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Top-down Calculus Workbook

Derivatives Intuition Practice

S. Gill Williamson

©S. Gill Williamson 2016

Preface

The book Top-down Calculus, Computer Science Press, 1987, evolved from work-books written for a UCSD program to help high school mathematics teachersreview calculus to improve their intuition and technical skills. Top-Down CalculusWorkbook, Derivatives Intuition Practice, reconstructs the original workbookformat.

Reviewing calculus is much more fun than learning it for the �rst time. Youcan go over the material in any order, including backwards. Make one passthrough the material fairly quickly, focusing on concepts and choosing carefullywhich problems to work. Additional problems can be worked later to improvetechnique. Excellent online software is available for calculus (di�erentiation andintegration) and graphing functions.

To facilitate navigation, important elements of each chapter (e.g., de�nition,lemma, theorem, exercise, �gure, etc.) are labeled in order of appearance using ashared numbering system. For example, 2.13 is a �gure and 2.14 a de�nition. Alist of all such labeled elements is in the index.

I wish to thank Professor Emeritus Michael Sharpe, Department of Mathematics,UCSD, for his LaTeX Scanpages Package which made this project feasible.

S. Gill Williamson, 2016http : \www.cse.ucsd.edu\∼gill

iii

Table of Contents

Chapter 1 (Linear Functions and Derivatives) ............................................. 1

Chapter 2 (Computing Derivatives)..............................................................31

Chapter 3 (Applications of Derivatives) ......................................................81

Index..................................................................................................................137

v

X

Y

5

X

Y

-1/2

0

2

6

1.8 NOTE: Intuitively, a function is a rule that assigns to each elementof a set X (called the domain) a unique element of a set Y (called thecodomain or range). In calculus, we usually take Y = R (the real num-bers) and D ⊆ R (a subset, possibly equal, of R). Common choicesof D are open intervals (a,b) B {x | a < x < b}, closed intervals[a,b] B {x | a ≤ x ≤ b} and intervals of the form [a,b) and (a,b]with the obvious de�nitions. Either a, b or both may be replaced by∞ (the in�nity symbol): (a,∞) = {x : a < x}, (∞,∞) = R.

9

Nancy workedhard to graphi-cally compose fand д to get h.Having done this,she can graph-ically computeh′(x) for any x .Lazy Larry, willonly composelinear functions.If Nancy asks himto compose f andд to get h he willunwittingly com-pute h′(3) for her.Forget Larry! Shecan just go alongthe graph of hcomputing slopesto get h′(x) forall x . Thinkingabout this, Nancysees that x = 3is not specialand [f (д(x))]′ =f ′(д(x))д′(x) forall x . She hasdiscovered thechain rule. This ishow good math isdone!

10

(k)

23

4/3

(cos(x))2

35

similar

37

(1) We computed

dx

(x − x1/2

x + x1/2

)1/2

as a �rst example. We use an on-

line derivative calculator. Search for “math di�erentiation integrationsolver.” Here is what we get as an answer from one:

DSd

dx

(x − x1/2

x + x1/2

)1/2

=

√x

2(x + √x)2√

x−√x

x+√x

.

We can work DS using �rst the chain rule and then the quotient rule:

CRd

dx

(x − x1/2

x + x1/2

)1/2

= 1/2(x − x1/2

x + x1/2

)−1/2d

dx

(x − x1/2

x + x1/2

)where

QRd

dx

(x − x1/2

x + x1/2

)=

(x − x1/2)′(x + x1/2) − (x − x1/2)(x + x1/2)′(x + x1/2)2 =

(1 − (1/2)x−1/2)(x + x1/2) − (x − x1/2)(1 + (1/2)x−1/2)(x + x1/2)2 =

x1/2

(x + x1/2)2 .

Substituting into CR gives

CRRd

dx

(x − x1/2

x + x1/2

)1/2

= 1/2(x − x1/2

x + x1/2

)−1/2x1/2

(x + x1/2)2 .

This is the same as equation DS (from the online derivative calculator).

47

We can also write CRR

1/2(x + x1/2

x − x1/2

)1/2x1/2

(x + x1/2)2 =x1/2

2(x − x1/2)1/2 (x + x1/2)3/2 =

x1/2

2(x2 − x)1/2 (x + x1/2) =1

2x1/2(x − 1)1/2 (x1/2 + 1) .

As you see, using online software can result in very di�erent analyticexpressions for the same answer. To check if these expressionsrepresent the same function you can use algebra (as here) or use agrapher to see of the two functions have the same graph.

