Example 2.4.1 Show that Theorem 2.4.2 guarantees the validity of the fol-

lowing method of bisection for ¬nding zeros of a continuous function f :

Bisection Method: We wish to solve f (x) = 0 for x.

Step 1. Locate two points such that f (a)f (b) < 0.

1

(a + b) .

Step 2. Determine the sign of f

2

86 CHAPTER 2. LIMITS AND CONTINUITY

1 1

(i) If f (a + b) = 0, stop the procedure; (a + b) is a zero of f .

2 2

1 1

(a + b) · f (a) < 0, then let a1 = a, b1 = (a + b).

(ii) If f

2 2

1 1

(a + b) · f (b) < 0, then let a1 = (a + b), b1 = b.

(iii) If f

2 2

1

Then f (a1 ) · f (b1 ) < 0, and |b1 ’ a1 | =

(b ’ a).

2

Step 3. Repeat Step 2 and continue the loop between Step 2 and Step 3 until

|bn ’ an |/2n < Tolerance Error.

Then stop.

This method is slow but it approximates the number c guaranteed by

Theorem 2.4.2. This method is used to get close enough to the zero. The

switchover to the faster Newton™s Method that will be discussed in the next

section.

Theorem 2.4.3 (Intermediate Value Theorem). Suppose that a function

is de¬ned and continuous on a closed and bounded interval [a, b]. Suppose

further that there exists some real number k such that either (i) f (a) < k <

f (b) or (ii) f (b) < k < f (a). Then there exists some c such that a < c < b

and f (c) = k.

Proof. Let g(x) = f (x) ’ k. Then g is continuous on [a, b] and either (i)

g(a) < 0 < g(b) or (ii) g(b) < 0 < g(a). By Theorem 2.4.2, there exists some

c such that a < c < b and g(c) = 0. Then

0 = g(c) = f (c) ’ k

and

f (c) = k

as required. This completes the proof.

2.4. PROPERTIES OF CONTINUOUS FUNCTIONS 87

Theorem 2.4.4 Suppose that a function f is de¬ned and continuous on a

closed and bounded interval [a, b]. Then there exist real numbers m and M

such that

[m, M ] = {f (x) : a ¤ x ¤ b}.

That is, a continuous function f maps a closed and bounded interval [a, b]

onto a closed and bounded interval [m, M ].

Proof. By Theorem 2.3.14, there exist two numbers c1 and c2 in [a, b] such

that for all x ∈ [a, b],

m = f (c1 ) ¤ f (x) ¤ f (c2 ) = M.

By the Intermediate Value Theorem (2.4.3), every real value between m and

M is in the range of f contained in the interval with end points c1 and c2 .

Therefore,

[m, M ] = {f (x) : a ¤ x ¤ b}.

Recall that m = absolute minimum and M = absolute maximum of f on

[a, b]. This completes the proof of the theorem.

Theorem 2.4.5 Suppose that a function f is continuous on an interval [a, b]

and f has an inverse on [a, b]. Then f is either strictly increasing on [a, b]

or strictly decreasing on [a, b].

Proof. Since f has in inverse on [a, b], f is a one-to-one function on [a, b].

So, f (a) = f (b). Suppose that f (a) < f (b). Let

A = {x : f is strictly increasing on [a, x] and a ¤ x ¤ b}.

Let c be the least upper bound of A. If c = b, then f is strictly increasing on

[a, b] and the proof is complete. If c = a, then there exists some d such that

a < d < b and f (d) < f (a) < f (b). By the intermediate value theorem there

must exist some x such that d < x < b and f (x) = f (a). This contradicts the

fact that f is one-to-one. Then a < c < b and there exists some d such that

c < d < b and f (a) < f (d) < f (c). By the intermediate value theorem there

exists some x such that a < x < c and f (x) = f (c) and f is not one-to-one.

It follows that c must equal b and f is strictly increasing on [a, b]. Similarly,

if f (a) > f (b), f will be strictly decreasing on [a, b]. This completes the

proof of the theorem.

88 CHAPTER 2. LIMITS AND CONTINUITY

Theorem 2.4.6 Suppose that a function f is continuous on [a, b] and f

is one-to-one on [a, b]. Then the inverse of f exists and is continuous on

J = {f (x) : a ¤ x ¤ b}.

Proof. By Theorem 2.4.4, J = [m, M ] where m and M are the absolute

minimum and the absolute maximum of f on [a, b]. Also, there exist numbers

c1 and c2 on [a, b] such that f (c1 ) = m and f (c2 ) = M . Since f is either

strictly increasing or strictly decreasing on [a, b], either a = c1 and b = c2

or a = c2 and b = c1 . Consider the case where f is strictly increasing and

a = c1 , b = c2 . Let m < d < M and d = f (c). Then a < c < b. We show

that f ’1 is continuous at d. Let > 0 be such that a < c ’ < c < c + 2b.

