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Chapter 2, , Rates of Change, and the, Chain Ru, The rate at which one variable is changing with respect to another can be, computed using differential calculus., , In Chapter 1, we learned how to differentiate algebraic functions and, thereby,, to find velocities and slopes. In this chapter, we will learn some applications, involving rates of change. We will also develop a new rule of differential, calculus called the chain rule. This rule is important for our study of related, rates in this chapter and will be indispensable when we come to use trigonometric and exponential functions., , 2.1 Rates of Change and the, Second Derivative, If y, , = f(x),, , then f'(x) is the rate of change of y with respect to x., , The derivative concept applies to more than just velocities and slopes. To, explain these other applications of the derivative, we shall begin with the, situation where two quantities are related linearly., Suppose that two quantities x and y are related in such a way that a, change Ax in x always produces a change Ay in y which is proportional to, Ax; that is, the ratio Ay/Ax equals a constant, m. We say that y changes, proportionally or linearly with x., For instance, consider a hanging spring to which objects may be attached. Let x be the weight of the object in grams, and let y be the resulting, length of the spring in centimeters. It is an experimental fact called Hooke's, law that (for values of Ax which are not too large) a change Ax in the weight, of the object produces a proportional change Ay in the length of the spring., (See Fig. 2.1.1.), If we graph y against x, we get a segment of a straight line with slope, Figure 2.1.1. Hooke's law, states that the change in, length Ay is proportional to, the change in weight Ax., , m = - A~, , Ax, as shown in Fig. 2.1.2. The equation of the line is y, , = rnx, , + b,, , and the, , Copyright 1985 Springer-Verlag. All rights reserved.

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100, , Chanter 2 Rates sf Change and the Chain Rule, , +, , function f(x) = mx b is a linear function. The slope m of a straight line, represents the rate of change o f y with respect to x . (The quantity b is the length, of the spring when the weight is removed.), , proportionally to x when, Ay/Ax is constant., , C, , v, , Suppose that y changes proportionally with x, and the rate of change is 3. If, = 2 when x = 0, find the equation relating y to x ., , Example 4, , y, Solution, , The rate of change is the slope: m = 3. The equation of a straight line with this, slope is y = 3x + b, where b is to be determined. Since y = 2 when x = 0, b, must be 2; hence y = 3x + 2. A, , The variable y changes proportionally with x when y is related to x by a, , Example 2, , Let S denote the supply of hogs in Chicago, measured in thousands, and let P, denote the price of pork in cents per pound. Suppose that, for S between 0, and 100, P changes linearly with S . On April 1, S = 50 and P = 163; on April, 3, a rise in S of 10 leads to a decline in P to 161. What happens if S falls, to 30?, , Solution, , Watch the words of and to! The rise in S of 10 means that AS = 10; the, decline of P to 161 means that A P = 16 1 - 163 = - 2. Thus the rate of change, is - 6 = - L . (The minus sign indicates that the direction of price change is, opposite to the direction of supply change.) We have P = - S b for some, b. Since P = 163 when S = 50, we have 163 = - f .50 + b, or b = 163 + 10, = 173, so P = - 15 S, 173. When S = 30, this gives P = -6, 173 = 167. At, this point, then, pork will cost $1.67 a pound. A, , ,, , 4 +, , +, , +, , If the dependence of p~ = j ( x ) on x is not linear, we can still introduce the, notion of the average rate of change of y with respect to x, just as we, introduced the average velocity in Section 1.1. Namely, the difference quotient, , is called the average rate of change of y with respect to x on the interval between, x , and x , + Ax. For functions f which are not linear, this average rate of, change depends on the interval chosen. If we fix x , and let Ax approach 0 , the, limit of the average rate of change is the derivative f'(x,), which we refer to as, the rate of change of y with respect to x at the point xo. This may be referred to, as an instantaneous rate of change, especially when the independent variable, represents time., , Copyright 1985 Springer-Verlag. All rights reserved.

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2.1 Rates of Change and the Second Derivative, , Example 3, , Solution, , 101, , An oil slick has area y = 30x3 + l00x square meters x minutes after a tanker, explosion. Find the average rate of change in area with respect to time during, the period from x = 2 to x = 3 and from x = 2 to x = 2.1. What is the, instantaneous rate of change of area with respect to time at x = 2?, The average rate of change from x = 2 to r = 3 is, (30 . 33, , + 100. 3 - 30 . z3 - 100 . 2) square meters, , square meters, = 670, 1 minute, minute, From x = 2 to x = 2.1, the average rate is calculated in a similar way to be, 47.83 = 478.3 square meters, 0.1, minute, Finally, the instantaneous rate of change is found by evaluating the, derivative 90x2 100 at x = 2 to obtain 460. Since the instantaneous rate of, change is a limit of average rates, it is measured in the same units, so the oil, slick is growing at a rate of 460 square meters per minute after 2 minutes. A, , +, , If two quantities x and y are related by y = f(x), the derivative f'(xo), represents the rate of change of y with respect to x at the point xo.It is, measured in (units of y)/(units of x)., , A positive rate of change is sometimes called a rate of increase., Example 4, , A circle with radius r millimeters has area A = a r 2 square millimeters. Find, the rate of increase of area with respect to radius at r, = 5. Interpret your, answer geometrically., , Solution, , Were A = f(r) = rr2. Since a is a constant, the derivative f'(r) is 2ar, and, f'(5) = 1071. Notice that the rate of change is measured in units of (square, millimeters)/millimeters, which are just millimeters. The value 2ar of the rate, of change can be interpreted as the circumference of the circle (Fig. 2.1.3). A, , Figure 2.1.3. The rate of, change of A with respect to, r is 271r, the circumference, of the circle., , In the next two examples, a negative rate of change indicates that one, quantity decreases when another increases. Since Ay = f(xo Ax) - f(x,),, it follows that Ay is negative when f(xo Ax) < f(xo). Thus, if Ay/Ax is, negative, an increase in x produces a decrease in y . This leads to our stated, interpretation of negative rates of change. If a rate of change is negative, its, absolute value is sometimes called a rate of decrease., , +, , Example 5, , +, , Suppose that the price of pork P depends on the supply S by the formula, , P = 160 - 3s + (0.01)S2. Find the rate of change of P with respect to S when, S, , = 50., , (See Example 2 for units.), , Copyright 1985 Springer-Verlag. All rights reserved.

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182, , Chapter 2 Rates of Change and the Chain Rule, , Sslartlon The rate of change is the derivative of f(S) = 0.01 S2- 3 S + 160 with respect, to x = S at x, = 50. The derivative is f'(S) = 0.02S - 3. When S = 50, we get, f'(50) = 1 - 3 = -2. Thus the price is decreasing by 2 cents per pound per, thousand hogs when S = 58. thousand. A, , +, , Example 6, , A reservoir contains lo8 - 104t - 80t2 - lot3 5t5 liters of water at time I ,, where t is the time in hours from when the gates are opened. How many liters, per hour are leaving the reservoir after one hour?, , Solution, , The rate of change of the amount of water in the reservoir is the derivative of, the polynomial 10' - 104t - 80t2 - lot3 5t5, namely - lo4 - 160t - 30t2, 25t4. At t = 1, this equals - lo4 - 160 - 30 + 25 = - 10,165 liters per hour., This is negative, so 10,165 liters per hour are leaving the reservoir after one, hour. A, , +, , +, , We now reconsider the velocity and acceleration of a particle moving on a, straight line. Suppose for the moment that the line is vertical and designate, one direction as "+"and the other as " - ". We shall usually choose the, but consistently using the other sign would give, upward direction as, equivalent results. We also choose some point as the origin, designated by, x = 0, as well as a unit of length, such as meters. Thus, if our designated origin, represents the level of the Golden Gate Bridge, x = 100 would designate a, location 100 meters\,,above the bridge along our vertical straight line, and, x = - 10 would indicate a location 10 meters beneath the bridge (Fig. 2.1.4)., , "+,", , Figure 2.1.4. A coordinate, upwards, system with ", and x = 0 at bridge level., , +", , I, Suppose that, at time t, a particle has location x = f(t) along our line. We, call [ f(t + At) - f(t)]/At the average velocity and dx/dt = f'(t) the instantaneous velocity; this can either be positive, indicating upward motion, or can, be negative, indicating downward motion., , Example 7, , Suppose that x = 0 represents the level of the Golden Gate Bridge and that, x = f(t) = 8 6t - 5t2 represents the position of a stone at time t in seconds., , +, , (a) Is the stone above the bridge, at the bridge, or below the bridge at t = 0?, How about at t = 2?, (b) Suppose that the average velocity during the interval from to to to + At is, negative; what can be said about the height at time to At?, (c) What is the instantaneous velocity at t = I?, , +, , Solullon, , (a) At t = 0, x = 8, so the stone is 8 meters above the level of the bridge. At, t = 2, x = 8 + 6 . 2 - 5 . 4 = 0, so the stone is at the level of the bridge., (b) It is less than that at time to., (c) We compute dx/dt = 6 - 10 t, which at t = 1 is -4. Thus, the instantaneous velocity is 4 meters per second downward. A, , Copyright 1985 Springer-Verlag. All rights reserved.

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2.1 Rates of Change and the Second Derivative, ( p o s i t ~ o n .V), , I, I, , t,, , ,(l,me,, , Figure 2.1.5. The velocity is, positive until = t , and, negative afterward., , Example 8, , 403, , Of course, these interpretations of positive and negative velocity also apply to, horizontal motion, like that which we discussed at the beginning of Chapter 1., In particular, we may let x denote the position along a road with larger values, of x corresponding to points further east, say. Then d x / d t > 0 indicates that, motion is eastward; d x l d t < 0 indicates westward motion. The magnitude, or, absolute value, of the velocity is called the speed., If we graph x = f(t) against t , the slope of the graph indicates whether, the velocity is positive or negative. (See Fig. 2.1.5.), Note that the instantaneous velocity v = d x l d t = f'(t) is usually itself, changing with time. The rate of change of v with respect to time is called, acceleration; it may be computed by differentiating v = f'(t) once again., Suppose that x, , = j(t) = $ t 2 -, , t, , + 2 denotes the position of a bus at time t., , (a) Find the velocity as a function of time; plot its graph., (b) Find and plot the speed as a function of time., (c) Find the acceleration., Solution, , (a) The velocity is v, 11, , =, , dx/dt, , =, , t t - 1 (see Fig. 2.1.6(a))., , = velocity, , lul = speed, , Figure 2.1.6. (a) Velocity, and (b) speed as functions, The acceleration dv/dt is, constant and positive., , I, (b), , (b) The speed 101 = I t t - 1I [see Fig. 2.1.6(b)]., (c) The acceleration is d v l d t = A, , 4., , In this example, the acceleration happens to be constant and positive, indicating that the velocily is increasing at a constant rate. Note, though, that the, speed decreases and then increases; it decreases when the velocity and acceleration have opposite signs and increases when the signs are the same. This may, be illustrated by an example. If your car is moving backwards (negative, velocity) but you have a positive acceleration, your speed decreases until your, car reverses direction, moves forward (positive velocity), and the speed increases., Since acceleration is the derivative of the velocity and velocity is already, , Copyright 1985 Springer-Verlag. All rights reserved.

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104, , Chapter 2 Rates of Change and the Chain Rule, , a derivative, we have an example of the general concept of the second, derivative, i.e., the derivative of the derivative. If y = f(x), the second derivative is denoted fU(x) and is defined to be the derivative of f'(x). In Leibniz, notation we write d?/dx2 for the second derivative of y = f(x). Note that, d?/dx2 is not the square of dy/dx, but rather represents the result of the, operation d/dx performed twice., If an object has position x (in meters) which is a function x = f(t) of time, t (in seconds), the acceleration is thus denoted by f"(t) or d2x/dt2. It is, measured in meters per second per second, i.e., meters per second2 or feet per, second2., , To compute the second derivative f"(x):, 1. Compute the first derivative f'(x)., 2. Calculate the derivative of f'(x); the result is f "(x)., , The second derivative of y = f(x) is written in Leibniz notation as, , The second derivative of position with respect to time is called accelera-, , If we plot the graph y =j(x), we know that f'(x) represents the slope of the, tangent line. Thus, if the second derivative is positive, the slope must be, increasing as we move to the right, as in Fig. 2.1.7(a). Likewise, a negative, second derivative means that the slope is decreasing as we increase x, as in, Fig. 2.1.7(b)., , change of slope with, respect to x is the second, derivative f "(x). (a)f " > 0., (b) f " < 0., , Example 9, , -, , I, , c, , Calculate the second derivative of, (a) f(x) = x4 + 2x3 - 8x,, (c), , Solution, , ( h ) decreasing slope means, trcgativc second d c r ~ v a t ~ v c, , ( a ) increasing slope tneans positlve, second der~vative, , d 2 (3x2 - 2x, dx2, , + I),, , (b) f(x), (d), , =, , x+l, -., , 6, , d 2 (8r2 + 21. + 10)., dr, , (a) By our rules for differentiating polynomials from Section 1.4,, f'(x) = 4x3 + 6x2 - 8., Now we differentiate this new polynomial:, f U ( x )= 12x2+ 12x - 0 = 12(x2+ x)., , Copyright 1985 Springer-Verlag. All rights reserved.