Removing common factors in the beginning can simplify calculations:d

dx

(x − x1/2

x + x1/2

)1/2

=d

dx

(x1/2 − 1x1/2 + 1

)1/2

. Let f (v) = ((v − 1)/(v + 1))1/2

where v = x1/2 so dvdx = 1/2v . Thus,

d f

dx=d f

dv

dv

dx=

d

dv

(v − 1v + 1

)1/2 dv

dx=

12

(v − 1v + 1

)−1/2 2(v + 1)2

12v

Setting v = x1/2 plus a bit of algebra gives CRR again.

48

x

(a)-x x

(b)f (x)f ′(x)

f (|x |)[f (|x |)]′

DAB

SUMMARY:Think of (a), to theleft, as the graphof ln(x) for x > 0.Then (b) is the graphof ln(|x |). Note that:[ln(| − x |)]′ = −1/x= 1/(−x). Thus, wehave for all x , 0,[ln(|x |)]′ = 1/x .

Think of the graph of (a), above, as the graph of ln(x) for x > 0. Then(b) is the graph of ln(|x |). We have shown that [ln(x)]′ = 1/x . Usingthe symmetry of (b) [ln(| − x |)]′ = −1/x = 1/(−x). Thus, we have forall x , 0, [ln(|x |)]′ = 1/x . More generally, the �gures (a) and (b) aboveshow a function f (x), domain {x | x > 0}, and h(x) = f (|x |), domain{x | x , 0}. Now think of f (x) as any (di�erentiable) function. UseDAB above and apply the chain rule:

CHR h′(x) = f ′(|x |)|x |′ =

f ′(x) if x > 0−f ′(−x) if x < 0

where we note x > 0 implies |x | = x and x < 0 implies −x = |x |.Assume that f ′ is an odd function (i.e., −f ′(t) = f ′(−t) for all t ). Check“even and odd functions” in Wikipedia for a discussion. The identityCHR then impliesh′(x) = f ′(x) for all x . For f (x) = ln(x), f ′(x) = 1/xis an odd function. Proving again that for all x , 0, [ln(|x |)]′ = 1/x .

The function xx is well de�ned for x > 0 (ln(xx ) = x ln(x)).

51

− +

− +

53

2 2

56

1/2

65

NOTE: Items 2.26 through 2.34 of the book Top-down Calculus (Com-puter Science Press, 1987) contain numerous variations on 2.23 EX-ERCISES. We omit these exercises here but retain the book’s originalnumbering for convenient referencing.

67

(Use tricks of Figure 240)arcsin(x )= π2 −arccos(x )(arcsin(x ))′=−(arccos(x ))′= 1

(1−x2)1/2

(arctan(x ))′= 1tan′(sec2(arctan(x ))=

1sec2arcsec((1+x2)1/2)=

1(1+x2)

arccot(x )= π2 −arctan(x )(arcsec(x ))′= 1

sec′(arcsec(x ))=1

sec(arcsec(x ))tan(arcsec(x ))

76

(e.g. Wikipedia).

106

0/02

110

The following exercises will give you practice in graphing and limits.

limx→−2

(x + 3) = 1

limx→9

(x + 9)/2 = 9

Use L’Hopital’s rule or sin(x) ≈ x − x3/6.

Use L’Hopital’s rule.

12+ lim

x→∞

sin(x)x=

12

limx→∞

ln(x)xa= 0 by L’Hopital’s rule

max. x = 3/4min. x = 2, , in�. pt. x = 3

f ′α (x) = 4αx − 6α2 = 0 at x = 3α/2 (min.)f ′α (x) = −2x−3 + αx−2

f ′α (x) = 0 if x = 2/α (min)∗f ′′α (x) = 6x−4 − 2αx−3

f ′′α (x) = 0 if x = 3/α∗ f ′′α (2/α) = α4/8 > 0

(check solutions by graphing)

112

+

+

+

(SOLUTIONS end of chapter)

113

+

+(0)

(0)

(π + 2) Let z = x − π/2 so cos(x) =cos(z + π/2) = − sin(z).

(1) l’Hopital

(0) l′Hopital or cos(x) ≈ 1 − x2/2

(1) z = ln(x) − 1, limz→0

ln(1 + z)z

(Use online graphers to check work.)

114

D01={x |y′(x )=0}={x |cos(x )=−1}D02={x |y′′(x )=0}={x |sin(x )=0} D01⊂D02

replace by ln |x|+ln |x2−3|, D01={−1,+1}, z=x2 helps for y′′

f ′α (x )=ln |αx |+1, f ′α (x )=0 if x=±(αe)−1; f ′′α (x )=1/x, x,0

z=x−α, fα (z)= (z+α )2+αz =z+2α+(α 2+α )z−1

f ′α (z)=1+(α 2+α )(−1)z−2, f ′α (z)=0 if z=±(α 2+α )1/2

row

115

row

118

д = 32ft/s2 or д = 9.8m/s2.

