x
= −
,
dx
4y
which is defined whenever y = 0 (corresponding to the points (−2, 0) and (2, 0),
at which, as we saw above, the slope of the tangent lines is undefined). For
54
CHAPTER 1. DERIVATIVES
2
y
√
1
3
y = − √ (x − 1) +
1
2 3
2
0
−3
−2
−1
0
1
2
3
x
−1
2
2
x + 4y
= 1
−2
√
Figure 1.11.2: The ellipse x2 + 4y2 = 4 with tangent line at
1,
3
2
example, we have
dy
1
= − √ ,
dx
√
“
”
(x,y)= 1,
3
2 3
2
√
and so the equation of the line tangent to the ellipse at the point
1,
3
is
2
√
1
3
y = − √ (x − 1) +
.
2 3
2
See Figure 1.11.2.
Example 1.11.2. Consider the hyperbola H with equation
x2 − 4xy + y2 = 4.
Differentiating both sides of the equation, remembering to treat y as a function
of x, we have
dy
dy
2x − 4x
− 4y + 2y
= 0.
dx
dx
Solving for dy , we see that
dx
dy
4y − 2x
2y − x
=
=
.
dx
2y − 4x
y − 2x
For example,
dy
2
1
=
=
.
dx
4
2
(x,y)=(2,0)
1.11. IMPLICIT DIFFERENTIATION AND RATES OF CHANGE
55
5
y
3
1
y =
(x
2
− 2)
1
−10
−5
0
5
10
−1
x
2
2
x − 4xy + y = 4
−3
−5
Figure 1.11.3: The hyperbola x2 − 4xy + y2 = 4 with tangent line at (2, 0)
Hence the equation of the line tangent to H at (2, 0) is
1
y =
(x − 2).
2
See Figure 1.11.3.
Exercise 1.11.1.
Find dy if y2 + 8xy − x2 = 10.
dx
Exercise 1.11.2.
Find dy
if x2y + 3xy − 12y = 2.
dx (x,y)=(2,−1)
Exercise 1.11.3.
Find the equation of the line tangent to the circle with equa-
tion x2 + y2 = 25 at the point (3, 4).
Exercise 1.11.4.
Find the equation of the line tangent to the ellipse with
equation x2 + xy + y2 = 19 at the point (2, 3).
The technique described above, known as implicit differentiation, is also
useful in finding rates of change for variables related by an equation. The next
examples illustrate this idea, with the first being similar to examples we saw
earlier while discussing the chain rule.
Example 1.11.3. Suppose oil is being poured onto the surface of a calm body
of water. As the oil spreads out, it forms a right circular cylinder whose volume
is
V = πr2h,
where r and h are, respectively, the radius and height of the cylinder. Now
suppose the oil is being poured out at a rate of 10 cubic centimeters per second
56
CHAPTER 1. DERIVATIVES
A
z
x
P
y
B
Figure 1.11.4: Ships A and B passing a point P
and that the height remains a constant 0.25 centimeters. Then the volume of
the cylinder is increasing at a rate of 10 cubic centimeters per second, so
dV = 10 cm3/sec
dt
at any time t. Now with h = 0.25,
V = 0.25πr2,
so
dV
1
dr
=
πr
.
dt
2
dt
Hence
dr
2 dV
20
=
=
cm/sec.
dt
πr dt
πr
For example, if r = 10 centimeters at some time t = t0, then
dr
20
2
=
=
≈ 0.6366 cm/sec.
dt
10π
π
t=t0
Example 1.11.4. Suppose ship A, headed due north at 20 miles per hour, and
ship B, headed due east at 30 miles per hour, both pass through the same point
P in the ocean, ship A at noon and ship B two hours later (see Figure 1.11.4).
If we let x denote the distance from A to P t hours after noon, y denote the
distance from B to P t hours after noon, and z denote the distance from A to
B t hours after noon, then, by the Pythagorean theorem,
z2 = x2 + y2.
1.12. HIGHER-ORDER DERIVATIVES
57
Differentiating this equation with respect to t, we find
dz
dx
dy
2z
= 2x
+ 2y
,
dt
dt
dt
or
dz
dx
dy
z
= x
+ y
.
dt
dt
dt
For example, at 4 in the afternoon, that is, when t = 4, we know that
x = (4)(20) = 80 miles,
y = (2)(30) = 60 miles,
and
z =
802 + 602 = 100 miles,
so
dz
dx
dy
100
= 80
+ 60
miles/hour.
dt
dt
dt
Since at any time t,
dx = 20 miles/hour
dt
and
dy = 30 miles/hour,
dt
we have
dz
(80)(20) + (60)(30)
=
= 34 miles/hour.
dt
100
t=4
Exercise 1.11.5.
Suppose the volume of a cube is growing at a rate of 150
cubic centimeters per second. Find the rate at which the length of a side of the
cube is growing when each side of the cube is 10 centimeters.
Exercise 1.11.6.
A plane flies over a point P on the surface of the earth at
a height of 4 miles. Find the rate of change of the distance between P and the
plane one minute later if the plane is traveling at 300 miles per hour.
Exercise 1.11.7.
Suppose the length of a rectangle is growing at a rate of 2
centimeters per second and its width is growing at a rate of 4 centimeters per
second. Find the rate of change of the area of the rectangle when the length is
10 centimeters and the width is 12 centimeters.