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2.1 Rates of Change and the Second Derivative, , 105, , (b) By the quotient rule from Section 1.5,, , By the quotient rule again,, , and so, , d2(3x2- 2x, , + 1), , =, , dx, , d, -(6x, dx, , - 2) = 6,, , a constant function., d 2 (8r2, (d) 7, dr, , 1, , + 2r + 10) = ( - f ) [ ( $ ) ( 8 r 2 + 2r + 10), , Next we consider a word problem involving second derivatives., Example 10, , A race car travels, seconds being f(t), , mile in 6 seconds, its distance from the start in feet after t, + 132t., , = 44t2/3, , (a) Find its velocity and acceleration as it crosses the finish line., (b) How fast was it going halfway down the track?, Solullon, , +, , (a) The velocity at time t is v = f'(t) = 88t/3, 132, and the acceleration is, a = f"(t) = y . Substituting t = 6, we get v = 308 feet per second ( = 210, miles per hour) and a x 29.3 feet per second2., (b) To find the velocity halfway down, we do not substitute t = 3.00 in, v = f'(t)-that, would be its velocity after half the time has elapsed. The, total distance covered is f(6) = (44)(36)/3 (132)(6) = 1320 feet ( =, mile). Thus, half the distance is 660 feet. To find the time t corresponding, to the distance 660, we write f(t) = 660 and solve for t using the quadratic, formula:, , +, , t2 + 9t - 45 = 0, t, , =, , -9rtJ8ii87j;, 2, , a, , (multiply by &),, , = - 12.58,3.58, , (quadratic formula)., , Since the time during the race is positive, we discard the negative root and, retain t = 3.58. Substituting into v = f'(t) = 88t/3 + 132 gives v x 237 feet, per second (e162 miles per hour). A, We end this section with a discussion of some concepts from economics,, where special names are given to certain rates of change., Imagine a factory in which x worker-hours of labor can produce y = f(x), dollars worth of output. First, suppose that y changes proportionally with x ., , Copyright 1985 Springer-Verlag. All rights reserved.

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906, , Chapter 2 Rates of Change and the Chain Rule, , +, , Then Ay = f i x , Ax) - f(x,) represents the amount of extra output produced if Ax extra worker-hours of labor are employed. Thus, A y / A x is the, output per worker-hour. This average rate of change is called the productivity, of labor., Next, suppose that f ( x ) is not necessarily linear. Then A y / A x is the extra, output per extra worker-hour of extra labor when Ax extra worker-hours are, employed. The limiting value, as Ax becomes very small, is f'(xo). This, instantaneous rate of change is called the marginal productivity of labor at the, level x,., In Fig. 2.1.8 we sketch a possible productivity curve y = f ( x ) . Notice that, as x , becomes larger and larger, the marginal productivity f'(x,) ( = dollars of, output per worker-hour at level x,) becomes smaller. One says that the law of, diminishing returns applies., f, , Figure 2.1.8. A possible, productivity curve; the, slope of the tangent line is, the marginal productivity., , Y = Output in dollars, , 1, , 1, , Example I 1, , Solution, , I~OO, , 20b0, , .x(,, , 1I, 3d00, , I, , C, , 4obo, , i=, , worker-hours, , A bagel factory produces 30x - 2 x 2 - 2 dollars worth of bagels for each x, worker hours of labor. Find the marginal pr~ductivitywhen 5 worker koqrs, .we smpto9eh., , The output is f i x ) = 30x - 2 x 2 - 2 dollars. The marginal productivity at, x , = 5 is f'(5) = 30 - 4 . 5 = 10 dollars per worker-hour. Thus, at xo = 5,, production would increase by 10 dollars per additional worker-hour. a, Next we discuss marginal cost and marginal revenue. Suppose that a company, makes x calculators per week and that the management is free to adjust x ., Define the following quantities:, C ( x ) = the cost of making x calculators (labor, supplies, etc.), R ( x ) = the revenue obtained by producing x calculators (sales)., P i x ) = R ( x ) - C ( x ) = the profit., , Even though C ( x ) , R ( x ) , and P ( x ) are defined only for integers x ,, economists find it useful to imagine them defined for all real x . This works, nicely if x is so large that a change of one unit, Ax = 1, can legitimately be, called "very small.", The derivative C f ( x )is called the marginal cost and R 1 ( x )is the marginal, revenue:, C ' ( x ) = marginal cost =, , the cost per calculator for producing, additional calculators at production level x., , the revenue per calculator obtained by, R ' ( x ) = marginal revenue = producing additional calculators at, production level x ., Since P i x ) = R ( x ) - C ( x ) , we get P ' ( x ) = R r ( x )- C 1 ( x ) ,the profit per additional calculator at production level x . This is the marginalprofit. If the price, , Copyright 1985 Springer-Verlag. All rights reserved.

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2.1 Rates of Change and the Second Derivative, , 107, , per unit is f ( x ) a n d x calculators a r e sold, then R ( x ) = xf(x). By the product, rule, the marginal revenue is R i ( x ) = x f ' ( x ) f(x)., , +, , Example 12, , Solution, , +, , +, , Suppose that it costs (30x 0.04x2)/(1 0.0003x3) dollars if x calculators, are made, where 0 < x < 100, a n d that calculators are priced a t 100 - 0 . 0 5 ~, dollars. If all x calculators are sold, what is the marginal profit?, The revenue is R = x(100 - 0.05x), so the profit is P = x(100 - 0 . 0 5 ~), (30x 0.04x2)/(1 0.0003x3).The marginal profit is therefore d P / d x , which, may be calculated using the sum a n d quotient rules as follows:, , +, , +, , Exercises for Seetion 2.1, In Exercises 1-4, assume that y changes proportionally, with x and the rate of change is r. In each case, find y, as a function of x, as in Example 1., 1. r = 5 , y = 1 when x = 4 ., 2. r = - 2 , y = 10 when x = 15., 3. r = $ , y = 1 when x = 3., 4. r = l O , y = 4 w h e n x = -1., 5. If the price of electricity changes proportionally, with time, and if the price goes from 2 cents per, kilowatt-hour in 1982 to 3.2 cents per kilowatthour in 1984, what is the rate of change of price, with respect to time? When will the price be 5, cents per kilowatt-hour? What will the price be, in 1991?, 6. It will take a certain woman seven bags of cement to build a 6-meter-long sidewalk of uniform, width and thickness. Her husband offers to contribute enough of his own labor to extend the, sidewalk to 7 meters. How much more cement, do they need?, 7. A rock is thrown straight down the face of a, vertical cliff with an initial velocity of 3 meters, per second. Two seconds later, the rock is falling, at a velocity of 22.6 meters per second. Assuming, that the velocity v changes proportionally with, time t , find the equation relating v to t. How fast, is the rock falling after 15 seconds?, 8. In November 1980, Mr. B used 302 kilowatthours of electricity and paid $18.10 to do so. In, December 1980, he paid $21.30 for 366 kilowatthours. Assuming that the cost of electricity, changes linearly with the amount used, how, much would Mr. B pay if he used no electricity, at all? Suppose that Mr. I3 can reduce his bill to, zero by selling solar-generated electricity back to, , the company. How much must he sell? Interpret, your answers on a graph., Find the average rate of change of the functions in, Exercises 9-12 on the specified interval., 9. f(t) = 400 - 20t - 16t2; t between to = 1 and, ti = t., , +, , 10. g ( t ) = 18t2+ 2t 3; t between to = 2 and t i =, 3.5., 11. f(x) = (x + i ) 2 ; x, = 2, Ax = 0.5., 12. g(s) = (3s 2)(s - 1) - 3s2; so = 0, As = 6., , +, , 13. The volume of a cone is f (area of base) x height., If the base has radius always equal to the height,, find the rate of change of the volume with respect to this radius., 14. Find the rate of change of the area of an equilateral triangle with respect to the length of one of, its sides., 15. During takeoff, a 747 has 25,000 - 80t 2t2 +, 0.2t3 gallons of fuel in its tanks t seconds after, starting its takeoff, 0 < t < 10. How many gallons per second are being burned 2 seconds into, the takeoff?, 16. A space shuttle's external tank contains lo5 1O4t - 103t3liters of fuel t minutes after blastoff., How many liters per minute are being burned, two minutes after blastoff?, 17. If the height N in feet of a certain species of tree, depends on its base diameter d in feet through, the formula N = 56d - 3d2, find the rate of, change of H with respect to d at d = 0.5., 18. Suppose that tension T of a muscle is related, to the time t of exertion by T = 5 3t - t 2 ,, 0 < t <;. Find the rate of change of T with, respect to t at t = 1., , +, , +, , Copyright 1985 Springer-Verlag. All rights reserved.

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108, , Chapter 2 Rates of Change and the Chain Rule, , 19. A flu epidemic has infected P = 30t2 + lOOt people by t days after its outbreak. How fast is the, epidemic spreading (in people per day) after 5, days?, 20. Find the rate of change of the area of a circle, with respect to its diameter when the diameter is, 10. Compare with Example 4., 21. A sphere of radius r has volume V = 4nr3. What, is the rate of change of the volume of the sphere, with respect to its radius? Give a geometric interpretation of the answer., 22. A balloon being blown up has a volume V =, 3t3 8t2 + 161 cubic centimeters after t minutes., What is the rate of change of volume (in cubic, centimeters per minute) at t = 0.5?, 23. Let x(t) = t3 + ct be the position of a particle at, time t. For which values of c does the particle, reverse direction, and at what times does the, reversal take place for each such value of c?, Does the value of c affect the particle's acceleration?, 24. An evasive moth has position j 3 - t + 2 at time, t. A hungry bat has positiony(t) = - $ t2 + t + 2, at time t. Now many chances does the bat have, to catch the moth? How fast are they going and, what are their accelerations at these times?, 25. If the position of a moving object at time t is, (t5 l)(t 2), find its velocity and acceleration, when t = 0.1., 26. Let h(t) = 2t3 be the position of an object moving along a straight line at time t. What are the, velocity and acceleration at t = 3?, Compute the second derivatives in Exercises 27-32., , +, , +, , +, , Find the second derivative of the functions in Exercises, 33-40., 33. f ( x ) = x 2 - 5, 34. f ( x ) = x - 2, 35. y = x5 + 7x4 - 2x + 3, 36. y = [(x - I) + x2][x3- 11, , In Exercises 41-46, find the velocities and accelerations, at the indicated times of the particles whose positions y, (in meters) on a line are given by the following func-, , tions of time t (in seconds):, 41. y = 3t 2; to = 1, 42. y = 5 t - 1 ; t o = O, 43. y = 8 t 2 + I; t , = O, 44. y = 1 8 t 2 - 2 t + 5 ; t o = 2, 45. y = 10 - 2t - 0.O1f4; to = 0, 46. y = 2 0 - 8t -0.02t6; t o = 1 ', , +, , 47. The height of a pebble dropped off a building at, time t = 0 is h(t) = 44.1 - 4.9t2 meters at time t., The pebble strikes the ground at t = 3.00 seconds., (a) What is its velocity and acceleration when it, strikes the ground?, (b) What is its velocity when it is halfway down, the building?, 48. The amount of rainy in inches at time x in hours, from the start of the September 3, 1975 Owens, Valley thunderstorm was given by y = 2x - x2,, Osxsl., (a) Find how many inches of rain per hour were, falling halfway through the storm., (b) Find how many inches of rain per hour were, falling after half an inch of rain has fallen., 49. A shoe repair shop can produce 20x - x2 - 3, dollars of revenue ever3 how wCen R 5~4rk*rs, are eapbo2rh. FLR&, .the mur$~nai p r o & ~ % t ~ i % g, when 5 workers are employes., 50. The owners of a restaurant find that they can, serve 300w - 2w2 - 14 dinners when w workerweeks of labor are employed. If an average dinner is worth $7.50, what is the marginal productivity (in dollars) of a worker when 10 workers, are employed?, 51. A factory employing w workers produces, 1OOw + w2/ 100 - (1 /5000)w4 dollars worth of, tools per day. Find the marginal productivity o i, labor when w = 20., 52. A farm can grow IOOOOx - 35x3 dollars worth of, tomatoes if x tons of fertilizer are used. Find the, marginal productivity of the fertilizer when, x = 10. Interpret the sign of your answer., 53. In a boot factory, the cost in dollars of making x, boots is ( 4 x + 0 . 0 2 x 2 ) / ( 1 + 0.002x3). If, boots are priced at 25 - 0 . 0 2 ~dollars, what is, the marginal profit, assuming that x boots are, sold?, 54. Tn a pizza parlor, the cost in dollars of making x, p i z z a s is ( 5 x + 0 . 0 1 x 2 ) / ( 1 + 0 . 0 0 1 x 3 ) ., The price per pizza sold is set by the rule:, price = 7 - 0 . 0 5 ~if x pizzas are made. If all x, pizzas are sold, what is the marginal profit?, In each of Exercises 55-58, what name would you give, to the rate of change of y with respect to x? In what, units could this rate be expressed?, 55. x = amount of fuel used; y = distance driven in, an automobile., 56. x = distance driven in an automobile; y =, amount of fuel used., , Copyright 1985 Springer-Verlag. All rights reserved.