Let d1 = f (c ’ ), d2 = f (c + ). Since f is strictly increasing, d1 < d <

d2 . Let δ = min(d ’ d1 , d2 ’ d). It follows that if 0 < |y ’ d| < δ, then

|f ’1 (y) ’ f ’1 (d)| < and f ’1 is continuous at d. Similarly, we can prove the

one-sided continuity of f ’1 at m and M . A similar argument will prove the

continuity of f ’1 if f is strictly decreasing on [a, b].

Theorem 2.4.7 Suppose that a function f is continuous on an interval I

and f is one-to-one on I. Then the inverse of f exists and is continuous on

I.

Proof. Let J = {f (x) : x is in I}. By the intermediate value theorem

J is also an interval. Let d be an interior point of J. Then there exists a

closed interval [m, M ] contained in I and m < d < M . Let c1 = f ’1 (m), c2 =

f ’1 (b), a = min{c1 , c2 } Since the theorem is valid on [a, b], f ’1 is continuous

at d. The end points can be treated in a similar way. This completes the

proof of the theorem. (See the proof of Theorem 2.4.6).

Theorem 2.4.8 (Fixed Point Theorem). Let f satisfy the conditions of

Theorem 2.4.4. Suppose further that a ¤ m ¤ M ¤ b, where m and M are

the absolute minimum and absolute maximum, respectively, of f on [a, b].

Then there exists some p ∈ [a, b] such that f (p) = p. That is, f has a ¬xed

point p on [a, b].

Proof. If f (a) = a, then a is a ¬xed point. If f (b) = b, then b is a ¬xed

point. Suppose that neither a nor b is a ¬xed point of f . Then we de¬ne

g(x) = f (x) ’ x

for all x ∈ [a, b].

2.4. PROPERTIES OF CONTINUOUS FUNCTIONS 89

We observe that g(b) < 0 < g(a). By the Intermediate Value Theorem

(2.4.3) there exists some p such that a < p < b and g(p) = 0. Then

0 = g(p) = f (p) ’ p

and hence,

f (p) = p

and p is a ¬xed point of f on [a, b]. This completes the proof.

Remark 9 The Fixed Point Theorem (2.4.5) is the basis of the ¬xed point

iteration methods that are used to locate zeros of continuous functions. We

illustrate this concept by using Newton™s Method as an example.

Example 2.4.2 Consider f (x) = x3 + 4x ’ 10.

Since f (1) = ’5 and f (2) = 6, by the Intermediate Value Theorem (2.4.3)

there is some c such that 1 < c < 2 and f (c) = 0. We construct a function g

whose ¬xed points agree with the zeros of f . In Newton™s Method we used

the following general formula:

f (x)

g(x) = x ’ .

slope(f (x), x)

Note that if f (x) = 0, then g(x) = x, provided slope (f (x), x) = 0. We ¬rst

compute

1

[f (x + h) ’ f (x)]

Slope(f (x), x) = lim

h’0 h

1

[{(x + h)3 + 4(x + h) ’ 10} ’ {x3 + 4x ’ 10}]

= lim

h’0 h

1

[3x2 h + 3xh2 + h3 + 4h]

= lim

h’0 h

= lim [3x2 + 3xh + h2 + 4]

h’0

2

= 3x + 4.

We note that 3x2 + 4 is never zero. So, Newton™s Method is de¬ned.

The ¬xed point iteration is de¬ned by the equation

f (xn )

xn+1 = g(xn ) = xn ’

slope(f (x), xn )

90 CHAPTER 2. LIMITS AND CONTINUITY

or

x3 + 4xn ’ 10

= xn ’ n 2

xn+1 .

3xn + 4

Geometrically, we draw a tangent line at the point (xn , f (xn )) and label the

x-coordinate of its point of intersection with the x-axis as xn+1 .

graph

y ’ f (xn ) = m(x ’ xn )

Tangent line:

0 ’ f (xn ) = m(xn+1 ’ xn )

f (xn )

xn+1 = xn ’ ,

m

where m = slope (f (x), xn ) = 3x2 + 4.

n

To begin the iteration we required a guess x0 . This guess is generally

obtained by using a few steps of the Bisection Method described in Example

36. Let x0 = 1.5. Next, we need a stopping rule. Let us say that we will

stop when a few digits of xn do not change anymore. Let us stop when

|xn+1 ’ xn | < 10’4 .

We will leave the computation of x1 , x2 , x3 , . . . as an exercise.

Remark 10 Newton™s Method is fast and quite robust as long as the initial

guess is chosen close enough to the intended zeros.

Example 2.4.3 Consider the same equation (x3 + 4x ’ 10 = 0) as in the

preceding example.

We solve for x in some way, such as,

1/2

10

= g(x).

x=

4+x

2.4. PROPERTIES OF CONTINUOUS FUNCTIONS 91

In this case the new equation is good enough for positive roots. We then

de¬ne

xn+1 = g(xn ), x0 = 1.5

and stop when

|xn+1 ’ xn | < 10’4 .

We leave the computations of x1 , x2 , x3 . . . as an exercise. Try to compare the

number of iterations needed to get the same accuracy as Newton™s Method

in the previous example.

Exercises 2.4

1. Perform the required iterations in the last two examples to approximate

the roots of the equation x3 + 4x ’ 10 = 0.

