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1 2

2

Moreover, we then have

sin2( )

1 − cos2( )

1 − cos( )

1

=

=

(1 + cos(t))

(2) = 1.

(1.5.35)

2

2

2

2

Since

and sin( ) have the same sign, it follows that

sin( )

1.

(1.5.36)

For real numbers t, (1.5.34) and (1.5.36) say that, for small values of t, t2

cos(t) ≈ 1 −

(1.5.37)

2

and

sin(t) ≈ t.

(1.5.38)

Figures 1.5.3 and 1.5.4 graphically display the comparisons in (1.5.37) and

20

CHAPTER 1. DERIVATIVES

y

1

0.8

0.6

y = t

0.4

y = sin(t)

0.2

0

−1

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1

−0.2

t

−0.4

−0.6

−0.8

−1

Figure 1.5.4: Comparison of y = sin(t) with y = t

Example 1.5.5. For a numerical comparison, note that for t = 0.1, cos(t) =

0.9950042, compared to 1 − t2 = 0.995, and sin(t) = 0.0998334, compared to

2

t = 0.1.

Exercise 1.5.3.

Verify that the triangle with vertices at A, B, and D in Figure

1.5.2 is an isosceles triangle with base angles of t at A and B.

2

Exercise 1.5.4.

Verify the half-angle formula,

1

cos(θ) =

(1 + cos(2θ)),

2

for any angle θ, using the identities cos(2θ) = cos2(θ) − sin2(θ) (a consequence

of the addition formula) and sin2(θ) + cos2(θ) = 1.

1.5.3

Compositions

Given functions f and g, we call the function

f ◦ g(x) = f (g(x))

(1.5.39)

the composition of f with g. If g is continuous at a real number c, f is continuous

at g(c), and

is an infinitesimal, then

f ◦ g(c + ) = f (g(c + ))

f (g(c))

(1.5.40)

since g(c + )

g(c).

1.5. PROPERTIES OF CONTINUOUS FUNCTIONS

21

Theorem 1.5.9. If g is continuous at c and f is continuous at g(c), then f ◦ g

is continuous at c.

Example 1.5.6. Since f (t) = sin(t) is continuous for all t and

3t2 + 1

g(t) = 4t − 8

is continuous at all real numbers except t = 2, it follows that

3t2 + 1

h(t) = sin

4t − 8

is continuous on the intervals (−∞, 2) and (2, ∞).

√

Note that if f (x) =

x and

is an infinitesimal, then, for any x = 0,

√

√

f (x + ) − f (x) =

x + −

x

√

√

√

√

x + +

x

= ( x + −

x)

√

√

x + +

x

x + − x

= √

√

x + +

x

= √

√ ,

x + +

x

which is infinitesimal. Hence f is continuous on (0, ∞). Moreover, if

is a

√

√

positive infinitesimal, then

must be an infinitesimal (since if a =

is not

an infinitesimal, then a2 =

is not an infinitesimal). Hence

√

f (0 + ) =

0 = f (0).

√

Thus f is continuous at 0, and so f (x) =

x is continuous on [0, ∞).

√

Theorem 1.5.10. The function f (x) =

x is continuous on [0, ∞).

√

Example 1.5.7. It now follows that f (x) =

4x − 2 is continuous everywhere

it is defined, namely, on [2, ∞).

Exercise 1.5.5.

Find the interval or intervals on which f (x) = sin 1

is

x

continuous.

Exercise 1.5.6.

Find the interval or intervals on which

1 + t2

g(t) =

1 − t2

is continuous.

22

CHAPTER 1. DERIVATIVES

1.5.4

Consequences of continuity

Continuous functions have two important properties that will play key roles

in our discussions in the rest of the text: the extreme-value property and the

intermediate-value property. Both of these properties rely on technical aspects

of the real numbers which lie beyond the scope of this text, and so we will not

attempt justifications.

The extreme-value property states that a continuous function on a closed

interval [a, b] attains both a maximum and minimum value.

Theorem 1.5.11. If f is continuous on a closed interval [a, b], then there exists

a real number c in [a, b] for which f (c) ≤ f (x) for all x in [a, b] and a real number

d in [a, b] for which f (d) ≥ f (x) for all x in [a, b].

The following examples show the necessity of the two conditions of the the-

orem (that is, the function must be continuous and the interval must be closed

in order to ensure the conclusion).