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2.1 Rates of Change and the Second Derivative, , 57. x = amount of fuel purchased; y = amount of, money paid for fuel., 58. x = distance driven in an automobile; y =, amount of money paid for fuel., 59. The cost c of fuel for driving, measured in cents, per kilometer, can be written as the product, c = rp, where r is the fuel consumption rate in, liters per kilometer and p is the price of fuel in, cents per liter. If r a n d p depend on time (the car, deteriorates, price fluctuates), so does c. The, rates of change are connected by the product, rule, , 60., , 61., , 62., 63., , Interpret in words each of the terms on the, right-hand side of this equation, and explain why, dc/dt should be their sum., If f(x) represents the cost of living at time x,, then f'(x) > 0 means that there is inflation., (a) What does f "(2)> 0 mean?, (b) A government spokesman says, "The rate at, which inflation is getting worse is decreasing." Interpret this statement in terms of, f'(x), fl'(x), and f"'(x)., Let y = 4x2 - 2x 7. Compute the average rate, of change of y with respect to x over the interval, from x, = 0 to x , = Ax for the following values, of Ax: 0.1,0.001,0.000001. Compare with the, derivative at x, = 0., Repeat Exercise 61 with Ax = -0.1, -0.001,, and - 0.00000 1., Find the average rate of change of the following, functions on the given interval. Compare with, the derivative at the midpoint., (a) f(x) = (X - +)(x 1) between x = - and, x = 0., (b) g(t) = 3(t 5)(t - 3) on [2,6]., (c) h ( r ) = 10r2-3r+6on[-0.1,0.4]., (d) r(t) = (2 - t)(t 4); t in [3,7]., (a) Let y = ax2 bx c, where a , b, and c are, constant. Show that the average rate of, change of y with respect to x on any interval, [ x I , x 2 ] equals the instantaneous rate of, change at the midpoint; i.e., at (x, x2)/2., (b) Let f(x) = ax2 bx + c, where a, b, and c, are constant. Prove that, for any x,,, , +, , +, , +, , +, , 64., , +, + +, , +, , +, , f(x) =f(xo) + f'(m>(x - xo>, where m = (x x,)/2., 65. The length I and width w of a rectangle are, functions of time given by 1 = (3 t2 t3) centimeters and w = (5 - t 2t2) centimeters at time, t (in seconds). What is the rate of change of area, with respect to time at time t?, 66. If the height and radius of a right circular cylin-, , +, , +, , + +, , 109, , +, , der are functions of time given by h = (1 + t2, t3 t4) centimeters and r = (1 2t - t2 t5), centimeters at time t (in seconds), what is the, rate of change with respect to time of the lateral, surface area, i.e., the total surface area minus the, top and bottom?, 67. Let f(t) = 2t2 - 5t + 2 be the position of object, A and let h ( t ) = -3t2 t 3 be the position of, object B., (a) When is A moving faster than B?, (b) How fast is B going when A stops?, (c) When does B change direction?, 68. Repeat Example 7 with the same data but with, the following two conventions changed. First, the, origin is now chosen at a point 20 meters above, the bridge. Second, we designate down as, and up as " - " rather than vice versa., 69. For which functions f(x) = ax2 + bx + c is the, second derivative equal to the zero function?, 70. How do the graphs of functions ax2 + bx c, whose second derivative is positive compare with, those for which the second derivative is negative, and those for which the second derivative is, zero?, 71. A particle is said to be accelerating (or decelerating) if the sign of its acceleration is the same as, (or opposite to) the sign of its velocity. (a) Let, f(t) = - t 3 be the position of a particle on a, straight line at time t. When is the particle accelerating and when is it decelerating? (b) If the, position of a particle on a line is given as a, quadratic function of time and the particle is, accelerating at time to, does the particle ever, decelerate?, 72. One summer day in Los Angeles, the pollution, index at 7:00 AM was 20 parts per milion, increasing linearly 15 parts per million each hour, until 5:00 PM. Let y be the amount of pollutants, in the air x hours after 7:00 AM., (a) Find a linear equation relating y and x., (b) The slope is the increase in pollution for, each hour increase in time. Find it., (c) Find the pollution level at 5:00 PM., 73. Straight-line depreciation means that the difference between current value and original value is, directly proportional to the time t. Suppose a, home office is presently furnished for $4000 and, salvaged for $500 after ten years. Assume, straight line depreciation., (a) Find a linear equation for the value V of the, office furniture after t years, for tax purposes., (b) The slope of the line indicates the decrease, in value each year of the office furniture, to, be usedfin preparing a tax return. Find it., , +, , +, , +, , + +, , Copyright 1985 Springer-Verlag. All rights reserved., , "+", +

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890, , Chapter 2 Rates of Change and the Chain Rule, , *74. Suppose that the acceleration of an object is, constant and equal to 9.8 meters/second2 and, that its velocity at time x = O is 2 meters/second., (a) Express the velocity as a function of x., (b) What is the velocity when x = 3?, (c) Express the position y of the object as a, function of x, if y = 4 when x = 0., (d) How far does the object travel between, x = 2 and x = 5?, , it75. Let, , f, , x, , ), x,, , x<O., x>o., , (a) Sketch a graph of f(x)., (b) Find f'(x). Sketch its graph., (c) Find f "(x) for x f 0. Sketch its graph. What, happens when x = O?, (d) Suppose that f(x) is the position of an object, at time x. What might have happened at, x = O?, , 2.2 The Chain Rule, The derivative of f(g(x)) is a product of derivatives., None of the rules which we have derived so far tell us how to differentiate, JE, = (x3 - 5)'12. The chain rule will. Before deriving this rule, though,, we shall look at what happens when we differentiate a function raised to an, integer power., If g(x) is any function, we can use the product rule to differentiate, [ g(x)12:, , If we write u = g(x), this can be expressed in Leibniz notation as, , In the same way, we may differentiate u3:, , Similarly, (d/dx)(u4) = 4u3(du/dx) (check it yourself); and, for a general, positive integer n, we have (dldx) un = nun-' (duldx). (This may be formally, proved by induction-see Exercise 52.), , To differentiate the nth power [ g ( x ) r of a function g(x), where n is a, positive integer, take out the exponent as a factor, reduce the exponent, by 1, and multiply by the derivative of g(x):, , ( gn)'(x) = nII g(x) In-'g'(x)*, , If u = x, then duldx, ordinary power rule., , =, , I, and the power of a function rule reduces to the, , Copyright 1985 Springer-Verlag. All rights reserved.

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2.2 The Chain Rule, , 111, , A common mistake made by students in applying the power of a function rule, is to forget the extra factor of gf(x)-that is, du/dx., , Example 1 Find the derivative of [g(x)13,where g ( x ) = x4 + 2x2,first by using the power, of a function rule and then by expanding the cube and differentiating directly., Compare the answers., , Solullon, , By the power of a function rule, with u = x4, , + 2x2 and n = 3,, +, , If we expand the cube first, we get ( x 4 2, 8x6, SO, , +, , ~= xI2, ~ + )6xI0~ 12x8 +, , = 12x" + 60x9 + 96x7 + 48x5., To compare the two answers, we expand the first one:, , which checks. A, Example 2, Solution, , d, (s4+ 2s3 + 3)8, ds, , Find, , We apply the power of a function rule, with u, x replaced by s):, , = 8(s4, , = s4 + 2s3, , + 3 (and the variable, , + 2s' + 3)7(4s3+ 6s')., , (You could also do this problem by expanding the eighth power and then, differentiating; obviously, this practice is not recommended.) A, Example 3, Solution, , + 1 ) 2 7 ( ~ 4+ 3x +, , If y, , =(x2, , find the rate of change of y with respect to x ., , First of all, by the power of a function rule,, , and, d ( x 4+ 3x, -, , +, , + 1 )8 = 8(x4 3x + I ) 7(4x3 + 3), dx, Thus, by the product rule, the rate of change of y with respect to x is, , + (xi + I, , ) ~ 8(x4, ~ . + 3x, , + 1 ) ~ ( 4 x+' 3)., , To simplify this, we can factor out the highest powers of x2 + 1 and x4 +, 3x 1 to get, , +, , (xi, , + 1 ) 2 6 ( ~ 4+ 3x + 1 ) ~ [ 2 72 x ( x 4+ 3x + I ) -+ ( x 2+ 1) . 8(4x3+ 3 ) ] ., , We can consolidate the expression in square brackets to a single polynomial of, , Copyright 1985 Springer-Verlag. All rights reserved.

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112, , Chapter 2 Rates of Change and the Chain Rule, , +, , + +, , +, , +, , degree 5, getting 2(x2 1 ) 2 6 ( ~ 4 3x 1)'(43x5 16x3 + 93x2 27x + 12), as our rate of change. [Note: Consult your instructor regarding the amount of, simplification required.] A, The power of a function rule is a special case of an important differentiation, rule called the chain rule. To understand this more general rule, we begin by, noting that the process of forming the power [ g ( x ) r can be broken into two, successive operations: first find u = g(x), and then find f(u), where y = f(u), = un. The chain rule will help us to differentiate any function formed from, two functions in this way., If f and g are functions defined for all real numbers, we define their, composition to be the function which assigns to x the number f(g(x)). The, composition is often denoted by f 0 g. Thus (f 0 g)(x) = f(g(x)). To evaluate, y = (f 0 g)(x), we introduce an intermediate variable u and write u = g(x) and, y = f(u). To evaluate y, we substitute g(x) for u in f(u). (If f(x) and g(x) are, not defined for all x, then (f 0 g)(x) is defined only when x is in the domain, of g and g(x) is in the domain off.), Example 4, , Solution, , +, , (a) If f(u) = u3 + 2 and g(x) = ( x 2 I ) ~what, ,, is h = f 0 g?, t, and g(x) = x 3 - 5. Find f 0 g and g 0 f., (b) ~ ef(x), /[2 + (1 + x')~] as a composition of simpler functions., (c) Write -/I, , =G, , (a) We calculate h(x) = f(g(x)) by writing u = g(x) and substituting in f(u)., We get u = (x2 1)' and so, , +, , The functions f 0 g and g 0 f a r e certainly different., u3). Then the given function can be, (c) Let g(x) = 1 + x2 and f(u) = &/(2, written as f 0 g. A, , +, , @ Gaflculator Discussion, , 6,, , On electronic calculators, several functions, such as l / x , x2,, and sinx, are, evaluated by the push of a single key. To evaluate the composite function f 0 g, on x, you first enter x, then push the key for g to get g(x), then push the key, for f to get f(g(x)). For instance, let f(x) = x2, g(x) = sinx. To calculate, (f 0 g)(x) = f(g(x)) = sin^)^ for x = 32 (degrees), we enter 32, then press the, sin key, then the x2 key. The result is 32 +0.52991926 +0.2808 1442. Notice, that ( g 0 f)(x) = sin(x2) is quite different: entering 32 and pressing the x 2 key, followed by the sin key, we get 3 2 3 1 0 2 4 3 -0.82903756. A, Do not confuse the composition of functions with the product. We have, (fs)(x> = f ( x >g(x>7, while, ( f o g > ( x > =f ( g(x))., , Copyright 1985 Springer-Verlag. All rights reserved.

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2.2 The Chain Rule, , 113, , In the case of the product we evaluate f(x) and g(x) separately and then, multiply the results; in the case of the composition, we evaluate g(x) first and, then apply f to the result. While the order off and g does not matter for the, product, it does for composition., , The composition f g is obtained by writing u = g(x) and evaluating, f(u). To break up a given function h(x) as a composition, find an, intermediate expression u = g(x) such that h(x) can be written in terms, 0, , The derivative of a composite function turns out to be the product of the, derivatives of the separate functions. The exact statement is given in the, following box., , To differentiate a composition f(g(x)), differentiate g at x, differentiate f, at g(x), and multiply the results:, , g)'(x> = f'( g(x)) . g'(x)., In Leibniz notation,, , (f, , O, , A complete proof of the chain rule can be given by using the theory of limits, (see Review Exercise 99, Chapter 11). The basic argument, however, is simple, and goes as follows. If x is changed by a small amount Ax and the, corresponding change in u = g(x) is Au,we know that, , Corresponding to the small change Au is a change Ay i n y = f(u), and, , f'(u), , =, , d~ = lim, . AY ., -, , du AU+O Au, To calculate the rate of change dy/dx, we write, , In going to the second line, we replace Ax 4 0 by Au -+0 because the, differentiable function g is continuous, i.e., Ax -+0 implies Au -+ 0, as we saw, in Section 1.3., There is a flaw in this proof: the Au determined by Ax could well be zero,, and division by zero is not allowed. This difficulty is fortunately not an, essential one, and the more technical proof given in Chapter 11 avoids it., Notice that the chain rule written in Leibniz notation is closely related to, , Copyright 1985 Springer-Verlag. All rights reserved.