123

For this problem we have:

125

row

126

minimal.

127

Assume the container is a parallelepiped.

128

+

+ +

+

+ +

+ +

+

SOLUTIONS to 3.22 (variations on exercise 3.20)(1)

(a) limx→3

3x3 + 4x2 − x + 12x2 + 4x − 5

=3 · 33 + 4 · 32 − 3 + 1

2 · 32 + 4 · 3 − 5

(b) limx→−1

x2 + 1x2 − 1

= +∞

(2)(a) lim

x→0+

ln(tan(x))x3 =

(limx→0+

ln(tan(x))) (

limx→0+

1x3

)= −∞

(b) limx→0+

sin(x)2 tan(x1/2) = lim

x→0+

x − x3/62(x1/2 + x3/2/3) = 0 (TP’s)

or limx→0+

sin(x)2 tan(x1/2) = lim

x→0+

cos(x)2 sec2(x1/2)(1/2)x−1/2 =

limx→0+

x1/2 cos(x) cos2(x1/2) = 0 (L’Hopital)

(3)(a) lim

x→0+

ex − ex1/2

x1/2 = −e

(b) limx→π /2

sin(cos(x))cos(x) = lim

z→0

sin(z)z= 1 (z = cos(x))

y = x ln |x |

y = x 2+1x 2−1

(4)(a) We have y = x ln |x |, y ′(x) = ln |x | + 1 and y ′′(x) = 1/x .

Thus, y(x) = 0 when x = ±1, y ′(x) = 0 when x = ±e .The maximum occurs at x = −e , the minimum at x = +e:

(b) y = (x2 + 1)/(x2 − 1), y ′ = −4x/(x2 − 1)2, y ′′ = −4.A maximum occurs at x = 0. There are no in�ection points.

132

α=3, β=1.5

(5)(a) fα (x) = x − α/x , α > 0, f ′(x) = 1 + α/x2, f ′′(x) = −2α/x3.

There are no maxima, minima or points of in�ection.

(b) Let f (x) = xα − x β , α > β > 0, f ′(x) = αxα−1 − βx β−1

f ′′(x) = α(α − 1)xα−2 − β(β − 1)x β−2. f ′(x) = 0 whenxα−β = β/α , x = (β/α)1/(α−β ) .f ′′(x) = α(α − 1)xα−2 − β(β − 1)x β−2 = 0

when x =

(β(β − 1)α(α − 1)

)1/(α−β )(take β = α/2 for the problem).

w

A

Bβ

rθ

SOLUTIONS to 3.31 (variations on exercise 3.25)

(1) Optimal can with �xed surface area s: Let x be the radius ofthe base, h the height. Then s = 2πx2 + 2πxh, h = (s − 2πx2)/2πx ,V = πx2h. Thus,

V = πx2(s − 2πx2

2πx

)= (x/2)(s − 2πx2) = (s/2)x − πx3

and V ′(x) = s/2 − 3πx2. Thus, V ′(x) = 0 when x2 = s/6π . The can ofarea s and maximal volume has radius r = (s/6π )1/2 and

h =s

2π·

1r− r = 3

( s

6π

) 1r− r = 3

�r 2� 1

r− r = 2r B d .

(2) Running and rowing to cross river: See solutions 3.26, Figure3.27 where we have β = 6, ρ = 10, σ = 5, w = 1. Here, we have

T (r ) = r

ρ+(w2 + (β − r )2)1/2

σ.

T ′(r ) = 1ρ−

β − r

σ (w2 + (β − r )2)1/2 = 0.

if w2 + (β − r )2 = ρ2(β − r )2/σ 2 or

(β − r ) = w

(ρ2/σ 2 − 1)1/2 .

133

w

A

Bβ

rθ

β

λt

ta

b

c

1

(1+γ 2)1/2

γ

tm

λ

λ

τ

τ

xs

-

SOLUTIONS to 3.31 continued (variations on exercise 3.25)

(2) Crossing river - alternative solution (see rules 2.24):The variable is the angle θ between the rowing direction and the bank.

T (θ ) = β −w cot(θ )ρ

+w csc(θ )

σ

T ′(θ ) = w csc2(θ )ρ

−w csc(θ ) cot(θ )

σ= 0

σ

ρ=

csc(θ ) cot(θ )csc2(θ ) = cos(θ ).

