Yet Another Calculus Text, 2007a

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34 CHAPTER 1. DERIVATIVES Note that if we want to evaluate dy dx when, for example, x = 2, we may either evaluate the final form, that is, dy dx∣ =(24x + 12)| x=2 = 48 + 12 = 60, x=2 or, noting that u =5whenx = 2, the intermediate form, that is, dy dx∣ =12u| u=5 =60. x=2 In other words, Exercise 1.7.11. dy dx∣ = dy du x=2 du∣ u=5 dx∣ . x=2 If y = u 3 + 5 and u = x 2 − 1, find dy ∣ . x=1 dx Example 1.7.14. If h(x) = √ x 2 +1, then h(x) =f(g(x)) where f(x) = √ x and g(x) =x 2 +1. Since it follows that f ′ (x) = 1 2 √ x and g′ (x) =2x, h ′ (x) =f ′ (g(x))g ′ (x) = 1 2 √ x 2 +1· 2x = x √ x2 +1 . Exercise 1.7.12. Find the derivative of f(x) = √ 4x +6. Exercise 1.7.13. Find the derivative of y =(x 2 +5) 10 . Example 1.7.15. As we saw in Example 1.2.5, ifM is the mass, in grams, of a spherical balloon being filled with water and r is the radius of the balloon, in centimeters, then M = 4 3 πr3 grams and dM dr =4πr2 grams/centimeter, a result which we may verify easily now using the power rule. Suppose water is being pumped into the balloon so that the radius of the balloon is increasing at the rate of 0.1 centimeters per second when the balloon has a radius of 10 centimeters. Since M is a function of time t, as well as a function of the radius
1.7. PROPERTIES OF DERIVATIVES 35 of the balloon r, wemightwishtoknowtherateofchangeofM with respect to t. Since we are given that dr dt ∣ =0.1 centimeters/second, r=10 we may use the chain rule to find that dM dt ∣ = dM dr r=10 dr ∣ r=10 dt ∣ = (400π)(0.1) = 40π grams/second. r=10 Exercise 1.7.14. Suppose A is the area and r is the radius of a circular wave at time t. Supposewhenr = 100 centimeters the radius of the circle is increasing at a rate of 2 centimeters per second. Find the rate at which the area of the circle is growing when r = 100 centimeters. Exercise 1.7.15. If water is being pumped into a spherical balloon at the rate of 100 grams per second, find the rate of change of the radius r of the balloon when the radius of the balloon is r =15centimeters. As an important special case of the chain rule, suppose n ≠ 0 is an integer, g is a differentiable function, and h(x) =(g(x)) n .Thenh is the composition of f(x) =x n with g, andso,usingthechainrule, h ′ (x) =f ′ (g(x))g ′ (x) =n(g(x)) n−1 g ′ (x). (1.7.43) If we let u = g(x), we could also express this result as d dx un n−1 du = nu dx . (1.7.44) Example 1.7.16. With n =10andg(x) =x 2 +3,wehave d dx (x2 +3) 10 = 10(x 2 +3) 9 (2x) =20x(x 2 +3) 9 . Example 1.7.17. If 15 f(x) = (x 4 +5) 2 , then we may apply the previous result with n = −2 andg(x) =x 4 + 5 to obtain f ′ (x) =−30(x 4 +5) −3 (4x 3 )=− 120x3 (x 4 +5) 3 . We may use the previous result to derive yet another extension to the power rule. If n ≠ 0 is an integer and y = x 1 n ,theny n = x, andso,assumingy is differentiable, d dx yn = d x. (1.7.45) dx
Page 1 and 2: Yet Another Calculus Text A Short I
Page 3 and 4: Preface I intend this book to be, f
Page 5 and 6: Contents Preface Contents iii v 1 D
Page 7 and 8: Chapter 1 Derivatives 1.1 The arrow
Page 9 and 10: 1.2. RATES OF CHANGE 3 determinate
Page 11 and 12: 1.2. RATES OF CHANGE 5 Exercise 1.2
Page 13 and 14: 1.3. THE HYPERREALS 7 To find the r
Page 15 and 16: 1.4. CONTINUOUS FUNCTIONS 9 the dis
Page 17 and 18: 1.4. CONTINUOUS FUNCTIONS 11 Simila
Page 19 and 20: 1.5. PROPERTIES OF CONTINUOUS FUNCT
Page 29 and 30: 1.6. THE DERIVATIVE 23 y 10 8 6 4 y
Page 31 and 32: 1.6. THE DERIVATIVE 25 y 1 y = x 2
Page 33 and 34: 1.7. PROPERTIES OF DERIVATIVES 27 1
Page 35 and 36: 1.7. PROPERTIES OF DERIVATIVES 29 a
Page 37 and 38: 1.7. PROPERTIES OF DERIVATIVES 31 T
Page 39: 1.7. PROPERTIES OF DERIVATIVES 33 1
Page 43 and 44: 1.7. PROPERTIES OF DERIVATIVES 37 E
Page 45 and 46: 1.8. A GEOMETRIC INTERPRETATION OF
Page 47 and 48: 1.9. INCREASING, DECREASING, AND LO
Page 49 and 50: 1.9. INCREASING, DECREASING, AND LO
Page 51 and 52: 1.10. OPTIMIZATION 45 y 10 y = x +2
Page 53 and 54: 1.10. OPTIMIZATION 47 y 6 5 4 3 2 y
Page 55 and 56: 1.10. OPTIMIZATION 49 y 5 4 3 2 1 4
Page 57 and 58: 1.10. OPTIMIZATION 51 y 4 3 y = x 2
Page 59 and 60: 1.11. IMPLICIT DIFFERENTIATION AND
Page 61 and 62: 1.11. IMPLICIT DIFFERENTIATION AND
Page 63 and 64: 1.12. HIGHER-ORDER DERIVATIVES 57 D
Page 65 and 66: 1.12. HIGHER-ORDER DERIVATIVES 59 1
Page 67 and 68: 1.12. HIGHER-ORDER DERIVATIVES 61 y
Page 69 and 70: Chapter 2 Integrals 2.1 Integrals W
Page 71 and 72: 2.1. INTEGRALS 65 From our rules fo
Page 73 and 74: 2.1. INTEGRALS 67 Exercise 2.1.2. F
Page 75 and 76: 2.2. DEFINITE INTEGRALS 69 2.2 Defi
Page 77 and 78: 2.2. DEFINITE INTEGRALS 71 integral
Page 79 and 80: 2.3. PROPERTIES OF DEFINITE INTEGRA
Page 81 and 82: 2.4. THE FUNDAMENTAL THEOREM OF INT
Page 83 and 84: 2.4. THE FUNDAMENTAL THEOREM OF INT
Page 85 and 86: 2.5. APPLICATIONS OF DEFINITE INTEG
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2.5. APPLICATIONS OF DEFINITE INTEG
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2.6. SOME TECHNIQUES FOR EVALUATING
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2.7. THE EXPONENTIAL AND LOGARITHM
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Answers to Exercises 1.2.1. (a) 32
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1.7.11. dy dx∣ = 216 x=1 131 1.7.
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1.11.2. dy dx∣ = − 7 (x,y)=(2,
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135 2.5.9. 8π 3 2.5.10. 2π 3 2.5.
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137 2.6.22. ∫ π 2 0 sin(x)sin(2x
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139 2.7.20. ∫ 1 0 x +14 x 2 dx =5
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INDEX 141 partial fractions, 123, 1
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Yet Another Calculus Text, 2007a

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