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114, , Chapter 2 Rates of Change and the Chain Rule, , our argument and is easy to remember. Although du does not really have an, independent meaning, one may "cancel" it informally from the product, ( d y / d u ) ( d u l d x ) to obtain d y / d x ., A physical model illustrating the chain rule is given at the end of this, section., , Example 5, Solution, , Verify the chain rule for f ( u ) = u2 and g ( x ) = x 3 + 1., , +, , Let h ( x ) = f ( g ( x ) ) = [ g ( ~ ) ]=2 ( x 3+ 1)2 = x6 + 2x3 1. Thus h l ( x )= 6 x 5 +, 6x2. On the other hand, since f'(u) = 2u and g t ( x ) = 3x2,, , I3ence the chain rule is verified in this case. A, Let us check that the power of a function rule follows from the chain rule: If, y = [ g ( x ) r ,we may write u = g ( x ) , y = f ( u ) = u ". Since dy/du = nu "- the, chain rule gives, , ',, , This calculation applies to negative or zero powers as well as positive ones., Thus the power of a function rule holds for all integer powers., , Example 6, Solullon, , + 1)Io0]. Find f'(x)., We write f ( x ) as ( 3 x 2- 2 x + 1)-loo. Thus, Let f ( x ), , =, , 1 / [ ( 3 x 2- 2 x, , The chain rule also solves the problem which began the section., , Example 7, Solution, , Differentiate J x 3 - 5 ., , J/n, , In Example 4(b) we saw that J, E, = li'E; if u = x 3 - 5. Thus, if y =, = 6 , then d y l d u = 1 / 2 6 (Example 4, Sect. 1.3), and d u / d x = 3x2, so, , Example 8, , If h ( x ) = f ( x 2 ) , find a formula for h'(x). Check your formula in the case, f ( u ) = u3., , Solution, , Let u = g ( x ) = x2, so h ( x ) = f(u). Then h t ( x )= f'(u) . g r ( x )= f'(x2) - 2x., Thus h'(x) = f ' ( x 2 ) . 2 x ., If f ( u ) = u3, then f'(u) = 3u2 and f'(x2) = 3x4. Thus, h'(x) = 3x4 2 x, = 6x5. In fact, h ( x ) = ( x ~=)x6, ~ in this case, so differentiating h directly gives, the same result. A, , Example 9, Solullon, , Use the chain rule to differentiate f ( x ), Let u = ( x 2+ 1)20, , + 1 and y, , + 1)20 + I ) ~ (Do, ., not expand!), , = ((x2, , = u4, s o y = f ( x ) . By the chain rule,, 20, , To calculate d u / d x we use the chain rule again (or the power of a function, rule):, , Copyright 1985 Springer-Verlag. All rights reserved.

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2.2 The Chain Rule, , 115, , Thus, , The following examples require us to translate words into equations before, using the chain rule., Example 10, , The population of Thin City is increasing at the rate of 10,000 people per day, on March 30, 1984. The area of the city grows to keep the ratio of 1 square, mile per 1000 people. How fast is the area increasing per day on this date?, , Solution, , Let A = area, p = population, t = time (days). The rate of increase of area, with respect to time is dA /dt = (dA /dp)(dp/dt) = & . 10000 = 10 square, miles per day. A, , Example 11, , A dog 2 feet high trots proudly away from a 10-foot-high light post,, When he is 8 feet from the post's base, ko iS maQxfig a*, 3 Gee% per SCCDD&- \iaw .Fa%.% t%4 2 &Lp, ~ ~05 kki shdclc~r n o ~ " y ?, , Solution, , Refer to Fig. 2.2.1. By similar triangles, y / ( y - x) = 10/2; solving, y, Then dy /dl = (dy /dx)(dx/ dt) = ($)3 = 3 3 feet per second. A, , Figure 2.2.1. Dog trotting, proudly away from lamp, post., , = 5x/4., , Figwe 2-2.2. The geometric, interpretation of the, shifting rule., , Another special case of the chain rule may help you to understand it., Consider h(x) = f(x + c), c a constant. If we let u = g(x) = x + c, we get, g'(x) = 1, so, h'(x) = f'(g(x)) . gf(x) = f'(x, c) . 1 = f'(x + c)., , +, , Note that the graph of h is the same as that off except that it is shifted c, units lo the left (see Fig. 2.2.2). It is reasonable, then, that the tangent line to, the graph of h is obtained by shifting the tangent line to the graph off. Thus,, in this case, the chain rule is telling us something geometrically obvious. One, might call this formula the shifting rule. In Lcibniz notation it reads, , Copyright 1985 Springer-Verlag. All rights reserved.

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1116, , Chapter 2 Rates of Change and the Chain Rule, , Supplement to SectEon 2.2, A Physical Model lor the Chain Rule, When you change altitude rapidly, as in a moving car or plane, a pressure, difference develops between the inside and outside of your eardrums, and, your ears "pop." Three variables relevant to this phenomenon are the time t ,, the altitude u, and the air pressure p. Ear popping occurs when the rate of, change dp/dt is too large., The rate d p / d t is hard to measure directly. On the other hand, d u / d t can, be determined if we know the altitude as a function of t. For instance, if we, are rolling down a hill at 100 kilometers per hour, we could have d u / d t = - 3, meters per second. The rate of change d p l d u is known to meteorologists; near, sea level, it is about -0.12 gsc per meter. (The unit "gsc" of pressure is, "grams per square centimeter"; the rate of change is negative because pressure, decreases as altitude increases.), Now the chain rule enables us to calculate how fast the pressure is, changing with time:, , 9, = 2., dt, du dt, , = (-0.12)(-, , 3) gsc per second = 0.36 gsc per second., , This rate of pressure increase is fast enough so that the ears' internal pressure, control system cannot keep up with it, and they "pop.", , Exercises for Section 2.2, Find the derivatives of the functions in Exercises 1-10., I . ( x 3)4, 2. ( ~ +23 ~ 1)5, 3. ( x 3 + 1 0 ~ ) ~ ~, 4. (s4 4s3 + 3s2 + 2s, 5. ( x 2 8 x p . x, 6. ( x 2 2 ) 3 ( ~+9 8), 7 . ( ~ +22 ) ( ~ +, 9 8)3, 8. ( ~ +32 ) 9 ( ~ * 2 ~ i)1°, 9. ( y + U 3 ( y + 2I2(y + 3), 10. ( ( x 2- 112 + 3)'O, , +, , +, , +, +, +, , +, , + +, , 11. Let g ( x ) = x + 1 and f ( u ) = u2. Find f 0 g and, g, 12. Let h ( x ) = x~~+ 3xI2 + 1 . Write h ( x ) as a composite function f ( g ( x ) ) with g ( x ) = x 12., Find f 0 g and g 0 f i n each of Exercises 13-16., 13. g ( x ) = ~ 3 f ;( ~ =) ( X - 2)3., 14. g ( x ) = x " ; f ( x ) = x m ., 1 ;f ( x ) = 1 - o x ., 15. g ( x ) = 1-x, 3x - 2, 2x - 7, 16. g ( x ) = ;f ( x ) = 4x + 1, 9x+3', Write the functions in Exercises 17-20 as compositions, of simpler functions., , +3., 18. h ( r ) =JI+J;., 17. h ( x ) = J4x3 + 5 x, , 20. h ( x ) = ( ( x 2+ 1 ), , + ( x 2+ I ) +~112., , Verify the chain rule for f ( u ) and g ( x ) given in Exercises 2 1-24., 21. f ( u ) = u2, g ( x ) = x 2 - 1, 22. f ( u ) = u3, g ( x ) = x + 1, 23. f ( u ) = u2, g ( x ) = 6, 24. f ( u ) = 6 ,, g(x)= x2, Use the chain rule rule to differentiate the functions in, Exercises 25-34., 26. ( X - 2 ~ ~ ) ~, 25. ( x 2 - 6 x 1)3, 9 +2 ~ 5, 1, 27. 28., 3 5x5, ( x 3+ 5Xl4, , +, , +, , 29. ( ( x 2+ 2)', , + 1)2, , +, , ( x 2 3)5, , 31., [l, , + ( x 2 + 3)18, , 30., , 3, [ ( x + 2)', , 32., , 2, [(2x +, 2x + I, , + 414, , + 51, , 35. If h ( x ) = x3S(2x2), find a formula for hl(x)., 36. If h ( x ) = f ( g ( x 2 ) ) ,find a formula for h'(x)., 37. Given three functions, f, g, and h :, (a) Wow would you define the composition, fogoh?, (b) Use the chain rule twice to obtain a formula, for the derivative off 0 g 0 h., 38. If h ( x ) = f ( g ( x 3 + 2)) + g( f ( x 2 ) ) ,find a formula, for h f ( x ) ., 39. Fat City occupies a circular area 10 miles in, diameter and contains 500,000 inhabitants. If the, population is growing now at the rate of 20,000, inhabitants per year, how fast should the diame-, , Copyright 1985 Springer-Verlag. All rights reserved.

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2.2 The Chain Rule, ter be increasing now in order to maintain the, circular shape and the same population density ( = number of people per square mile). If, the population continues to grow at the rate of, 20,000 per year, how fast should the diameter be, increasing in 5 years? Give an intuitive explanation of the relation between the two answers., 40. The radius at time t of a sphere S is given by, r = t 2 - 2t + 1. How fast is the volume V of S, changing at time t = f , 1,2?, 41. The kinetic energy K of a particle of mass m, moving with speed u is K = f mu2. A particle, with mass 10 grams has, at a certain moment,, velocity 30 centimeters per second and acceleration 5 centimeters per second per second. At, what rate is the kinetic energy changing?, 42. (a) At a certain moment, an airplane is at an, altitude of 1500 meters and is climbing at, the rate of 5 meters per second. At this, altitude, pressure decreases with altitude at, the rate of 0.095 gsc per meter. What is the, rate of change of pressure with respect to, time?, (b) Suppose that the airplane in (a) is descending rather than climbing at the rate of 5, meters per second. What is the rate of, change of pressure with respect to time?, 43. At a certain moment, your car is consuming, gasoline at the rate of 15 miles per gallon. If, gasoline costs 75 cents per gallon, what is the, cost per mile? Set the problem up in terms of, functions and apply the chain rule., 44. The price of eggs, in cents per dozen, is given by, the formula p = 55/(s where s is the supply of eggs, in units of 10,000 dozen, available to, the wholesaler. Suppose that the supply on July, 1, 1986 is s = 2.1 and is falling at a rate of 0.03, per month. How fast is the price rising?, 45. If an object has position (t2 + 4)' at time t, what, is its velocity when t = - I ?, 46. If an object has position (t2 + l)/(t2 - 1) at time, t, what is its velocity when t = 2?, Find the second derivatives of the functions in Exercises 47-50., 47. (X 1113, 48. (x3 - I)', 49. ( ~ +4 lox2 + 119', 50. (x2 113(x3+ 1)', +51. (a) Find a "stretching rule" for the derivative of, ~ ( c x )c, a constant., (b) Draw the graphs of y = 1 x 2 and of y =, 1 + ( 4 ~ and, ) ~ interpret the stretching rule, geometrically., , +, , 117, , +52. Prove that (d/dx)(un) = n u n ' d u / d x for all natural numbers n as follows:, (a) Note that this is established for n = 1,2,3 at, the beginning of this section., (b) Assume that the result is true for n - 1, and, write un = u(un-'). Now differentiate using, the product rule to establish the result for n., (c) Use induction to conclude the result for, all n. (See Exercise 65, p. 69.), *53. Find a general formula for (d2/dx2)(un), where, u = f(x) is any function of x., s54. (a) Let i be the "identity function" i(x) = x., Show that i 0 f = f and f 0 i = f for any function, f. (b) Verify the chain rule for f = f 0 i., *55. Let f and g be functions such that f 0 g = i,, where i is the function in Problem 54. Find a, formula for f'(x) in terms of the derivative of g., *56. Use the result of Exercise 55 to find the derivative of f(x) = 3& by letting g(x) = x3., *57. Find a formula for the second derivative off 0 g, in terms of the first and second derivatives o f f, and g., *58. Show that the power of a function rule for negative powers follows from that rule for positive, powers and the reciprocal rule., *59. For reasons which will become clear in Chapter, 6, the quotient f'(x)/ f(x) is called the logarithmic, derivative of f(x)., (a) Show that the logarithmic derivative of the, product of two functions is the sum of the, logarithmic derivatives of the functions., (b) Show that the logarithmic derivative of the, quotient of two functions is the difference of, their logarithmic derivatives., (c) Show that the logarithmic derivative of the, nth power of a function is n times the logarithmic derivative of the function., (d) Develop a formula for the logarithmic derivative of, , in terms of the logarithmic derivatives of f l, through S,., (e) Using your formula in part (d), find the, ordinary (not logarithmic) derivative of, , +, , +, , If you have enough stamina, compute f'(x), without using the formula in part (d)., *60. Differentiate ( 1 + (1 + (1 x~)')~)'., , +, , Copyright 1985 Springer-Verlag. All rights reserved.

