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