π

2. Let f (x) = x ’ cos x. Then slope (f (x), x) = 1 + sin x > 0 on 0, .

2

π

Approximate the zeros of f (x) on 0, by Newton™s Method:

2

xn ’ cos xn

xn+1 = xn ’ , x0 = 0.8

1 + sin xn

and stop when

|xn+1 ’ xn | < 10’4 .

π

3. Let f (x) = x ’ 0.8 ’ 0.4 sin x on 0, . then slope (f (x), x) = 1 ’

2

π

0.4 cos x > 0 on 0, . Approximate the zero of f using Newton™s

2

Iteration

xn ’ 0.8 ’ 0.4 sin(xn )

xn+1 = xn ’ , x0 = 0.5

1 ’ 0.4 cos(xn )

4. To avoid computing the slope function f , the Secant Method of iter-

ation uses the slope of the line going through the previous two points

92 CHAPTER 2. LIMITS AND CONTINUITY

(xn , f (xn )) and (xn+1 , f (xn+1 )) to de¬ne xn+2 as follows: Given x0 and

x1 , we de¬ne

f (xn+1 )

xn+2 = xn+1 ’

f (xn+1 )’f (xn )

xn+1 ’xn

f (xn+1 )(xn+1 ’ xn )

xn+2 = xn+1 ’

f (xn+1 ’ f (xn )

This method is slower than Newton™s Method, but faster than the Bi-

section. The big advantage is that we do not need to compute the slope

function for f . The stopping rule can be the same as in Newton™s Method.

Use the secant Method for Exercises 2 and 3 with x0 = 0.5, x1 = 0.7 and

|xn+1 ’ xn | < 10’4 . Compare the number of iterations needed with

Newton™s Method.

5. Use the Bisection Method to compute the zero of x3 +4x’10 on [1, 2] and

compare the number of iterations needed for the stopping rule |xn+1 ’

xn | < 10’4 .

6. A set S is said to be connected if S is not the union of two non-empty

sets A and B such that A contains no limit point of B and B contains

no limit point of A. Show that every closed and bounded interval [a, b]

is connected.

(Hint: Assume that [a, b] is not connected and [a, b] = A ∪ B, a ∈ A, B =

… as described in the problem. Let m = lub(A), M = glb(B). Argue

1

that m ∈ A and m ∈ B. Then (m + M ) ∈ (A ∪ B). The contradiction

/

2

proves the result.

7. Show that the Intermediate Value Theorem (2.4.3) guarantees that con-

tinuous functions map connected sets onto connected sets. (Hint: Let

S be connected and f be continuous on S. Let Rf = {f (x) : x ∈ S}.

Suppose Rf = A ∪ B, A = …, B = …, such that A contains no limit point

of B and B contains no limit point of A. Let U = {x ∈ S : f (x) ∈ A},

V = {x ∈ S : f (x) ∈ B}. Then S = U ∪ V, U = … and V = …. Since S

is connected, either U contains a limit point of V or V contains a limit

point of U . Suppose p ∈ V and p is a limit point of U . Then choose a

2.4. PROPERTIES OF CONTINUOUS FUNCTIONS 93

sequence {un } that converges to p, un ∈ U . By continuity, {f (un )} con-

verges to f (p). But f (un ) ∈ A and f (p) ∈ B. This is a contradiction.)

8. Find all of the ¬xed points of the following:

(a) f (x) = x2 , ’4 ¤ x ¤ 4

(b) f (x) = x3 , ’2 ¤ x ¤ 2

(c) f (x) = x2 + 3x + 1

(d) f (x) = x3 ’ 3x, ’4 ¤ x ¤ 4

(e) f (x) = sin x

9. Determine which of the following sets are

(i) bounded, (ii) closed, (iii) connected.

(a) N = {1, 2, 3, . . . , }

(b) Q = {x : x is rational number}

(c) R = {x : x is a real number}

(d) B1 = {sin x : ’π ¤ x ¤ π}

(e) B2 = {sin x : ’π < x < π}

’π π

<x<

(f) B3 = sin x :

2 2

’π π

(g) B4 = tan x : <x<

2 2

(h) C1 = [(’1, 0) ∪ (0, 1]

sin x

f (x) : ’π ¤ x ¤ π, f (x) =

(i) C2 = , x = 0; f (0) = 2

x

1 ’ cos x

g(x) : ’π ¤ x ¤ π, g(x) =

(j) C3 = , g(0) = 1

x

10. Suppose f is continuous on the set of all real numbers. Let the open

interval (c, d) be contained in the range of f . Let

A = {x : c < f (x) < d}.

94 CHAPTER 2. LIMITS AND CONTINUITY

Show that A is an open set.

(Hint: Let p ∈ A. Then f (p) ∈ (c, d). Choose > 0 such that c <

p ’ < p + < d. Since f is continuous at p, there is δ > 0 such that

|f (x)’f (p)| < whenever |x’p| < δ. This means that the open interval

(p ’ δ, p + δ) is contained in A. By de¬nition, A is open. This proves

that the inverse of a continuous function maps an open set onto an open

set.)