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118, , Chapter 2 Rates of Change and the Chain Rule, , 2.3 Fractional Powers and, Implicit Differentiation, , The power rule still holds when the exponent is a fraction., In this section we extend the power rule to include fractional exponents by, using a method called implicit differentiation, which can be applied to many, other problems as well., Let us begin by trying to find d y / d x when y = x ' j n = " 6 , where n is a, positive integer.' At the moment, we shall simply assume that this derivative, exists and try to calculate its value. This assumption will be justified in Section, 5.3, in connection with inverse functions., We may rewrite the relation y = x ' l n as y n = x , so we must have, d ( y " )= d (x)., -, , dx, dx, Recalling that y is a function of x , we may evaluate the left-hand side of (1), by the chain rule (or the power of a function rule) to get, , The right-hand side of (1) is simply, , dx-- 1., , dx, Substituting (2) and (3) into ( I ) gives, , which we may solve for d y / d x to obtain, , Thus, , Note that this rule reads the same as the ordinary power rule: "Bring down, the exponent as a multiplier and then decrease the exponent by one." The, special case ( d / d x ) ( x ' l 2 )= x - ' l 2 has already been considered in Example, 4, Section 1.3., , 4, , Example 1, Solution, , Differentiate f ( x ) = 3 5&, d, -35&=, dx, , d, 3-x1/5=, dx, , 3X(1/5)-1, 5, = 35x - 4 / 5 =, , 3, 5x4/5 ., , Next, we consider a general rational power f ( x ) = x r , where r = p / q is a ratio, of integers. Thinking of X P / ~ as (x1/4)p,we set g ( x ) = x l / q , so that f ( x ), = [ g ( x ) J PThen,, ., by the (integer) power of a function rule,, , 'Note that x ' / " is defined for all x if n is odd but only for nonnegative x if n is even. A brief, review of fractional exponents may be found in Section R.3., , Copyright 1985 Springer-Verlag. All rights reserved.

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2.3 Fractional Powers and Implicit Differentiation, , 119, , so by formula ( 4 ) with 1/ n replaced by 1/ q , we have, , -, , P x c ~ - I ) / ~ x ( I - 4 ) =/ ~ P x ( p - 4 ) / 4, , = Px(p/4)-1., , 4, 4, 4, We conclude that differentiation of rational powers follows the same rule, as integer powers., , (The formula is valid for all x for which the right-hand side makes, , Example 2, , Solution, , Differentiate f ( x ) = 3 x 2 + ( x 2, d ( 3 x 2+ x 3 j 2, ~ ' ( x=) dx, , +x'/~)/&., , +x-':~), , We can combine the rational power rule with the chain rule to prove a, rational power of a function rule. Let y = [ f ( x ) y and let u = f ( x ) so that, y = u r . Then, , To differentiate a power [ f(x)]' ( r a rational number), take out the, exponent as a factor, reduce the exponent by 1, and multiply by f ' ( x ) :, , Example 3, , Solution, , Differentiate g ( x ) = ( 9 x 3, , + 10)~/~., , Here f ( x ) = 9 x 3 + 10, r = 3 , and f ' ( x ), , = 27x2. Thus, , g r ( x ) = 3 ( 9 x 3 + 1 0 ) ~ ' ~ .27x2 = 45x2(9x3+ 10), , 2/3, , .A, , The rules for rational powers can be combined with the quotient rule of, differentiation, as in the next example., , Copyright 1985 Springer-Verlag. All rights reserved.

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120, , Chapter 2 Rates of Change and the Chain Rule, , Example 4, , Solution, , Differentiate, , + x3/2, x3/2 1, , ,y 1/2, , +, , We use the quotient and rational power rules:, , The method which we used to differentiate y = x ' / " , namely differentiating, the relation y n = x and then solving for dy/dx, is called implicit differentiation., This method can be applied to more complicated relationships such as, x 2 y 2 = 1 or x4 + xy + y5 = 2 which define y as a function of x implicitly, rather than explicitly. In general, such a relationship will not define? uniquely, as a function of x ; it may define two or more functions. For example, the, circle x2 y2 = 1 is not the graph of a function, but the upper and lower, semicircles are graphs of functions (see Fig. 2.3.1)., , +, , +, , Figure 2.3.1. Parts of the, circle x2 + Y 2 = 1 are the, graphs of functions., , Example 5, Solution, , If y, , = f(x), , and x2 + y2 = 1, express dy/dx in terms of x and y ., , Thinking of y as a function of x, we differentiate both sides of the relation, x2 y2 = 1 with respect to x. The derivative of the left-hand side is, , +, , while the right-hand side has derivative zero. Thus, , The result of Example 5 can be checked, since in this case we can solve for y, directly:, , Notice that the given relation then defines two functions: fl(x)= J1 - x 2 and, , Copyright 1985 Springer-Verlag. All rights reserved.

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2.3 Fractional Powers and Implicit Differentiation, , f 2 ( x )= -, , J-., , 121, , Taking the plus case, with u = 1 - x 2 and y = &, the, , chain rule gives, dY-1, _dY_du, _---(-2x1, dx, dudx, 2&, , so it checks. The minus case gives the same answer., From the form of the derivative given by implicit differentiation, dy/dx, = - x / y , we see that the tangent line to a circle at ( x , y ) is perpendicular to, the line through ( x , y ) and the origin, since their slopes are negative reciprocals of one another. (See Fig. 2.3.2.) Implicit differentiation often leads, directly to such striking results, and for this reason it is sometimes preferable, to use this method even when y could be expressed in terms of x., , +, , Figure 2.3.2. If x 2 y2 = I,, the formula dyldx =, - x / y means that the, tangent line to a circle at a, point on the circle is, perpendicular to the line, from that point to the, center of the circle., , I, There is a device which may help you to remember that the chain rule, must be used. In Example 5, if we keep the notation f ( x ) for y , then the, relation x2 + y 2 = 1 becomes x2 + [f(x)I2 = 1 , and differentiating with respect, to x gives 2x + 2f(x)f'(x) = 0. Now we solve for f ' ( x ) to get f ' ( x ) = - x / f ( x ), or, in Leibniz notation, dy/dx = - x / y , just as before. Once you have done a, few examples in this long-winded way, you should be able to go back toy and, dy/dx without the f., The following is an example in which we cannot solve for y in terms of x., , Example 6, , +, , Find the equation of the tangent line to the curve 2x6 y 4 = 9xy at the point, (L2)., , Solution, , We note first that (1,2) lies on the curve, since 2 ( 1 )+, ~ 24 = 9(1)(2). Now, suppose that y = f ( x ) and differentiate both sides of the defining relation. The, left-hand side gives, , while the right-hand side gives, , Equating both sides and solving for dy/dx, we have, , Copyright 1985 Springer-Verlag. All rights reserved.

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122, , Chapter 2 Rates of Change and the Chain Rule, , When x = 1 and y = 2,, , Thus, the slope of the tangent line is &; by the point-slope formula, the, equation of the tangent line is y - 2 = & ( x - I), or ,y = A, 23 X + 40., 23, , To calculate d y / d x if x and y are related by an equation:, 1. Differentiate both sides of the equation with respect to x, thinking of, y as a function of x and using the chain rule., 2. Solve the resulting equation for d y l d x ., , Exercises far Section 2.3, Differentiate the functions in Exercises 1-24., 1., 2. x3/5, 3. g x ' / 4 - x - 2 / 3, 4. 8 ~ -4 3~ - 5 / 4, 5. 3 ~ - ( 5, ~ x ) ' l/2 ~, 6. x 2 - 3 x i i 2, 7. x2(x'13 x 4 l 3 ), 8. ( x 2 l 3 I 2 m, 10. ( x i / ' x ~ / ' ) ' / ~, 9. ( x 5 I ) ~ / ~, , +, , +, , +, , +, , 35. If x 2 + y 2 = 3, compute d y / d x when x = O, andy, 36. If x3 + y 3 = xy, compute dx/dy in terms of x, and y ., 37. Suppose that x4 + y 2 y - 3 = 0., (a) Compute dy/dx by implicit differentiation., (b) What is dy/dx when x = I , y = I ?, (c) Solve for y in terms of x (by the quadratic, formula) and compute dy/dx directly. Compare with your answer in part (a)., 38. Suppose that xy +, = 7., (a) Find dy/dx., ( b ) Find dx/dy., (c) What is the relatlon between dy/dx and, dx / dy ?, 39. Suppose that x 2 / ( x y 2 )= y 2 / 2 ., (a) Find dy/dx when x = 2, y =, (b) Find dy/dx when x = 2, y = - Q., 40. Let ( u 2 6)(u2 1) = IOuu. Find du/du and, dv/du when u = 2 and u = 1., 41. Find the equation of the tangent line to the curve, x 4 + y 4 = 2 w h e n x = y = 1., 42. Find the equation of the tangent line to the curve, 2x2+2xy+y2=8 whenx=2andy=0., 43. Find ( d 2 / d ~ 2 ) ( x i / x2 2 i 3 ) ., 44. Find ( d Z / d X 2 ) ( X / J G 7 ) ., 45. Find the equation of the tangent lice to the, at the point, / 2 , 1 /2)., graph of y =, 46. Find the equation of the line tangent to y, - ( X ' / 2 + X ' / 3 ) ' / 3 at x =: 1,, 47. Let x 4 y 4 = 1. Find dy/dx as a function of x, in two ways: by implicit differentiation and by, solving for y in terms of x., , =o., , +, , 6, +, , 0., , Find the indicated derivatives in Exercises 25-28., , where r is rational, dx, 27. f'(7), where f ( x ) = 7 3\1;;, , Find the derivatives of each of the functions in Exercises 29-34., 1, 29. f ( x ) = x 3 / " - x i / ', 30. k ( s ) =, 31. h ( y ) =, , -, , Y-2, , 32. g ( t ) = t ( t 2 I 3+ t 7 ), , +, , +, , -d, , (0, , +, , Copyright 1985 Springer-Verlag. All rights reserved.

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2.4 Related Rates and Parametric Curves, , 48. Differentiate the function (x3 + 2)/ Jx)+l in at, least two different ways. Be sure that the answers, you get are equivalent. (Doing the same problem, in several ways is a good method for checking, your calculations, useful on examinations as well, as in scientific work.), Find linear approximations for the expressions in Exercises 49-52., 50. (4.0213/", 49. 4~, 51. ( - 26.98)'13, 52. 2JIZZ, 53. The mass M of the first x meters of a concrete, b e a m is M = 24(2 + ~ ' 1 k~i l o)g r a~m s ., Find the density dM/dx., 54. The average pulse rate for persons 30 inches to, 74 inches tall can be approximated by y =, 589/& beats per minute for a person x inches, tall., (a) Find dy/dx, and show that it is always, negative., (b) The value of Idy/dxl at x = 65 is the decrease in beats per minute expected for a, 1-inch increase in height. Explain., (c) Do children have higher pulse rates than, adults according to this model?, 55. Let y = 2 4 6 be the learning curve for learning y, items in x hours, 0 < x < 5. Apply the linear, approximation to estimate the number of new, items learned in the 12-minute period given by, l < x g 1.2., , 123, , 56. The daily demand function for a certain manufactured good is p(x) = 75 + 4\12x - (x2/2),, where x is the production level., (a) Find dp/dx. Interpret., (b) Find the production level x for which dp/dx, - ., = 0. Interpret., 57. The fundamental period for vibration P and the, tension T in a certain string are related by, P = J31/-T seconds. Find the rate of change of, period with respect to tension when T = 9 lbs., 58. Lions in a small district in an African game, preserve defend an exclusive region of area A, which depends on their body weight W by the, formula A = w ' . ~ ' ., (a) Find dA /dW., (b) By what percentage should the defended, area increase after a 200-lb lion undergoes a, 20-lb weight gain? (Use the linear approximation.), 59. The object distance x and image distance y satl / y = 1/f,, isfy the thin lens equation l / x, where f is the focal length., (a) solve for y as a function of x when f = 50, millimeters., (b) Find dy/dx., (c) Find all (x, y ) such that (d/dx)(x y ) = 0., *60. Using implicit differentiation, find the equation, of the tangent line at the point (x,, yo) on the, circle (x - a)* ( y - b12 = r2. Interpret your result geometrically. ( a , b, and r are constants.), , +, , +, , +, , 2.4 Related Rates, and ParametrSc Curves, If two quantities satisb an equation, their rates of change can be related by, implicit differentiation., , Suppose that we have two quantities, x and y , each of which is a function of, time t . We know that the rates of change of x and y are given by d x / d t and, dy/dt. If x and y satisfy an equation, such as x 2 y 2 = 1 or x 2 y6 2y = 5 ,, then the rates d x / d t and dy/dt can be related by differentiating the equation, with respect to t and using the chain rule., , +, , Example d, Solution, , Suppose that x and y are functions of t and that x 4, d x / d t and dy/dt., , + +, , + xy + y4 = 1. Relate, , Differentiate the relation between x and y with respect to t , thinking of x and y, as functions of t:, , Copyright 1985 Springer-Verlag. All rights reserved.