Example 1.5.8. The function f (x) = x2 attains neither a maximum nor a

minimum value on the interval (0, 1). Indeed, given any point a in (0, 1), f (x) >

f (a) whenever a < x < 1 and f (x) < f (a) whenever 0 < x < a. Of course, this does not contradict the theorem because (0, 1) is not a closed interval. On the

closed interval [0, 1], we have f (1) ≥ f (x) for all x in [0, 1] and f (0) ≤ f (x) for

all x in [0, 1], in agreement with the theorem.

In this example the extreme values of f occurred at the endpoints of the

interval [−1, 1]. This need not be the case. For example, if g(t) = sin(t), then,

on the interval [0, 2π], g has a minimum value of −1 at t = 3π and a maximum

2

value of 1 at t = π .

2

Example 1.5.9. Let

1

,

if − 1 ≤ x < 0 or 0 < x ≤ 1,

f (x) =

x

0,

if x = 0.

See Figure 1.5.5. Then f does not have a maximum value: if a ≤ 0, then f (x) > f (a) for any x > 0, and if a > 0, then f (x) > f (a) whenever 0 < x < a.

Similarly, f has no minimum value: if a ≥ 0, then f (x) < f (a) for any x < 0,

and if a < 0, then f (x) < f (a) whenever a < x < 0. The problem this time is

that f is not continuous at x = 0. Indeed, if

is an infinitesimal, then f ( ) is

infinite, and, hence, not infinitesimally close to f (0) = 0.

Exercise 1.5.7.

Find an example of a continuous function which has both a

minimum value and a maximum value on the open interval (0, 1).

Exercise 1.5.8.

Find an example of function which has a minimum value and

a maximum value on the interval [0, 1], but is not continuous on [0, 1].

1.6. THE DERIVATIVE

23

y

10

8

6

y = f (x)

4

2

0

−1

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1

−2

x

−4

−6

−8

−10

Figure 1.5.5: A function with no minimum or maximum value on [−1, 1]

The intermediate-value property states that a continuous function attains

all values between any two given values of the function.

Theorem 1.5.12. If f is continuous on the interval [a, b] and m is any value

betwen f (a) and f (b), then there exists a real number c in [a, b] for which

f (c) = m.

The next example shows that a function which is not continuous need not

satisfy the intermediate-value property.

Example 1.5.10. If H is the Heaviside function, then H(−1) = 0 and H(1) =

1, but there does not exist any real number c in [−1, 1] for which H(c) = 1 ,

2

even though 0 < 1 < 1.

2

1.6

The derivative

We now return to the problem of rates of change. Given y = f (x), for any

infinitesimal dx we let

dy = f (x + dx) − f (x).

(1.6.1)

If y is a continuous function of x, then dy is infinitesimal and, if dx = 0, the

ratio

dy

f (x + dx) − f (x)

=

(1.6.2)

dx

dx

is a hyperreal number. If dy is finite, then its shadow, if it is the same for all

dx

values of dx, is the rate of change of y with respect to x, which we will call the

derivative of y with respect to x.

24

CHAPTER 1. DERIVATIVES

Definition 1.6.1. Given y = f (x), suppose

dy

f (x + dx) − f (x)

=

(1.6.3)

dx

dx

is finite and has the same shadow for all nonzero infinitesimals dx. Then we call

dy

sh

(1.6.4)

dx

the derivative of y with respect to x.

Note that the quotient in (1.6.3) will be infinite if f (x + dx) − f (x) is not an infinitesimal. Hence a function which is not continuous at x cannot have a

derivative at x.

There are numerous ways to denote the derivative of a function y = f (x).

One is to use dy to denote, depending on the context, both the ratio of the

dx

infinitesimals dy and dx and the shadow of this ratio, which is the derivative.

Another is to write f (x) for the derivative of the function f . We will use both

of these notations extensively.

Example 1.6.1. If y = x2, then, for any nonzero infinitesimal dx,

dy = (x + dx)2 − x2 = (x2 + 2xdx + (dx)2) − x2 = (2x + dx)dx.

Hence

dy = 2x + dx 2x,

dx

and so the derivative of y with respect to x is

dy = 2x.

dx

Example 1.6.2. If f (x) = 4x, then, for any nonzero infinitesimal dx,

f (x + dx) − f (x)

4(x + dx) − 4x

4dx

=

=

= 4.

dx

dx

dx

Hence f (x) = 4. Note that this implies that f (x) has a constant rate of change:

every change of one unit in x results in a change of 4 units in f (x).

Exercise 1.6.1.

Find dy if y = 5x − 2.

dx

Exercise 1.6.2.

