−x

→ 0asx →∞.

Describe the graph of x

−x

in terms of this tug-of-war.

3. Suppose that a hockey player is shooting at a 6-foot-wide net

from a distance of d feet away from the goal line and 4 feet to

the side of the center line. (a) Find the distance d that maxi-

mizes the shooting angle. (b) Repeat part (a) with the shooter

2 feet to the side of the center line. Explain why the answer is

so different. (c) Repeat part (a) with the goalie blocking all but

the far 2 feet of the goal.

6.6 THE HYPERBOLIC FUNCTIONS

The Gateway Arch in Saint Louis, Missouri, is one of the most distinctive and recognizable

architectural structures in the United States. Most people think that it is taller than it is

wide, but this is the result of a common optical illusion. In fact, the arch has the same width

as height. A slightly less mysterious misconception is that the arch’s shape is not that of

a parabola. Rather, its shape corresponds to the graph of the hyperbolic cosine function

(called a catenary). This function and the other ﬁve hyperbolic functions are introduced in

this section.

Saint Louis Gateway Arch

The hyperbolic functions are not entirely new, as they are simply common combina-

tions of exponentials. We study them because of their usefulness in applications and their

convenience in solving equations (in particular, differential equations).

The hyperbolic sine function is deﬁned by

sinh x =

− e

−x

for all x ∈ (−∞, ∞). The hyperbolic cosine function is deﬁned by

cosh x =

+ e

−x

again for all x ∈ (−∞, ∞). We leave it as an exercise to use the deﬁnitions to verify the

important identity

cosh

x −sinh

x = 1, (6.1)

for all x. Notice that if we take x = coshu and y = sinhu, then from (6.1) with x replaced

by u,

− y

= cosh

u − sinh

u = 1,

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CONFIRMING PAGES

412 CHAPTER 6

Exponentials, Logarithms and Other Transcendental Functions 6-38

which you should recognize as the equation of a hyperbola. This identity is the source of

the name “hyperbolic” for these functions. You should also notice some parallel with the

trigonometric functions cos x and sin x.

The remaining four hyperbolic functions are deﬁned in terms of the hyperbolic sine and

hyperboliccosine functions, inamanner analogous totheirtrigonometriccounterparts. That

is, we deﬁne the hyperbolic tangent function tanh x, the hyperbolic cotangent function

coth x, the hyperbolic secant function sech x and the hyperbolic cosecant function csch x

as follows:

tanh x =

sinh x

cosh x

, coth x =

cosh x

sinh x

sech x =

cosh x

, csch x =

sinh x

These functions are remarkably easy to deal with, and we can readily determine their

behavior, using what we already know about exponentials. First, note that

sinh x =



− e

−x



+ e

−x

= cosh x.

Similarly, we can establish the remaining derivative formulas:

cosh x = sinh x,

tanh x = sech

coth x =−csch

sech x =−sech x tanh x

and

csch x =−csch x coth x.

These are all elementary applications of earlier derivative rules and are left as exercises. As

it turns out, only the ﬁrst three of these are of much signiﬁcance.

EXAMPLE 6.1 Computing the Derivative of a Hyperbolic Function

Compute the derivative of f (x) = sinh

(3x).

Solution From the chain rule, we have



(x) =

sinh

(3x) =

[sinh(3x)]

= 2 sinh(3x)

[sinh(3x)]

= 2 sinh(3x) cosh(3x)

(3x)

= 2 sinh(3x) cosh(3x)(3)

= 6 sinh(3x) cosh(3x).



Of course, every new differentiation rule gives us a corresponding integration rule. In

particular, we now have



cosh xdx = sinh x + c,



sinh xdx = cosh x + c

and



sech

xdx= tanhx + c.