2.5 Limits and In¬nity

The convergence of a sequence {an }∞ depends on the limit of an as n tends

n=1

to ∞.

De¬nition 2.5.1 Suppose that a function f is de¬ned on an open interval

(a, b) and a < c < b. Then we de¬ne the following limits:

(i) lim f (x) = +∞

’

x’c

if and only if for every M > 0 there exists some δ > 0 such that f (x) > M

whenever c ’ δ < x < c.

(ii) lim f (x) = +∞

+

x’c

if and only if for every M > 0 there exists some δ > 0 such that f (x) > M

whenever c < x < c + δ.

(iii) lim f (x) = +∞

x’c

if and only if for every M > 0 there exists some δ > 0 such that f (x) > M

whenever 0 < |x ’ c| < δ.

(iv) lim f (x) = ’∞

x’c

if and only if for every M > 0 there exists some δ > 0 such that f (x) <

’M whenever 0 < |x ’ c| < δ.

(v) lim f (x) = ’∞

+

x’c

if and only if for every M > 0 there exists some δ > 0 such that f (x) <

’M whenever c < x < c + δ.

2.5. LIMITS AND INFINITY 95

(vi) lim f (x) = ’∞

’

x’c

if and only if for every M > 0 there exists some δ > 0 such that f (x) <

’M whenever c ’ δ < x < c.

De¬nition 2.5.2 Suppose that a function f is de¬ned for all real numbers.

(i) lim f (x) = L

x’+∞

if and only if for every > 0 there exists some M > 0 such that |f (x) ’

L| < whenever x > M .

(ii) lim f (x) = L

x’’∞

if and only if for every > 0 there exists some M > 0 such that |f (x) ’

L| < whenever x < ’M .

(iii) lim f (x) = ∞

x’+∞

if and only if for every M > 0 there exists some N > 0 such that

f (x) > M whenever x > N .

(iv) lim f (x) = ’∞

x’+∞

if and only if for every M > 0 there exists some N > 0 such that

f (x) < ’M whenever x > M .

(v) lim f (x) = ∞

x’’∞

if and only if for every M > 0 there exists some N > 0 such that

f (x) > M whenever x < ’N .

(vi) lim f (x) = ’∞

x’’∞

if and only if for every M > 0 there exists some N > 0 such that

f (x) < ’M whenever x < ’N .

De¬nition 2.5.3 The vertical line x = c is called a vertical asymptote to

the graph of f if and only if either

(i) lim f (x) = ∞ or ’∞; or

x’c

(ii) lim f (x) = ∞ or ’∞; or both.

’

x’c

96 CHAPTER 2. LIMITS AND CONTINUITY

De¬nition 2.5.4 The horizontal line y = L is a horizontal asymptote to the

graph of f if and only if

lim f (x) = L or lim f (x) = L, or both.

x’∞ x’’∞

Example 2.5.1 Compute the following limits:

sin x cos x

(i) lim (ii) lim

x x

x’∞ x’∞

x2 + 1 x3 ’ 2

(iii) lim (iv) lim

x’∞ 3x3 + 10 3x3 + 2x ’ 3

x’’∞

3x3 + 4x ’ 7 ’x4 + 3x ’ 10

(v) lim (vi) lim

2x2 + 5x + 2 2x2 + 3x ’ 5

x’’∞ x’’∞

(i) We observe that ’1 ¤ sin x ¤ 1 and hence

’1 sin x 1

¤ lim ¤ lim

0 = lim = 0.

x’∞ x x’∞ x x’∞ x

Hence, y = 0 is the horizontal asymptote and

sin x

lim = 0.

x

x’∞

(ii) ’1 ¤ cos x ¤ 1 and, by a similar argument as in part (i),

cos x

lim = 0.

x

x’∞

(iii) We divide the numerator and denominator by x2 and then take the limit

as follows:

1 + 1/x2

x2 + 1

= lim = 0.

lim

x’∞ 3x3 + 10 x’∞ 3x + 10/x2

2.5. LIMITS AND INFINITY 97

(iv) We divide the numerator and denominator by x3 and then take the limit

as follows:

x3 ’ 2 1 ’ 2/x3 1

lim = lim =.

3x3 + 2x ’ 3 x’’∞ 3 + 2/x2 ’ 3/x3 3

x’’∞

(v) We divide the numerator and denominator by x2 and then take the limit

as follows:

3x + 4/x ’ 7/x2

3x3 + 4x ’ 7

= ’∞.

= lim

lim

2x2 + 5x ’ 2 x’’∞ 2 + 5/x + 2/x2

x’’∞

(vi) We divide the numerator and denominator by x2 and then take the limit

as follows:

’x2 + 3/x ’ 10/x2

’x4 + 3x ’ 10

= ’∞.