(3) Longest board problem: For �xed λ, β , the length of the board isL(t) = β sec(t)+ λ csc(t). Thus, L′(t) = β sec(t) tan(t)− λ csc(t) cot(t).Setting L′(t) = 0, we get

λ

β=

sec(t) tan(t)csc(t) cot(t) = tan3(t).

Thus, tm = arctan(γ ), γ B (λ/β)1/3, minimizes L(t).The longest board that slides between the two hallways haslength Lλβ = β sec(arctan(γ )) + λ csc(arctan(γ )). Note

arctan(γ ) = arcsec((1 + γ 2)1/2

)arctan(γ ) = arccsc

((1 + γ 2)1/2/γ)= arccsc

((1 + γ−2)1/2)

so

.

Lβλ = Lλβ = β *,1 +

(λ

β

)2/3+-

1/2

+ λ *,1 +

(β

λ

)2/3+-

1/2

.

(4) Changing shadow problem: By similar triangles,

x

λ − τ=

s

τso s =

( τ

λ − τ

)x .

s =( τ

λ − τ

)ρ

no matter what x is.

134

SOLUTIONS to 3.31 continued (variations on exercise 3.25)

(5) Weight loss with height: As a function of height r abovethe surface, the weight of an object weighing ω pounds on thesurface is W (r ) = ω(1 + .00025r )−2. We use z := d

dt z. Then,W (r ) = −2ω(1 + .00025r )−3.00025r and set ω = R.

(6) Vertical motion and gravity: Recall that д = 32ft/s2 orд = 9.8m/s2 and h(t) = −(1/2)дt2 + V0t . Then, h′(t) = 0 implies−дt + V0 = 0 or t = V0/д. Thus, hmax = (−1/2)д(V0/д)2 + V 2

0 /д andh(t) = 0 when t = 0 or t = 2V0/д. Also, h′(t) = −V0 when t = 2V0/д.

(7) Rotating tangent: Review Figure 3.30 and the discussion there.Recall di�erentiation rules 2.41. We have (x ,y) = (a + b cos(t), c +d sin(t)) and

dy

dx=

(c + d sin(t))′(a + b cos(t))′ =

d cos(t)−b sin(t) = −

d

bcot(t).

Thus,

ϕ(t) = arctan(−d

bcot(t)

)and

ϕ ′(t) = 1

1 +(db cot(t))2

(−d

b

) �− csc2(t)� = bd csc2(t)

b2 + d2 cot2(t) .

(8) Arclength: The curve (x(t),y(t)) = (a +b cos(t), c +d sin(t)) is anellipse. Referring to 3.26 solution (8), let L(t) be the arclength and

dL

dt=

�(x ′(t))2 + (y ′(t))2�1/2=

�b2 sin2(t) + d2 cos2(t)�1/2

.

135

Index

Contents, v1 Chapter, 11.1 FIGURE Linear f, g, and h, 21.2 FIGURE Derivative at Point, 41.3 SLOPPY DEFINITION derivative, 51.4 FIGURE Derivative f ′, 61.5 DEFINITION derivative f ′ of f , 61.6 EXERCISES derivative graphical, 61.7 LINEAR FUNCTIONS, 61.8 DEFINITION composition f ,д, 81.8 NOTE function, 91.9 FIGURE Graphical Chain Rule, 101.10 THEOREM chain rule, 111.11 FIGURE Derivative of Sum, 121.12 RULES DERIVATIVES, 131.13 FIGURE Derivative of x2, 141.14 Di�erence Quotient, 151.15 EXAMPLES rules 1.12, 161.16 EXERCISES compositions, 211.17 SOLUTIONS 1.16, 242 Chapter, 312.1 LEMMA (xn)′ = nxn−1,n ≥ 0, 322.2 LEMMA (xr )′ = rxr−1, r = 1/m, 322.3 LEMMA (xr )′ = rxr−1, r = n/m, 322.4 LEMMA (xr )′ = rxr−1, r = −1, 332.5 THEOREM (xr )′ = rxr−1, r real, 332.6 EXAMPLES QUOTIENT RULE, 352.7 DIFFERENTIATION RULES, 362.8 Derivative sin(x) is cos(x), 372.9 THEOREM (sin(x))′ = cos(x), 382.10 BASIC TRIG DERIVATIVES, 382.11 FIGURE Exponential Functions, 402.12 DEFINITION e = 2.7182818 . . ., 41