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Chapter 2 Rates of Change and the Chain Rule, , 124, , We can simplify this to, , which is the desired relation. A, If you have trouble remembering to use the chain rule, you can use a device, like that following Example 5 in Section 2.3. Namely, write f ( t ) for x and g ( t ), for y, then differentiate the relation (such as [f(t)14 f ( t ) g ( t ) tg(t)l4= 1), with respect to t. This will give a relation between f'(t), gl(t),f ( t ) , and g(t)., Once you have done a few examples in this long-winded way, you should be, ready to go back to the d/dt's., , +, , Y, , I, , +, , +, , To relate the rates d x / d t and d y / d t if x and y satisfy a given equation:, , I, , 1. Differentiate both sides of the equation with respect to t , thinking of, x and y as functions of t., 2. Solve the resulting equation for dy/dt in terms of d x / d t (or vice versa, , I, , There is a useful geometric interpretation of related rates. (This topic is treated, in more detail in Section 10.4.) If x and y are each functions of t , say x = f ( t ), and y = g(t), we can plot the points ( x , y) for various values of t. As t varies,, the point ( x , y ) will move along a curve. When a curve is described this way,, it is called aparametric curve (see Fig. 2.4.1)., , ~i~, 2.4.1. ~f andy are, functions of t , the point, (x, y) follows a curve as t, varies., , Example 2, , Y, , If x = t4 and y, , =, , t2, what curve does ( x ,y ) follow for - w, , < t < oo?, , S ~ l u l l s n We notice that y 2 = x , so the point ( x ,y ) lies on a parabola. As t ranges from, - w to w, y goes from + oo to zero and back to + w , so ( x , Y ) stays on the, half of the parabola with y > 0 and traverses it twice (see Fig. 2.4.2). A, It may be possible to describe a parametric curve in other ways. For instance,, it may be described by a relation between x and y. Specifically, suppose that, the parametric curve x = f(t), y = g ( t ) can be described by an equation, y = h ( x ) (the case x = k ( y ) will be similar). Then we can differentiate by the, chain rule. Using Leibniz notation:, , Figure 2.4.2. As t ranges, from - oo to + oo,the, point (t4,t 2 ) traverses the, parabola twice in the, directions shown., , dy ,, dy / d l, SO, dt, dx dt, dx, dx/dt ', This shows that the slope of the tangent line to a parametric curve is given by, (dy/dt)/(dx/dt)d ~ = d ~ dx, -, , As t varies, two equations x = f ( t ) and y = g ( t ) describe a curve in the, plane called a parametric curve. The slope of its tangent line is given by, , Copyright 1985 Springer-Verlag. All rights reserved.

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2.4 Related Rates and Parametric Curves, , 125, , Example 3, , Find the equation of the line tangent to the parametric curve given by the, equations x = (1 + t3)4+ t2,y = t5 t 2 + 2 at t = 1., , Solution, , Here the relation between x and y is not clear, but we do not need to know it., (We tacitly assume that the path followed by ( x ,y ) can be described by a, function y = h(x).) We have, , +, , +, , = 4(1 + t 3 ) . 3t2 2r = I2(1, dt, so the slope of the tangent line is, 3, , At t, , =, , + t 3 ) t 2 + 2t, 3, , and, , 3, = 5t4 + 21,, dt, , 1, we get, , Since x = 17 and y = 4 at t = 1, the equation of the tangent line is given by, the point-slope formula:, , y - 4 = & ( x - 17),, , Example 4, Solution, , Example 5, , Show that the parametric equations x = at + b and y = ct, straight line if a and c are not both zero. What is its slope?, , +, , describe a, , +, , Multiplying x = U P b by c, multiplying y = ct d by a, and subtracting, we, get, cx - ay = bc - ad,, so y = ( c / a ) x + ( l / a ) ( a d - bc), which is the equation of a line with slope, c / a . (If a = 0, x is constant and the line is vertical; if c were also zero the line, would degenerate to a point.) Note that the slope can also be obtained as, ( d y / d t ) / ( d x / d t ) , since dy/dt = c and d x / d t = a. a, Suppose that x and y are functions of time and that ( x , y ) moves on the circle, Y 2 = 1. If x is increasing at 1 centimeter per second, what is the rate of, change of y when x = l / f i and y =, , x2, , Solullow, , +d, , +, , 1/a?, , +, , Differentiating x 2 + y2 = 1 gives 2 x ( d x / d t ) 2y(dy/dt) = 0; so d y / d t =, (- x / y ) ( d x / d t ) . If x = y = I / @ , dy/dt = - d x / d t = - 1 centimeter per second. A, In word problems involving related rates, the hardest job may be to translate, the verbal problem into mathematical terms. You need to identify the variables which are changing with time and to find relations between them. If, some geometry is involved, drawing a figure is essential and will often help, you to spot the important relations. Similar triangles and Pythagoras' theorem, are frequently useful in these problems., , Example 6, , A light L is being raised up a pole (see Fig. 2.4.3). The light shines on the, object Q, casting a shadow on the ground. At a certain moment the light is 40, meters off the ground, rising at 5 meters per minute. How fast is the shadow, shrinking at that instant?, , Copyright 1985 Springer-Verlag. All rights reserved.

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126, , Chapter 2 Rates of Change and the Chain Rule, , Solution, , Let the height of the light b e y at time t and the length of the shadow be x . By, similar triangles, x / 1 0 = ( x 2 0 ) / y ; i.e., xy = 10(x + 20). Differentiating, with respect to t , x ( d y / d t ) + ( d x / d t ) y = 10(dx/dt). At the moment in question y = 4 0 , and so x . 4 0 = l O ( x + 2 0 ) or x = ? . Also, dy/dt = 5 and so, 9 - 5 40(dx/dt) = lO(dx/dt). Solving for ( d x / d t ) , we get d x / d t = - 9., Thus the shadow is shrinking at 9 meters per minute. A, , +, , +, , Example 7, , A spherical balloon is being blown up by a child. At a certain instant during, inflation, air enters the balloon to make the volume increase at a rate of 50, cubic centimeters a second. At the same instant the balloon has a radius of 10, centimeters. Mow fast is the radius changing with time?, , Solution, , Let the radius of the balloon be denoted by r and the volume by V . Thus, v = $ nr3 and so, , dr ., d!? = 477r 2 -, , dt, dt, At the instant in question, d V / d t = 50 and r = 10. Thus, , and so, , dr=A!L=-- 0.04 centimeters per second., dt, , @, , l -1, , 40071, , 8n, , Example 8, , A thunderstorm is dropping rain at the rate of 2 inches per hour into a conical, tank of diameter 15 feet and height 30 feet. At what rate is the water level, rising when the water is 20 feet deep?, , Solution, , Figure 2.4.4 shows top and side views of the partially filled tank, both of, which will be useful for our solution., We denote by h the height of the water in the tank, so that dh/dt is the, rate to be found. To proceed, we need to use the fact that the rate of rainfall is, 2 inches per hour. What this means is that the water level in a cylindrical tank, would rise uniformly at the rate of 2 inches per hour, so that the volume of the, water pouring every hour into a circle of diameter 15 feet is TI . (7.5)2. cubic, feet = 9TI cubic feet. It is useful, then, to introduce the variable V representing the volume of water in the tank; we have d V / d t = 7 5 n / 8 cubic feet per, hour., Now V and h are related by the formula for the volume of a cone:, V = nr2h, where r is the radius of the "base" of the cone, in this case, the, radius of the water surface. From Fig. 2.4.4, we see, using similar triangles,, that r / h = 7.5/30 = 1/ 4 , so r = h, and hence V = & nh3. Differentiating, and using the chain rule gives d V / d t = &nh2dh/dt. Inserting the specific, n = & n . 400dh/dt =, data h = 20 and d V / d t = TI gives the equation, 2 5 ~ 1 d h / d twhich, ,, we may solve for dh/dt to get dh/dt = $. This is in feet per, hour, so the water level is rising at the rate of 44 inches per hour. A, , 5,-, , 4, , +, , Figure 2.4.4. A conical, tank partially filled with, water., , Copyright 1985 Springer-Verlag. All rights reserved.

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2.4 Related Rates and Parametric Curves, , 127, , Exercises For Section 2.4, In Exercises 1-8, assume that x and y are functions of, t . Relate dx/dt and dy/dt using the given relation., 1. ~ 2 - ~ 2 = 3, 2. x + y = 4, 3. x - Y 2 - Y 3 = 4, 4. 8x2 + gy2 = 5, 5. x + Y 2 = Y, 6. m, =, x, 7.6-fi=5, 8. ( x 2 + Y 2 + y3)3'2 = 5, 9. Sketch the curve defined by the parametric equat i o n s ~ =t2, y = 1 - t, -oo < t < oo., 10. Sketch the curve described by x = 3t + 2, y, 11. What curve do the parametric equations x = t2, and y = t6 describe?, 12. If x = (1 + t)2 and y = (1 + t14, what curve does, (x, y ) follow for - co < t < co?, 13. Find the equation of the tangent line to the, parametric curve x = t2, y = t3 at t = 5., 14. Find the equation of the line tangent to the, parametric curve x = t2 + I, y = l / ( t 4 1) at, t = 2., 15. Find the equation of the tangent line to the, parametric curve, , 23., , 24., , 25., , +, , 26., at t = 3., 16. (a) Find the slope of the parametric curve y, = t4 2t, x = 8t at t = 1., (b) What relationship between x and y is satisfied by the points on this curve?, (c) Verify that dy/dx = (dy/dt)/(dx/dt) for, this curve., 17. Suppose that xy = 4. Express dy/dt in terms of, dx/dt when x = 8 and y = f ., 18. I f x 2 y 2 = x5/y and dy/dt = 3 when x =, y =\/Z, what is dx/dt at that point?, 19. Suppose that x 2 + y 2 = t and that x = 3, y = 4,, and dx/dt = 7 when t = 25. What is dy/dt at, that moment?, 20. Let x and y depend on t in such a way that, ( x + ~ ) ~t 2+= 2 t and such t h a t x = O a n d y = 1, when t = I . If dx/dt = 4 at that moment, what is, dy/dt?, 21. The radius and height of a circular cylinder are, changing with time in such a way that the volume remains constant at 1 liter (= 1000 cubic, centimeters). If, at a certain time, the radius is 4, centimeters and is increasing at the rate of f, centimeter per second, what is the rate of change, of the height?, 22. A hurricane is dropping 10 inches of rain per, hour into a swimming pool which measures 40, feet long by 20 feet wide., (a), . , What is the rate at which the volume of, water in the pool is increasing?, (b) If the pool is 4 feet deep at the shallow end, , +, , +, , 27., , 28., , 29., , 30., , +31., , +32., , and 8 feet deep at the deep end, how fast is, the water level rising after 2 hours? (Suppose, the pool was empty to begin with.) How fast, after 6 hours?, Water is being pumped from a 20-meter square, pond into a round pond with radius 10 meters., At a certain moment, the water level in the, square pond is dropping by 2 inches per minute., How fast is the water rising in the round pond?, A ladder 25 feet long is leaning against a vertical, wall. The bottom is being shoved along the, ground, towards the wall a t 1 t feet per second., How fast is the top rising when it is 15 feet off, the ground?, A point in the plane moves in such a way that it, is always twice as far from (0,O) as it is from, (0,l)., (a) Show that the point moves on a circle., (b) At the moment when the point crosses the, segment between (0,O) and (0, l), what is, dy / dt?, (c) Where is the point when dy/dt = dx/dt?, (You may assume that dx/dt and dy/dt are, not simultaneously zero.), Two quantities p and q depending on t are, subject to the relation I / p + I / q = 1., (a) Find a relation between dp/dt and dq/dt., (b) At a certain moment, p = $ and dp/dt = 2., What are q and dq/dt?, Suppose the quantities x, y , and z are related by, the equation x 2 + Y 2 + z2 = 14. If dx/dt = 2 and, dy/dt = 3 when x = 2, y = 1, and z = 3, what is, dz / dt?, The pressure P, volume V, and temperature T of, a gas are related by the law P V / T = constant., Find a relation between the time derivatives of, P, V, and T., The area of a rectangle is kept fixed a t 25 square, meters while the length of the sides varies. Find, the rate of change of the length of one side with, respect to the other when the rectangle is a, square., The surface area of a cube is growing at the rate, of 4 square centimeters per second. How fast is, the length of a side growing when the cube has, sides 2 centimeters long?, (a) Give a rule for determining when the tangent line to a parametric curve x = f(t),, y = g(t) is horizontal and when it is vertical., (b) When is the tangent line to the curve x = t2,, y = t3 - t horizontal? When is it vertical?, (a) At which points is the tangent line to a, parametric curve parallel to the line y = x ?, (b) When is the tangent line to the curve in part, (b) of Exercise 3 1 parallel to the line y = x ?, (c) Sketch the curve of Exercise 3 1., , Copyright 1985 Springer-Verlag. All rights reserved.