Find dy if y = x3.

dx

Exercise 1.6.3.

Find f (x) if f (x) = 4x2.

To denote the rate of change of y with respect to x at a particular value of

x, say, when x = a, we write

dy

.

(1.6.5)

dx x=a

If y = f (x), then, of course, this is the same as writing f (a).

1.6. THE DERIVATIVE

25

y

1

2

y = x 3

0

2

1

0

1

2

−

−

x

2

Figure 1.6.1: y = x 3 is continuous, but not differentiable, at x = 0

Example 1.6.3. If y = x2, then we saw above that dy = 2x. Hence the rate

dx

of change of y with respect to x when x = 3 is

dy

= (2)(3) = 6.

dx x=3

2

Example 1.6.4. If f (x) = x 3 , then for any infinitesimal dx,

2

f (0 + dx) − f (0) = f (dx) = (dx) 3 ,

which is infinitesimal. Hence f is continuous at x = 0. Now if dx = 0, then

2

f (0 + dx) − f (0)

(dx) 3

1

=

=

.

dx

dx

1

(dx) 3

Since this is infinite, f does not have a derivative at x = 0. In particular, this

shows that a function may be continuous at a point, but not differentiable at

that point. See Figure 1.6.1.

√

Example 1.6.5. If f (x) =

x, then, as we have seen above, for any x > 0 and

any nonzero infinitesimal dx,

√

√

f (x + dx) − f (x) =

x + dx −

x

√

√

√

√

x + dx +

x

= ( x + dx −

x) √

√

x + dx +

x

(x + dx) − x

= √

√

x + dx +

x

dx

= √

√ .

x + dx +

x

26

CHAPTER 1. DERIVATIVES

It now follows that

f (x + dx) − f (x)

1

1

1

= √

√

√

√

=

√ .

dx

x + dx +

x

x +

x

2 x

Thus

1

f (x) =

√ .

2 x

For example, the rate of change of y with respect to x when x = 9 is

1

1

f (9) =

√ =

.

2 9

6

We will sometimes also write

d f(x)

(1.6.6)

dx

for f (x). With this notation, we could write the result of the previous example

as

d √

1

x =

√ .

dx

2 x

Definition 1.6.2. Given a function f , if f (a) exists we say f is differentiable

at a. We say f is differentiable on an open interval (a, b) if f is differentiable at

each point x in (a, b).

Example 1.6.6. The function y = x2 is differentiable on (−∞, ∞).

√

Example 1.6.7. The function f (x) =

x is differentiable on (0, ∞). Note that

f is not differentiable at x = 0 since f (0 + dx) = f (dx) is not defined for all

infinitesimals dx.

2

Example 1.6.8. The function f (x) = x 3 is not differentiable at x = 0.

1.7

Properties of derivatives

We will now develop some properties of derivatives with the aim of facilitating

their calculation for certain general classes of functions.

To begin, if f (x) = k for all x and some real constant k, then, for any

infinitesimal dx,

f (x + dx) − f (x) = k − k = 0.

(1.7.1)

Hence, if dx = 0,

f (x + dx) − f (x) = 0,

(1.7.2)

dx

and so f (x) = 0. In other words, the derivative of a constant is 0.

Theorem 1.7.1. For any real constant k,

d k = 0.

(1.7.3)

dx

d

Example 1.7.1.

4 = 0.

dx

1.7. PROPERTIES OF DERIVATIVES

27

1.7.1

Sums and differences

Now suppose u and v are both differentiable functions of x. Then, for any

infinitesimal dx,

d(u + v) = (u(x + dx) + v(x + dx)) − (u(x) − v(x))

= (u(x + dx) − u(x)) + (v(x + dx) − v(x))

= du + dv.

(1.7.4)

Hence, if dx = 0,

d(u + v)

du

dv

=

+

.

(1.7.5)

dx

dx

dx

In other words, the derivative of a sum is the sum of the derivatives.

Theorem 1.7.2. If f and g are both differentiable and s(x) = f (x) + g(x),

then

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

(1.7.6)

√

Example 1.7.2. If y = x2 +

x, then, using our results from the previous

section,

dy

d

d √

1

=

(x2) +

( x) = 2x + √ .

dx

dx

dx

2 x

A similar argument shows that

d

du

dv

(u − v) =

−

.

(1.7.7)

dx

dx

dx

Exercise 1.7.1.

Find the derivative of y = x2 + 5.

√

Exercise 1.7.2.

Find the derivative of f (x) =

x − x2 + 3.

