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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

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’∞


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

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