= lim

lim

2x2 + 3x ’ 5 2 + 3/x ’ 5/x2

x’’∞

x’’∞

Example 2.5.2

(’1)n + 1

=0

(i) lim

n

n’∞

n2 n2 n3 + 4n2 ’ n3 ’ 3n2

’

(ii) lim = lim

n2 + 7n + 12

n+3 n+4

n’∞ n’∞

n2

= lim 2

n’∞ n + 7n + 12

1

= lim

1 + 7/n + 12/n2

n’∞

=1

√ √√ √

√ ( n + 4 ’ n)( n + 4 + n)

√

(iii) lim ( n + 4 ’ n) = lim √

( n + 4 + n)

n’∞ n’∞

4

= lim √ √

n’∞ ( n + 4 + n)

=0

98 CHAPTER 2. LIMITS AND CONTINUITY

n2 nπ

(vi) lim sin does not exist because it oscillates:

n2

1+± 2

n’∞

0 if n = 2m

nπ

1 if n = 2m + 1

sin =

2

’1 if n = 2m + 3

3n 1

(v) lim = lim =1

h’∞ 4 · e’n + 1

n’∞ 4 + 3n

(vi) lim {cos(nπ)} = lim (’1)n does not exist.

n’∞ n’∞

Exercises 2.5 Evaluate the following limits:

x x

1. lim 2. lim

x’2 x2 ’ 4 x2 ’ 4

+

x’2

x

3. lim 4. lim’ tan(x)

x2 ’ 1

’ π

x’1 x’ 2

5. lim+ sec x 6. lim cot x

+

π x’0

x’ 2

3x2 ’ 7x + 5

7. lim csc x 8. lim

x’∞ 4x2 + 5x ’ 7

’

x’0

x2 + 4 ’x4 + 2x ’ 1

9. lim 10. lim 2

x’’∞ 4x3 + 3x ’ 5 x’∞ x + 3x + 2

1 + (’1)n

cos(nπ)

11. lim 12. lim

n2 n3

x’∞ x’∞

1 ’ cos n

sin(n)

13. lim 14. lim

n n

x’∞ x’∞

nπ

cos nπ

2

15. lim 16. lim tan

n n

x’∞ x’∞

Chapter 3

Di¬erentiation

In De¬nition 2.2.2, we de¬ned the slope function of a function f at c by

f (x) ’ f (c)

slope(f (x), c) = lim

x’c

x’c

f (c + h) ’ f (c)

= lim .

h

h’0

The slope (f (x), c) is called the derivative of f at c and is denoted f (c).

Thus,

f (c + h) ’ f (c)

f (c) = lim .

h

h’0

Link to another ¬le.

3.1 The Derivative

De¬nition 3.1.1 Let f be de¬ned on a closed interval [a, b] and a < x < b.

Then the derivative of f at x, denoted f (x), is de¬ned by

f (x + h) ’ f (x)

f (x) = lim

h

h’0

whenever the limit exists. When f (x) exists, we say that f is di¬erentiable

at x. At the end points a and b, we de¬ne one-sided derivatives as follows:

f (x) ’ f (a) f (a + h) ’ f (a)

(i) f (a+ ) = lim+ = lim .

x’a h

h’0+

x’a

99

100 CHAPTER 3. DIFFERENTIATION

We call f (a+) the right-hand derivative of f at a.

f (x) ’ f (b) f (b + h) ’ f (b)

(ii) f (b’ ) = lim = lim .

x’b h

’ h’0’

x’b

We call f (b) the left-hand derivative of f at b.

Example 3.1.1 In Example 28 of Section 2.2, we proved that if f (x) = sin x,

then f (c) = slope (sin x, c) = cos c. Thus, f (x) = cos x if f (x) = sin x.

Example 3.1.2 In Example 29 of Section 2.2, we proved that if f (x) =

cos x, then f (c) = ’ sin c. Thus, f (x) = ’ sin x if f (x) = cos x.

Example 3.1.3 In Example 30 of Section 2.2, we proved that if f (x) = xn

for a natural number n, then f (c) = ncn’1 . Thus f (x) = nxn’1 , when

f (x) = xn , for any natural number n.

In order to ¬nd derivatives of functions obtained from the basic elemen-

tary functions using the operations of addition, subtraction, multiplication

and division, we state and prove the following theorem.

Theorem 3.1.1 If f is di¬erentiable at c, then f is continuous at c. The

converse is false.

Proof. Suppose that f is di¬erentiable at c. Then

f (x) ’ f (c)

= f (c)

lim

x’c

x’c

and f (c) is a real number. So,

f (x) ’ f (c)

(x ’ c) + f (c)

lim f (x) = lim

x’c

x’c x’c

f (x) ’ f (c)

· lim(x ’ c) + f (c)

= lim

x’c

x’0 x’c

= f (c) · 0 + f (c)

= f (c).

3.1. THE DERIVATIVE 101

Therefore, if f is di¬erentiable at c, then f is continuous at c.

To prove that the converse is false we consider the function f (x) = |x|.

This function is continuous at x = 0. But

|x + h| ’ |x|

f (x) = lim

h

h’0

(|x + h| ’ |x|)(|x + h| + |x|)

= lim

h(|x + h| + |x|)

h’0

x2 + 2xh + h2 ’ x2

= lim

h(|x + h| + |x|)

h’0

2x + h

= lim

h’0 |x + h| |x|

x

=

|x|

±

1 for x > 0

= ’1 for x < 0

unde¬ned for x = 0.