1.7.2

Constant multiples

If c is any real constant and u is a differentiable function of x, then, for any

infinitesimal dx,

d(cu) = cu(x + dx) − cu(x) = c(u(x + dx) − u(x)) = cdu.

(1.7.8)

Hence, if dx = 0,

d(cu)

du

= c

.

(1.7.9)

dx

dx

In other words, the derivative of a constant times a function is the constant

times the derivative of the function.

Theorem 1.7.3. If c is a real constant, f is differentiable, and g(x) = cf (x),

then

g (x) = cf (x).

(1.7.10)

28

CHAPTER 1. DERIVATIVES

Example 1.7.3. If y = 5x2, then

dy

d

= 5

(x2) = 5(2x) = 10x.

dx

dx

Exercise 1.7.3.

Find the derivative of y = 8x2.

√

Exercise 1.7.4.

Find the derivative of f (x) = 4 x + 15.

1.7.3

Products

Again suppose u and v are differentiable functions of x. Note that, in partic-

ular, u and v are continuous, and so both du and dv are infinitesimal for any

infinitesimal dx. Moreover, note that

u(x + dx) = u(x) + du and v(x + dx) = v(x) + dv.

(1.7.11)

Hence

d(uv) = u(x + dx)v(x + dx) − u(x)v(x)

= (u(x) + du)(v(x) + dv) − u(x)v(x)

= (u(x)v(x) + u(x)dv + v(x)du + dudv) − u(x)v(x)

= udv + vdu + dudv,

(1.7.12)

and so, if dx = 0,

d(uv)

dv

du

dv

dv

du

= u

+ v

+ du

u

+ v

(1.7.13)

dx

dx

dx

dx

dx

dx

Thus we have, for any differentiable functions u and v,

d

dv

du

(uv) = u

+ v

,

(1.7.14)

dx

dx

dx

which we call the product rule.

Theorem 1.7.4. If f and g are both differentiable and p(x) = f (x)g(x), then

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

(1.7.15)

Example 1.7.4. We may use the product rule to find a formula for the deriva-

tive of a positive integer power of x. We first note that if y = x, then, for any

infinitesimal dx,

dy = (x + dx) − x = dx,

(1.7.16)

and so, if dx = 0,

dy

dx

=

= 1.

(1.7.17)

dx

dx

Thus we have

d x = 1,

(1.7.18)

dx

1.7. PROPERTIES OF DERIVATIVES

29

as we should expect, since y = x implies that y changes at exactly the same

rate as x.

Using the product rule, it now follows that

d

d

d

d

x2 =

(x · x) = x

x + x

x = x + x = 2x,

(1.7.19)

dx

dx

dx

dx

in agreement with a previous example. Next, we have

d

d

d

x3 = x

x2 + x2

x = 2x2 + x2 = 3x2

(1.7.20)

dx

dx

dx

and

d

d

d

x4 = x

x3 + x3

x = 3x3 + x3 = 4x3.

(1.7.21)

dx

dx

dx

At this point we might suspect that for any integer n ≥ 1,

d xn = nxn−1.

(1.7.22)

dx

This is in fact true, and follows easily from an inductive argument: Suppose we

have shown that for any k < n,

d xk = kxk−1.

(1.7.23)

dx

Then

d

d

d

xn = x

xn−1 + xn−1

x

dx

dx

dx

= x((n − 1)xn−2) + xn−1

= nxn−1.

(1.7.24)

We call this result the power rule.

Theorem 1.7.5. For any integer n ≥ 1,

d xn = nxn−1.

(1.7.25)

dx

We shall see eventually, in Theorems 1.7.7, 1.7.10, and 2.7.2, that the power rule in fact holds for any real number n = 0.

Example 1.7.5. When n = 34, the power rule shows that

d x34 = 34x33.

dx

Example 1.7.6. If f (x) = 14x5, then, combining the power rule with our result

for constant multiples,

f (x) = 14(5x4) = 70x4.

30

CHAPTER 1. DERIVATIVES

Exercise 1.7.5.

Find the derivative of y = 13x5.

Example 1.7.7. Combining the power rule with our results for constant mul-

tiples and differences, we have

d (3x2 − 5x) = 6x − 5.

dx

Exercise 1.7.6.

Find the derivative of f (x) = 5x4 − 3x2.

Exercise 1.7.7.

Find the derivative of y = 3x7 − 3x + 1.

1.7.4

Polynomials

As the previous examples illustrate, we may put together the above results