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128, , Chapter 2 Rates of Change and the Chain Rule, , *33. Read Example 8. Show that for any conical tank,, the ratio of dhldt to the rate of rainfall is equal, , to the ratio between the area of the tank's open-, , ing to the area of the water surface. (In fact, this, result is true for a tank of any shape; see Review, Exercise 32, Chapter 9.), , A n antiderivative off is a function whose derivative is f., Many applications of calculus require one to find a function whose derivative, is given. In this section, we show how to solve simple problems of this type., , Example 1, , Solution, , Find a function whose derivative is 2 x, , + 3., , We recall that the derivative of x 2 is 2 x and that the derivative of 3 x is 3, so, the unknown function could be x 2 3 x . We may check our answer by, differentiating: ( d / d x ) ( x 2 + 3 x ) = 2 x + 3. A, , +, , +, +, , The function x 2 3 x is not the only possible solution to Example 1; so are, x 2 + 3 x + 1 , x 2 3 x + 2, etc. In fact, since the derivative of a constant, function is zero, x 2 3 x + C solves the problem for any number C., A function F for which F' = f is called an antiderivative off. Unlike the, derivative, the antiderivative of a function is never unique. Indeed, if F is a n, antiderivative off, so is F + C for an arbitrary constant C. In Section 3.6 we, will show that all the antiderivatives are of this form. For now, we take this, fact for granted. We can make the solution of an antidifferentiation problem, unique by imposing an extra condition on the unknown function (see Fig., 2.5.1). The following example is a typical application of antidifferentiation., , +, , Figure 2.5.1. All these, , functions have the same, derivative. Specifying, F(xo) = y o picks out one of, them., , Example 2, , Solution, , The velocity of a particle moving along a line is 3t, particle is at position 4. Where is it at time lo?, , + 5 at time t . At time 1 , the, , Let F ( t ) denote the position of the particle at time t. We will determine the, function F. Since velocity is the rate of change of position with respect to time,, we must have F'(t) = 3t + 5; that is, F is an antiderivative of f(t) = 3t + 5. A, function whose derivative is 3t is 4 t2, since ( d / d t ) $ t 2 = 42t = 3t. Similarly, a, function whose derivative is 5 is 5t. Therefore, we take, F ( t ) = i t 2 + 5t, , + C,, , where C is a constant to be determined. To find the value of C, we use the, , Copyright 1985 Springer-Verlag. All rights reserved.

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2.5 Antiderivatives, , 129, , information that the particle is at position 4 at time I ; that is, F(1) = 4 ., Substituting 1 for t and 4 for F ( t ) in the equation F ( t ) = 3 t 2 5t C gives, , + +, , or C = - 3, and so F ( t ) = 3 t 2 + 5t - 4. Finally, we substitute 10 for t ,, obtaining the position at time 10: F(10) = 2 - 100 + 5 . 10 - 4 = 197;. A, At this point in our study of calculus, we must solve antidifferentiation, problems by guessing the answer and then checking and refining our guesses if, necessary. More systematic methods will be given shortly., Example 3, , Solution, , Find the general antiderivative for the function f ( x ) = x4, , + 6., , We may begin by looking for an antiderivative for x4. If we guess x5, the, derivative is 5x4, which is five times too big, so we make a new guess, x 5 ,, which works. An antiderivative for 6 is 6 x . Adding our two results gives, ix 5 6 x ; differentiating x 5 6 x gives x4 + 6 , so x5 + 6 x is an antiderivative for x4 + 6 . We may add an arbitrary constant to get the general, antiderivative 4 x 5 + 6 x + C . A, , +, , +, , 4 +, , 4, , Example 4, , The acceleration of a falling body near the earth's surface is 9.8 meters per, second per second. If the body has a downward velocity v , at t = 0 , what is its, velocity at time t? If the position is x , at time 0 , what is the position at time t?, (See Fig. 2.5.2.), , Solution, , We measure the position x in the downward direction. Let v be the velocity., Then d v / d t = 9.8; since an antiderivative of 9.8 is 9.8t, we have v = 9.8t C ., At t = 0 , v = v,, so v , = (9.8)0 C = C, and so v = 9.81 v,. Now d x / d t, = v = 9.8t + v,. Since an antiderivative of 9.8t is ( 9 . 8 / 2 ) t 2= 4.9t2 and an, antiderivative of v , is v,t, we have x = 4.9t2 uot + D . At t = 0 , x = x,, so, x , = 4.9(0)2 + v , . O D = D,and so x = 4.9t2 v,t + x,. A, , +, , Velocity = v o, I, , I, , +, , +, , +, , +, , +, , The most commonly used notation for the antiderivative is due to Eeibniz., The symbol, ff(x)dx, , Figu~e2.5.2. The body is, moving downward at 1 = 0, with velocity vo., , J, , denotes the class of all antiderivatives off; thus, if F is a particular antiderivative, we may write, I f ( x )dx = F ( x ) + C ,, , where C is an arbitrary constant. For instance, the result of Example 3 may be, written, I ( x 4 t 6 ) d x = -1x 5 + 6 x + C ., 5, The elongated S, called an integral sign, was introduced by Leibniz, because antidifferentiation (also called integration) turns out to be a form of, continuous Jummation. In Chapter 4 , we will study this aspect of integration, in detail. There and also in the supplement to this section, we explain the, presence of the "dx" in the notation. For now, we simply think of dx as, indicating that the independent variable is x . Its presence should also serve as, a reminder that integrating is inverse to differentiating, where the dx occurs in, the denominator of d y / d x ., The function f ( x ) in Jf ( x ) dx is called the integrand, and Jf ( x )dx is called, the inde8nite integral of f ( x ) . One traditionally refers to f ( x ) d x as being, "under" the integral sign, even though this is typographically inaccurate., , Copyright 1985 Springer-Verlag. All rights reserved.

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430, , Chapter 2 Rates of Change and the Chain Rule, , An antiderivative for f is a function F such that F'(x) = f ( x ) . We write, F(x) =, , J f ( x ) dx., , The function J f ( x ) d x is also called the indefinite integral off, and f is, called the integrand., If F ( x ) is an antiderivative of f ( x ) , the general antiderivative has, the form F ( x ) C for an arbitrary constant C., , +, , Some of the differentiation rules lead directly to systematic rules for antidifferentiation. The rules in the following box can be proved by differentiation, of their right-hand sides (see Example 5 below)., , Example 5, Solution, , Prove the power rule, , n+l, , +C, , ( n # -1), , By definition, F ( x ) = J x n d x is a function such that F f ( x )= f ( x ) = x n . I-lowever, F ( x ) = [ x n +' / ( n + l)] C is such a function since its derivative is, F t ( x )= ( n 1 ) . x, -'/(n, 1 ) = x " , by the power rule for derivatives. A, , +, , ", , +, , ', , +, +, , The exclusion n # - 1 in the power rule arises because the formula, x n + ' / ( n 1) makes no sense for n = - 1; the denominator is zero. (It turns, out that l / x = x - ' does have an antiderivative, but it is a logarithm function, rather than a power of x . We will study logarithms in Chapter 6.), , +, , Example 6, , Find, , 1', , J 1+ 3x + 2 x, , $+3x+2--, , 8, , 6, , Copyright 1985 Springer-Verlag. All rights reserved.

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2.5 Antiderivatives, , 631, , We write C only once because the sum of four constants is a constant. A, , Exampk 7, Solutlon, , Find, , dx, ( 3 x + I)'., , We are looking for an antiderivative of 1 / ( 3 x + 1)'. The power of a function, ., we have, rule suggests that we guess 1/(3x + I ) ~ Differentiating,, , Comparing with 1 / ( 3 x + I)', we see that we are off by a factor of - 12, so, , Using the same method as in Example 7, we find that, , where a and b are constants, a # 0, and n is a rational number, n # - 1., Example 8, Solution, , Example 9, , Find, , I/+, , dx., , By the formula for J(ax, , Find, , I xl-8, x -2, , + b)"dx with a = 3 , b = 2, and n = +,we get, , dx., , Solution, , Here we simplify first. Dividing x3 - 8 by x - 2 gives ( x 3 - 8 ) / ( x - 2), = x 2 + 2x + 4. Thus, , Example 10, , Let x = position, v = velocity, a = acceleration, 1 = time. Express the relations, between these variables by using the indefinite integral notation., , Solution, , By the definitions of velocity and acceleration, we have v = d x / d t and, a = d v / d t . It follows that, o=ladt, , Example 11, , and x =, , I, , vdt. A, , +, , Water is flowing into a tub at 3t + 1 / ( t 1)2 gallons per minute after t, minutes. How much water is in the tub after 2 minutes if it started out empty?, , Copyright 1985 Springer-Verlag. All rights reserved.

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132, , Chapter 2 Rates of Change and the Chain Rule, , Solution, , Let f(t) be the amount of water (in gallons) in the tub at time t . We are given, , This equation means that j is the antiderivative of 3t + 1/ ( t, , + I ) ~thus,, ;, , Since the rub started out empty, f(0) = 0; so 0 = - I + C,and thus C = 1., Therefore f(t) = 3t2/2 - l / ( t + 1) + 1. Setting t = 2 gives j(2) = 3 . 4 / 2 $ + 1 = 6 3 gallons after 2 minutes. A, , Supplement lo Section 2.5, The Notation f(x) dx, , I, , The Leibniz notation Jf(x)dx for the antiderivative of a function f(x) may, seem strange at this point, but it is really rather natural and remarkably, functional. To motivate it, let us study the velocity-distance relationship, again. As in Section 1.1, we imagine a bus moving on a straight road with, position y = F ( x ) in meters from a designated starting point at time x in, seconds (see Fig. 1.1.1). There we showed that v = d y l d x is the velocity of the, bus. As in Section 1.3, we may motivate this notation by writing the velocity, as the limit, , A y- - distance travelled, Ax, , elapsed time, , as Ax + 0., , Conversely, to reconstruct y from a given velocity function v = f(x), we, notice that in a short time interval from x to x + Ax, the bus has gone, approximately A y = f (x) Ax meters (distance travelled = velocity x time, elapsed). The total distance travelled is thus the jum of f ( x ) Ax over all the, little Ax's making up the total time of the trip. This abbreviates to jf(x) dx., On the other hand, the distance travelled is y = F(x),assuming F(0) = 0,, and we know that dy/dx = v = f(x), i.e., F is an antiderivative of f. Thus, ff(x) dx is a reasonable notation for this antiderivative. The arbitrariness in, the starting position F(0) corresponds to the arbitrary constant that can be, added to the antiderivative., , Exercises for Section 2.5, Find antiderivatives for each of the functions in Exercises 1-8., 1. x + 2, 2. x6 9, 3. s ( s + l)(s 2), 4. 4x8 3 x 2, 1, 2, 5. 6. x 5 + t3, x4, 1 +x3/2, 7 . x ~ -/6~, 8. x 4 - -, , +, +, , +, , 5, , In Exercises 9-12, v is the velocity of a particle on the, line, and F ( t ) is the position at time r ., 9. v = 8r 2; F(0) 0 ; find F(1)., 10. u = -2t, 3 ; F(1) = 2; find F(3)., , +, , +, , -, , 1 1 . v = t 2 + f i ; F ( l ) = 1 ; findF(+)., 12. v = t 3 l 2 - t 2 ; F(2) = 1; find E(1)., Find the general antiderivatives for the functions f, given in Exercises 13-20., 13. f ( x ) = 3 x, 14. f ( x ) = 3 x 4 4x3, , +, , x+l, 15. f ( x ) = -16. f ( t ) = ( I I ) ~, x, 17. f ( x ) =, 18. f ( x ) =,(G+ 112, 19. f ( t ) = ( t 1)3/2, 20. f ( s ) = ( s, In Exercises 21-24, the velocity v, of a falling body (in, meters per second) near the earth's surface is given at, time t = 0. Find the velocity at time t and the position, , +, , +, , +, , Copyright 1985 Springer-Verlag. All rights reserved.