Thus, |x| is continuous at 0 but not di¬erentiable at 0. This completes the

proof of Theorem 3.1.1.

Theorem 3.1.2 Suppose that functions f and g are de¬ned on some open

interval (a, b) and f (x) and g (x) exist at each point x in (a, b). Then

(i) (f + g) (x) = f (x) + g (x) (The Sum Rule)

(ii) (f ’ g) (x) = f (x) ’ g (x) (The Di¬erence Rule)

(iii) (kf ) (x) = kf (x), for each constant k. (The Multiple Rule)

(iv) (f · g) (x) = f (x) · g(x) + f (x) · g (x) (The Product Rule)

g(x)f (x) ’ f (x)g (x)

f

(x) = , if g(x) = 0. (The Quotient Rule)

(v)

(g(x))2

g

Proof.

102 CHAPTER 3. DIFFERENTIATION

[f (x + h) + g(x + h)] ’ [f (x) + g(x)]

Part (i) (f + g) (x) = lim

h

h’0

f (x + h) ’ f (x) g(x + h) ’ g(x)

= lim + lim

h h

h’0 h’0

= f (x) + g (x).

[f (x + h) ’ g(x + h)] ’ [f (x) ’ g(x)]

Part (ii) (f ’ g) (x) = lim

h

h’0

f (x + h) ’ f (x) g(x + h) ’ g(x)

’ lim

= lim

h h

h’0 h’0

= f (x) ’ g (x).

kf (x + h) ’ kf (x)

Part (iii) (kf ) (x) = lim

h

h’0

f (x + h) ’ f (x)

= k · lim

h

h’0

= kf (x).

Part (iv)

f (x + h)g(x + h) ’ f (x)g(x)

(f · g) (x) = lim

h

h’0

1

[(f (x + h) ’ f (x))g(x + h) + f (x)(g(x + h) ’ g(x))]

= lim

h’0 h

f (x + h) ’ f (x) g(x + h) ’ g(x)

· lim g(x + h) + f (x) lim

= lim

h h

h’0 h’0 h’0

= f (x)g(x) + f (x)g (x).

3.1. THE DERIVATIVE 103

f 1 f (x + h) f (x)

’

Part (v) (x) = lim

g h g(x + h) g(x)

h’0

f (x + h) · g(x) ’ g(x + h)f (x)

1

= lim

h g(x + h)g(x)

h’0

(g(x + h) ’ g(x))

(f (x + h) ’ f (x))

1

g(x) ’ f (x)

= lim

(g(x))2 h’0 h h

1

· [f (x)g(x) ’ f (x)g (x)]

=

(g(x))2

g(x)f (x) ’ g(x)g (x)

= , if g(x) = 0.

(g(x))2

To emphasize the fact that the derivatives are taken with respect to the

independent variable x, we use the following notation, as is customary:

d

(f (x)).

f (x) =

dx

Based on Theorem 3.1.2 and the de¬nition of the derivative, we get the

following theorem.

Theorem 3.1.3

d(k)

(i) = 0, where k is a real constant.

dx

d

(xn ) = nxn’1 , for each real number x and natural number n.

(ii)

dx

d

(iii) (sin x) = cos x, for all real numbers (radian measure) x.

dx

d

(cos x) = ’ sin x, for all real numbers (radian measure) x.

(iv)

dx

d π

(tan x) = sec2 x, for all real numbers x = (2n + 1) , n = integer.

(v)

dx 2

104 CHAPTER 3. DIFFERENTIATION

d

(cot x) = ’ csc2 x, for all real numbers x = nπ, n = integer.

(vi)

dx

d π

(vii) (sec x) = sec x tan x, for all real numbers x = (2n + 1) , n = integer.

dx 2

d

(csc x) = ’ csc x cot x, for all real numbers x = nπ, n = integer.

(viii)

dx

Proof.

k’k

d(k)

Part(i) (k) = lim

dx h

h’0

0

= lim

h

h’0

= 0.

Part (ii) For each natural n, we get

(x + h)n ’ xn

dn

(x ) = lim (Binomial Expansion)

dx h

h’0

n(n ’ 1) n’2 2

1

xn + nxn’1 h + x h + · · · + hn ’ xn

= lim

h’0 h 2!

n(n ’ 1) n’2

= lim nxn’1 + x h + · · · + hn’1

2!

h’0

= nxn’1 .

Part (iii) By de¬nition, we get

sin(x + h) ’ sin x

d

(sin x) = lim

dx h

h’0

sin x cos h + cos x sin h ’ sin x

= lim

h

h’0

1 ’ cos h

sin h

’ sin x

= lim cos x

h h

h’0

= cos x · 1 ’ sin x · 0

= cos x

3.1. THE DERIVATIVE 105

since

1 ’ cos h

sin h

lim = 1, lim = 0. (Why?)

h h

h’0 h’0

Part (iv) By de¬nition, we get

cos(x + h) ’ cos x

d

(cos x) = lim

dx h

h’0

1

[cos x cos h ’ sin x sin h ’ cos x]

= lim

h’0 h

1 ’ cos h

sin h

’ cos x

= lim ’ sin x ·

h h

h’0

= ’ sin x · 1 ’ cos x · 0 (Why?)