2.13 FIGURE Log Functions, 422.14 DEFINITION loga(y),y > 0, 432.15 loga(x), ax inverses, 432.16 THEOREM f ′a (x) = f ′a (0)ax , 452.17 THEOREM f ′a (x) = ln(a)ax , 462.18 THEOREM (loga x)′ = (ln(a)x)−1, 462.19 EXP AND LOG DERIVATIVES, 472.20 EXAMPLES DERIVATIVES, 472.21 BASIC LOG FACTS, 572.22 BASIC TRIG FACTS, 582.23 EXERCISES di�erentiation, 582.24 TRIG AND LOG DIFF RULES, 602.25 SOLUTIONS EXERCISE 2.23, 612.35 INV. TRIG HYPERBOLIC FNCS, 682.36 Four Functions, 702.37 FIGURE Function not One-to-One, 712.38 FIGURE Compositional Inverses, 722.39 RULE (h−1(x))′ = 1/(h′(h−1(x)), 732.40 Inverse Trig Functions, 742.41 DIFF RULES INV TRIG FNCS, 752.42 EXERCISES derive rules, 762.43 FIGURE Inverse secant, cosecant, 762.44 DEFINITION hyperbolic functions, 772.45 DIFF RULES HYPERBOLIC FNCS, 772.46 Graphs hyperbolic functions, 782.47 INV HYPERBOLIC FNCS, 792.48 DIFF INV HYPERBOLIC FNCS, 793 Chapter, 813.1 FIGUREy = x−(1/3)x3 andy ′, 823.2 GRAPHING x − (1/3)x3, 833.3 GRAPH 1+2x+3x2+4x3+5x5, 863.4 DATA 1+2x+3x2+4x3+5x5, 873.5 EXAMPLE fa(x) = ax(x−a)2, 89

137

3.6 FIGURE Parameterized fa , 903.7 SECOND DERIVATIVE TEST, 913.8 REVIEW OF GRAPHING, 913.9 LIMITS from above and below, 933.10 PROGRAM SEQUENCE, 943.11 DEFINITION limit, 973.12 DEFINITION continuous function, 973.13 PROGRAM INFINITE LIMIT, 993.14 RULES FOR LIMITS, 993.15 L’HOPITAL’S RULE, 1013.16 L’HOPITAL’S FAILURE, 1043.17 TAYLOR POLYNOMIALS, 1063.18 DATA TAYLOR POLYNOMIAL, 1073.19 EXERCISES limits, 1093.20 EXERCISES limits, 1123.21 VARIATIONS ON 320, 1123.22 VARIATIONS ON 320, 1133.23 VARIATIONS ON 320, 1143.24 CLASSICAL APPLICATIONS, 1153.25 EXERCISES word problems, 1153.26 SOLUTIONS EXERCISE 3.25, 1163.27 FIGURE Straight Slow River, 1183.28 FIGURE Board Around Corner, 1203.29 FIGURE Walking from Lampost, 1213.30 FIGURE Tangent to Ellipse, 1243.31 VARIATIONS ON 3.25, 1253.32 VARIATIONS ON 3.25, 1263.33 VARIATIONS ON 3.25, 1283.34 VARIATIONS ON 3.25, 1293.35 SOLUTIONS: partial for 3.22, 1323.36 SOLUTIONS: VAR. 3.31 (ON 3.25), 133

chain ruleand inverses, 73examples, 17–19graphical def., 10limited information, 55stated, 11

completing square, 25compositional inverses, 68critical points, 92

derivativecsc(x), tan(x), cot(x), 49loga(| x |), 51loga(x), ax , 44–47sec(x), 48sin, cos, 37xx , 50as limit, 15examples of rules, 16basic log rules, 57basic rules, 13basic trig and log rules, 60basic trig rules, 58basic trig., 38constant function, 16function, 5higher order, 52hyperbolic, 77implicit, 56inverse hyperbolic, 79inverse trig functions, 75, 76logarithmic, 54microsope, 3, 4

di�erential notation, 16

functionscomposition, 8exponential, 40, 43exponents, logs, 39graph, 69hyperbolic, 77, 78image, domain, 68, 69inverse hyperbolic, 79inverse trig, 74linear, 1logarithmic, 42, 43parameterized family, 89Sally, 71unreasonable, 103

graphing functions, 90

138

in�ection points, 92interval notation, 68inverses

and chain rule, 73basic properties, 72compositional, 24functional, 71trig functions, 73–75

L’Hopital’s Rule, 93L’Hopital’s Rule examples, 101–103limit terminology, 97limits and graphing, 112limits of ratios, 100linear functions

basic properties, 7, 8derivatives, 1

local max, min, 90local maximum, 84local minimum, 84

max-min problems, 115

numberse = 1.7182818 . . ., 41rational, irrational, 33

parmeterized family, 89point of in�ection, 85

quotient rule, 34

related rate problems, 115

second derivative test, 90

Taylor polynomials, 93, 106, 107

139

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