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2.5 Antiderivatives, , a t time t with the given initial positions xo. (The x axis, is oriented downwards as in Fig. 2.5.2.), 21. v, = I ; x, = 2, 22. 0, = 3; xo = - 1, 24. oO= - 2 ; x,= - 6, 23. v, = - 2 ; X, = 0, 25. Is it true that I f ( x ) g ( x ) d x is equal to, tJf(x) dxl g ( x ) + f ( x ) [ J g ( x ) dxl?, 26. Prove the constant multiple rule for antidifferentiation., 27. Prove the sum rule for antiderivatives., 28. Prove that Jf(x)f'(x)dx = f [ f(x)I2 + C for any, function f ., Find the indefinite integrals in Exercises 29-40., 30. j 4 n r 2 d r, 29. f ( x 2 3x 2)dx, , +, , +, , 133, , 58. Suppose that the marginal cost of producing, grumbies at production level p is 1 0 0 / ( p 20)*., If the cost of production is 100 when p = 0 (setup, costs), what is the cost when p = 80?, , +, , (b) Find, , dx., , ' x ', , ( x 3 - 1), 60. (a) Find ( d / d x ) J m ., , (b) Find, , +, , I, , dz., , J7T-z, , +, , 61. (a) Differentiate ( x 4 1)"., (b) Find / [ ( x 4 l)I9x3 + 3 x 2 / ' ] d x, , +, , I, 62. (a) Differentiate 3 x 9 I 2', x7/2, (b) Find, dx ., (3 x9/2l2, 63. Find a function F ( x ) such that x'F'(x), x3, 2 x = 3., 64. Find a function f ( x ) whose graph passes through, ( 1 , I) and such that the slope of its tangent line at, ( x , f ( x ) ) is 3x, 1., 65. Find the antiderivative F ( x ) of the function j ( x ), = x 3 3 x 2 + 2 which satisfies F(0) = 1., 66. Find the antiderivative G ( y ) of g ( y ) = ( y 412, which satisfies G(1) = 0., *67. (a) What integration formula can you derive, from the general power of a function rule? (See, Exercises 60 and 61.) (b) Find j ( x 3 4 ) - 3 x 2 d x ., +68. (a) What integration formula can you derive, from the chain rule?, , +, , J, , Find the indicated antiderivatives in Exercises 41-52., 41. ( x 3 + 3 x ) d x, 42. ( t 3 + t - 2 ) d t, , I, , J, , +, , + +, , +, , +, , +, , +, , (b) Find, , 53. A ball is thrown downward with a velocity of 10, meters per second. How long does it take the ball, to fall 150 meters?, 54. A particle moves along a line with velocity, u ( t ) = f t 2 I. If it is at x = 0 when t = 0, find, its position as a function of t ., 55. The population of Booneville increases at a rate, of r ( t ) = (3.62)(1 0.8 t 2 ) people per year, where, t is the time in years from 1970. The population, in 1976 was 726. What was it in 1984?, 56. A car accelerates from rest to 55 miles per hour, in 12 seconds. Assuming that the accerleration is, constant, how far does the car travel during, those 12 seconds?, 57. A rock is thrown vertically ~ p w a r dwith velocity, 19.6 meters per second. After how long does it, return to the thrower? (The acceleration due to, gravity is 9.8 meters per second per second; see, Example 4 and Fig. 2.5.3.), , { [ d m + ( x 2 + 20x)](2x + 20) dx., , +, , +, , Figure 2.5.3. The path of a, rock thrown upwards from, the earth., , Copyright 1985 Springer-Verlag. All rights reserved.

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134, , Chapter 2 Rates of Change and the Chain Rule, , Review Exerckses for Chapter 2, Differentiate each of the functions in Exercises 1-10., 1 . ( 6 x + 113, 2. ( x 2 9 x 1 0 ) ~, 3. ( x 3 x 2 - 1)1°, 4. ( ~ 2 1 ) - 1 3, 5. 6 / x, 6 . 9 x 9 - x 8 14x7 x6 5 x 4 x 2 2, , + +, +, +, , +, , + +, , + +, , Find the first and second derivatives of the functions in, Exercises 29-40., 29.f(x)=, X - a, ( a , b , c constants)., x2+2bx c, , +, , 30. S ( Z )=, , -( a , b, c, d constants)., +, , cz, , +d, , 31. x ( t ) =, , "2, , ( A , B, C constants)., , [(x', 10., , + 6)2- ( 2 x 4+ I ) ~ ], ( x 5 + 8), , Exercises 1 1-20, let, , Differentiate the given functions in Exercises 1 1- 16., , Find the equation of the tangent line to the graph of the, given function in Exercises 17-20, where'A, B, C , and, D are given above., 18. [B(x)I2 at x = 0, 17. [ A ( x ) ] ' /at, ~x =1, , 19.[c(x)12atx=-2, , 20.JD(x)atx=-l, , Differentiate the functions in Exercises 21-28., 21. f ( x ) = x5I3, 22. h ( . ~=) ( I, x3/2, 23. g ( x ) = ---, , +, , JG-7, , 35. f ( x ) = ( x - I ) ~ ~ ((here, x ) g ( x ) is some differentiable function)., 36. V ( r ) = $ v r 2 + 2vrh(r), where h ( r ) = 2r - 1 ., 37. h ( ~=) ( X - 2 ) 4 ( ~+ ~2), , 41. The volume of a falling spherical raindrop, grows at a rate which is proportional to the, surface area of the drop. Show that the radius of, the drop increases at a constant rate., 42. The temperature of the atmosphere decreases, with altitude at the rate of 2°C per kilometer at, the top of a certain cliff. A hang glider pilot finds, that the outside temperature is rising at the rate, of, degrees Centigrade per second. How fast, is the glider falling?, 43. For temperatures in the range [ - 5 0 , 1501 (degrees Celsius), the pressure in a certain closed, container of gas changes linearly with the temperature. Suppose that a 40" increase in temperature causes the pressure to increase by 30 millibars (a millibar is one thousandth of the average, atmospheric pressure at sea level). (a) What is, the rate of change of pressure with respect to, temperature? (b) What change of temperature, would cause the pressure to drop by 9 millibars?, 44. Find the rate of change of the length of an edge, of a cube with respect to its surface area., 45. The organism amoebus rectilineus always maintains the shape of a right triangle whose area is, square millimeters. Find the rate of change, of the perimeter at a moment when the organism, is isosceles and one of the legs is growing at, millimeters per second., , Copyright 1985 Springer-Verlag. All rights reserved.

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Review Exercises for Chapter 2, , rate of 756, 46. The price of calculus books riscss, per year. The price of books varies with weight at, a rate of $2.00 per pound. How fast is the weight, of books rising? (Ignore the effect of inflation)., 47. Two ships, A and B, leave San Francisco together and sail due west. A sails at 20 miles per, hour and B at 25 miles per hour. Ten miles out, to sea, A turns due north and B continues due, west. How fast are they moving away from each, other 4 hours after departing San Francisco?, 48. At an altitude of 2000 meters, a parachutist, jumps from an airplane and falls 4.9t2 meters in r, seconds. Suppose that the air pressure p decreases with altitude at the constant rate of 0.095, gsc per meter. The parachutist's ears pop when, dp/dt reaches 2 gsc per second. At what time, does this happen?, In Exercises 49-52, let A represent the area of the, shaded region in Fig. 2.R.1 ., 49. Find d A / d x and d 2 ~ / d x 2 ., 50. Find d A / d r and d 2 / d~r 2 ., 51. Find d A / d y and d 2 ~ / d y 2 ., 52. Find d A / d x and d 2 ~ / d x 2 ., , 135, , The number 20 in the formula represents the, fixed cost for placing the order, regardless of, size. The other terms represent the variable costs., (a) Find the marginal cost., (b) Find the cost for the 85th case of solvent,, i.e., the marginal cost for a purchase of 84, cases., (c) Explain in the language of marginal cost the, statement "the more you buy, the cheaper it, gets.", (d) Find a large value of x, beyond which it is, unreasonable for the given formula for C ( x ), to be applicable., 58. In Exercise 57, suppose that the solvent is priced, at 8 - [(8x 100)/(x + 300)] dollars per case at, production level x. Calculate the marginal revenue and the marginal profit., Find the equation of the line tangent to the graph of the, function at the indicated point in Exercises 59-62., 59. f ( x ) = ( x 3 - 6 x 1 ~(0,O), ;, , +, , b x 4 + 2 ~ ~; ( 1 , -2), x2+ 1, 63. If x2 + y2 + xy3 = 1 , find d y / d x when x = 0 ,, y = 1., 64. If x and y are functions of 1, x4 xy + y4 = 2,, and d y / d t = 1 at x = 1, y = 1, find d x / d t at, x = l , y = 1., 65. Let a curve be described by the parametric equations, 62. f ( x ) =, , X~ -, , +, , 49 Find d A / d x and d 2 A / d x 2, , SO Find dA/dr and d 2 ~ / d r 2, , T, , Find the equation of the tangent line at t = 2., 66. The speed of an object traveling on a parametric, , +, , 1, , (Rectangle with ;i of, disk removed), , 5 1 Find dAldy and d 2~, , /, , d 5 2 ~Find~dA/dx and d 2 ~ l d x 2, , Figure 2.R.1. Find the, indicated rates of change of, the areas., In Exercises 53-56 let P represent the perimeter of the, shaded region in Figure 2.R. 1., 53. For Exercise 49, find d A / d P and d P / d x ., 54. For Exercise 50, find d A / d P and d P / d r ., 55. For Exercise 51, find d A / d P and d P / d y ., 56. For Exercise 52, find d A / d P and d P / d x ., 57. The total cost C in dollars for producing x cases, of solvent is given by C ( x ) = 20 5x - (0.01)x2., , +, , curve is given by u = \l(dx/dt)' ( d y / d t ) 2 ., (a) Find the speed at t = 1 for the motion x, = t 3 - 3t2, I , y = t S - t7., (b) Repeat f a x = t 2 - 3 , y = + t 3 - t a t t = 1 ., 67. Find the linear approximations for: (a) 3,07n;, (b) v 7 m K ., 68. Find the linear approximations for (a), and (b), 69. (a) Find the linear approximation to the function, ( X ~ O, - 1 ) / ( ~ 2 9 1) at xo = 1., (b) Calculate [(l.021)40- 1 ] / [ ( 1 . 0 2 1 ) ~11, ~ approximately., , +, , 5m, , 6m., +, , +, , \I-., , 70. Find an approximate value for, 71. Find a formula for ( d 2 / d x 2 )~[ ( x ) ~ ( x ) ] ., 72. If f is a given differentiable function and g ( x ), = f(&), what is gJ(x)?, , Copyright 1985 Springer-Verlag. All rights reserved.

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136, , Chapter 2 Rates of Change and the Chain Rule, , 73. Differentiate both sides of the equation, , where m and n are positive integers and show, that you get the same result on each side., 74. Find a formula for (d/d~)[f(x)~g(x)"],where f, and g are differentiable functions and rn and n, are positive integers., Find the antiderivatives in Exercises 75-94., 75. (10dx, , +, , Differentiate each of the functions in Exercises 95- 102, and write the corresponding antidifferentiation formula., , 103. The [emniscate 3(x2 y2)2= 25(x2 - y2) is a planar curve which intersects itself at the origin., (a) Show by use of symmetry that the entire, lemniscate can be graphed by (1) reflecting, the first quadrant portion through the x, axis, and then (2) reflecting the right halfplane portion through the origin to the left, half-plane., (b) Find by means of implicit differentiation the, value of dy/dx at (2, 1)., (c) Determine the equation of the tangent line, to the lemniscate through (2, 1)., 104. The drag on an automobile is the force opposing, its motion down the highway, due largely to air, resistance. The drag in pounds D can be approximated for velocities u near 50 miles per hour by, D = kv2. Using k = 0.24, find the rate of increase of drag with respect to time at 55 miles, per hour when the automobile is undergoing an, acceleration of 3 miles per hour each second., 105. The air resistance of an aircraft fuel tank is given, approximately by D = 980 + 7(u - 700) lbs for, the velocity range of 700 ,< v < 800 miles per, hour. Find the rate of increase in air resistance, with respect to time as the aircraft accelerates, past the speed of sound (740 miles per hour) at a, constant rate of 12 miles per hour each second., 106. A physiology experiment measures the heart rate, R(x) in beats per minute of an athlet$i%rnbing a, vertical rope of length x feet. The experiment, produces two graphs: one is the heart rate R, versus the length x; the other is the length x, versus the time t in seconds it took to climb the, rope (from a fresh start, as fast as possible)., (a) Give a formula for the change in heart rate, in going from a 12-second climb to a 13second climb using the linear approximation., (b) Explain how to use the tangent lines to the, two graphs and the chain rule to compute, the change in part (a)., a107. (a) Find a formula for the second and third, derivatives of xn., (b) Find a formula for the rth derivative of x n if, n > r., (c) Find a formula for the derivative of the, product f(x)g(x)h(x) of three functions., , Copyright 1985 Springer-Verlag. All rights reserved.

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Review Exercises for Chapter 2, , *log. (a) Prove that if f/g is a rational function (i.e.,, a quotient of polynomials) with derivative, zero, then f/g is a constant., (b) Conclude that if the rational functions F, , 137, , and G are both antiderivatives for a function h, then F and G differ by a constant., *109. Prove that if the kth derivative of a rational, function r ( x ) is zero for some k, then r ( x ) is a, polynomial., , Copyright 1985 Springer-Verlag. All rights reserved.