= ’ sin x.

Part (v) Using the quotient rule and parts (iii) and (iv), we get

d d sin x

(tan x) =

dx dx cos x

cos x(sin x) ’ sin x(cos x)

=

(cos x)2

cos2 x + sin2 x

=

cos2 x

1

= (Why?)

cos2 x

π

= sec2 x, x = (2n + 1) , n = integer.

2

106 CHAPTER 3. DIFFERENTIATION

Part (vi) Using the quotient rule and Parts (iii) and (iv), we get

d d cos x

(cot x) =

dx dx sin x

(sin x)(cos x) ’ (cos x)(sin x)

=

(sin x)2

’ sin2 x ’ cos2 x

= (Why?)

(sin x)2

’1

= (why?)

(sin x)2

= ’ csc2 x, x = nπ, n = integer.

Part (vii) Using the quotient rule and Parts (iii) and (iv), we get

d 1

d

(sec x) =

dx dx cos x

(cos x) · 0 ’ 1 · (cos x)

=

(cos x)2

1 sin x

·

= (Why?)

cos x cos x

π

= sec x tan x, x = (2n + 1) , n = integer.

2

Part (viii) Using the quotient rule and Parts (iii) and (iv), we get

d d 1

(csc x) =

dx dx sin x

sin x · 0 ’ 1 · (sin x)

=

(sin x)2

’ cos x

1

·

= (Why?)

sin x sin x

= ’ csc x cot x, x = nπ, n = integer.

This concludes the proof of Theorem 3.1.3.

3.1. THE DERIVATIVE 107

Example 3.1.4 Compute the following derivatives:

d d

(4x3 ’ 3x2 + 2x + 10) (4 sin x ’ 3 cos x)

(i) (ii)

dx dx

x3 + 1

d d

(x sin x + x2 cos x)

(iii) (iv)

x2 + 4

dx dx

Part (i) Using the sum, di¬erence and constant multiple rules, we get

d d3 d d

(4x3 ’ 3x2 + 2x + 10) = 4 (x2 ) + 2

(x ) ’ 3 +0

dx dx dx dx

= 12x2 ’ 6x + 2.

d d d

(4 sin x ’ 3 cos x) = 4 (sin x) ’ 3

Part (ii) (cos x)

dx dx dx

= 4 cos x ’ 3(’ sin x)

= 4 cos x + 3 sin x.

Part (iii) Using the sum and product rules, we get

d d d

(x sin x + x2 cos x) = (x2 cos x) (Sum Rule)

(x sin x) +

dx dx dx

d d

= sin x + x (sin x)

dx dx

d d

(x2 ) cos x + x2

+ (cos x)

dx dx

= 1 · sin x + x cos x + 2x cos x + x2 (’ sin x)

= sin x + 3x cos x ’ x2 sin x.

108 CHAPTER 3. DIFFERENTIATION

Part (iv). Using the sum and quotient rules, we get

d d

(x2 + 4) dx (x3 + 1) ’ (x3 + 1) (x2 + 4)

x3 + 1

d dx

= (Why?)

x2 + 4 (x2 + 4)2

dx

(x2 + 4)(3x2 ) ’ (x3 + 1) = x

= (Why?)

(x2 + 4)2

3x4 + 12x2 ’ 2x3 ’ 2x

= (Why?)

(x2 + 4)2

3x4 ’ 2x3 + 12x2 ’ 2x

= .

(x2 + 4)2

Exercises 3.1

d

(x3 ) = 3x2 .

1. From the de¬nition, prove that

dx

’1

d 1

2. From the de¬nition, prove that = .

x2

dx x

Compute the following derivatives:

d

d

(x5 ’ 4x2 + 7x ’ 2) (4 sin x + 2 cos x ’ 3 tan x)

4.

3.

dx dx

x4 + 2

d 2x + 1 d

5. 6.

x2 + 1

dx dx 3x + 1

d d

(3x sin x + 4x2 cos x) (4 tan x ’ 3 sec x)

7. 8.

dx dx

d d

(x2 tan x + x cot x)

9. (3 cot x + 5 csc x) 10.

dx dx

Recall that the equation of the line tangent to the graph of f at (c, f (c)) has

slope f (c) and equations.

y ’ f (c) = f (c)(x ’ c)

Tangent Line:

The normal line has slope ’1/f (c), if f (c) = 0 and has the equation:

3.1. THE DERIVATIVE 109

’1

y ’ f (c) = (x ’ c).

Normal Line:

f (c)

In each of the following, ¬nd the equation of the tangent line and the equation

of the normal line for the graph of f at the given c.

11. f (x) = x3 + 4x ’ 12, c = 1 12. f (x) = sin x, c = π/6

13. f (x) = cos x, c = π/3 14. f (x) = tan x, c = π/4

15. f (x) = cot x, c = π/4 16. f (x) = sec x, c = π/3

17. f (x) = csc x, c = π/6 18. f (x) = 3 sin x + 4 cos x, c = 0.

Recall that Newton™s Method solves f (x) = 0 for x by using the ¬xed point

iteration algorithm:

f (xn )

xn+1 = g(xn ) = xn ’ , x0 = given,

f (xn )

with the stopping rule, for a given natural number n,

|xn+1 ’ xn | < 10’n .

In each of the following, set up Newton™s Iteration and perform 3 calculations

for a given x0 .

19. f (x) = 2x ’ cos x , x0 = 0.5

20. f (x) = x3 + 2x + 1 , x0 = ’0.5

21. f (x) = x3 + 3x2 ’ 1 = 0, x0 = 0.5

22. Suppose that f (c) exists. Compute each of the following limits in terms

of f (c)

f (c + h) ’ f (c)

f (x) ’ f (c)

(b) lim

(a) lim

x’c h

x’c h’0

f (c ’ h) ’ f (c) f (c) ’ f (t)

(c) lim (d) lim

t’c

h t’c

h’0

f (c + h) ’ f (c ’ h) f (c + 2h) ’ f (c ’ 2h)

(e) lim (f) lim

2h h

h’0 h’0

110 CHAPTER 3. DIFFERENTIATION

23. Suppose that g is di¬erentiable at c and

g(t)’g(c)

if t = c

t’c

f (t) =

g (c) if t = c.

Show that f is continuous at c.

Suppose that a business produces and markets x units of a commercial

item. Let

C(x) = The total cost of producing x-units.

p(x) = The sale price per item when x-units are on the market.

R(x) = xp(x) = The revenue for selling x-units.

P (x) = R(x) ’ C(x) = The gross pro¬t for selling x-items.

C (x) = The marginal cost.

R (x) = The marginal revenue.

P (x) = The marginal pro¬t.

In each of the problems 24“26, use the given functions C(x) and p(x) and

compute the revenue, pro¬t, marginal cost, marginal revenue and marginal

pro¬t.

24. C(x) = 100x ’ (0.2)x2 , 0 ¤ x ¤ 5000, p(x) = 10 ’ x

2 1

25. C(x) = 5000 + , 1 ¤ x ¤ 5000, p(x) = 20 +

x x

1

26. C(x) = 1000 + 4x ’ 0.1x2 , 1 ¤ x ¤ 2000, p(x) = 10 ’

x

In exercises 27“60, compute the derivative of the given function.

27. f (x) = 4x3 ’ 2x2 + 3x ’ 10 28. f (x) = 2 sin x ’ 3 cos x + 4

29. f (x) = 3 tan x ’ 4 sec x 30. f (x) = 2 cot x + 3 csc x

31. f (x) = 2x2 + 4x + 5 32. f (x) = x2/3 ’ 4x1/3 + 5

√

33. f (x) = 3x’4/3 + 3x’2/3 + 10 34. f (x) = 2 x + 4

3.2. THE CHAIN RULE 111

2 4 3 2

’ 2 + +1

35. f (x) = 36. f (x) =

x2 x3 x x

37. f (x) = x4 ’ 4x2 38. f (x) = (x2 + 2)(x2 + 1)

40. f (x) = (x3 + 1)(x3 ’ 1)

39. f (x) = (x + 2)(x ’ 4)

41. y = (x2 + 1) sin x 42. y = x2 cos x

43. y = (x2 + 1)(x10 ’ 5) 44. y = x2 tan x

45. y = (x1/2 + 4)(x1/3 ’ 5) 46. y = (2x + sin x)(x2 + 4)

47. y = x5 sin x 48. y = x4 (2 sin x ’ 3 cos x)

49. y = x2 cot x ’ 2x + 5 50. y = (x + sin x)(4 + csc x)

52. y = x2 (2 cot x ’ 3 csc x)

51. y = (sec x + tan x)(sin x + cos x)

x2 + 1 1 + sin x

53. y = 2 54. y =

x +4 1 + cos x

x1/2 + 1 sin x ’ cos x

55. y = 3/2 56. y =

3x + 2 sin x + cos x

t2 + 3t + 2 x 2 ex

57. y = 58. y =

t3 + 1 1 + ex

t2 sin t

3 + sin t cos t

60. y =

59. y =

4 + t2

4 + sec t tan t

3.2 The Chain Rule

Suppose we have two functions, u and y, related by the equations:

u = g(x) and y = f (u).

112 CHAPTER 3. DIFFERENTIATION

Then y = (f —¦ g)(x) = f (g(x)).

The chain rule deals with the derivative of the composition and may be

stated as the following theorem:

Theorem 3.2.1 (The Chain Rule). Suppose that g is de¬ned in an open

interval I containing c, and f is de¬ned in an open interval J containing

g(c), such that g(x) is in J for all x in I. If g is di¬erentiable at c, and f is

di¬erentiable at g(c), then the composition (f —¦ g) is di¬erentiable at c and

(f —¦ g) (c) = f (g(c)) · g (c).

In general, if u = g(x) and y = f (u), then

dy du