1.12
Higher-order derivatives
Given two quantities, y and x, with y a function of x, we know that the derivative
dy is the rate of change of y with respect to x. Since dy is then itself a function
dx
dx
of x, we may ask for its rate of change with respect to x, which we call the
second-order derivative of y with respect to x and denote d2y .
dx2
58
CHAPTER 1. DERIVATIVES
Example 1.12.1. If y = 4x5 − 3x2 + 4, then
dy = 20x4 − 6x,
dx
and so
d2y = 80x3 − 6.
dx2
Of course, we could continue to differentiate: the third derivative of y with
respect to x is
d3y = 240x2,
dx3
the fourth derivative of y with respect to x is
d4y = 480x,
dx4
and so on.
If y is a function of x with y = f (x), then we may also denote the second
derivative of y with respect to x by f (x), the third derivative by f (x), and
so on. The prime notation becomes cumbersome after awhile, and so we may
replace the primes with the corresponding number in parentheses; that is, we
may write, for example, f
(x) as f (4)(x).
Example 1.12.2. If
1
f (x) =
,
x
then
1
f (x) = −
,
x2
2
f (x) =
,
x3
6
f (x) = −
,
x4
and
24
f (4)(x) =
.
x5
Exercise 1.12.1.
Find the first, second, and third-order derivatives of y =
sin(2x).
Exercise 1.12.2.
Find the first, second, and third-order derivatives of f (x) =
√4x + 1
1.12. HIGHER-ORDER DERIVATIVES
59
1.12.1
Acceleration
If x is the position, at time t, of an object moving along a straight line, then we
know that
dx
v =
(1.12.1)
dt
is the velocity of the object at time t. Since acceleration is the rate of change
of velocity, it follows that the acceleration of the object is
dv
d2x
a =
=
.
(1.12.2)
dt
dt2
Example 1.12.3. Suppose an object, such as a lead ball, is dropped from a
height of 100 meters. Ignoring air resistance, the height of the ball above the
earth after t seconds is given by
x(t) = 100 − 4.9t2 meters,
as we discussed in Section 1.2. Hence the velocity of the object after t seconds
is
v(t) = −9.8t meters/second
and the acceleration of the object is
a(t) = −9.8 meters/second2.
Thus the acceleration of an object in free-fall near the surface of the earth, ignor-
ing air resistance, is constant. Historically, Galileo started with this observation
about acceleration of objects in free-fall and worked in the other direction to
discover the formulas for velocity and position.
Exercise 1.12.3.
Suppose an object oscillating at the end of a spring has
position x = 10 cos(πt) (measured in centimeters from the equilibrium position)
at time t seconds. Find the acceleration of the object at time t = 1.25.
1.12.2
Concavity
The second derivative of a function f tells us the rate at which the slope of
the graph of f is changing. Geometrically, this translates into measuring the
concavity of the graph of the function.
Definition 1.12.1. We say the graph of a function f is concave upward on an
open interval (a, b) if f is an increasing function on (a, b). We say the graph of
a function f is concave downward on an open interval (a, b) if f is a decreasing
function on (a, b).
To determine the concavity of the graph of a function f , we need to determine
the intervals on which f is increasing and the intervals on which f is decreasing.
Hence, from our earlier work, we need identify when the derivative of f is
positive and when it is negative.
60
CHAPTER 1. DERIVATIVES
20
y
10
3
2
y = 2x
3
12
−
x −
x + 1
0
2
1
0
1
2
3
−
−
x
10
−
20
−
Figure 1.12.1: Graph of f (x) = 2x3 − 3x2 − 12x + 1
Theorem 1.12.1. If f is twice differentiable on (a, b), then the graph of f is
concave upward on (a, b) if f (x) > 0 for all x in (a, b), and concave downward
on (a, b) if f (x) < 0 for all x in (a, b).
Example 1.12.4. If f (x) = 2x3 − 3x2 − 12x + 1, then
f (x) = 6x2 − 6x − 12
and
f (x) = 12x − 6.
Hence f (x) < 0 when x < 1 and f (x) > 0 when x > 1 , and so the graph
2
2
of f is concave downward on the interval −∞, 1
and concave upward on the
2
interval
1 , ∞ . One may see the distinction between concave downward and
2
concave upward very clearly in the graph of f shown in Figure 1.12.1.
We call a point on the graph of a function f at which the concavity changes,
either from upward to downward or from downward to upward, a point of in-
flection. In the previous example,
1 , − 11 is a point of inflection.
2
2
Exercise 1.12.4.
Find the intervals on which the graph of f (x) = 5x3 − 3x5
is concave upward and the intervals on which the graph is concave downward.
What are the points of inflection?
1.12.3
The second-derivative test
Suppose c is a stationary point of f and f (c) > 0. Then, since f
is the
derivative of f and f (c) = 0, for any infinitesimal dx = 0,
f (c + dx) − f (c)
f (c + dx)
=
> 0.
(1.12.3)
dx
dx
1.12. HIGHER-ORDER DERIVATIVES
61
2
y
1
4
3
y = x − x
0
1
0
1
2
−
x
Figure 1.12.2: Graph of f (x) = x4 − x3
It follows that f (c + dx) > 0 when dx > 0 and f (c + dx) < 0 when dx < 0.
Hence f is decreasing to the left of c and increasing to the right of c, and so f
has a local minimum at c. Similarly, if f (c) < 0 at a stationary point c, then
f has a local maximum at c. This result is the second-derivative test .
Example 1.12.5. If f (x) = x4 − x3, then
f (x) = 4x3 − 3x2 = x2(4x − 3)
and
f (x) = 12x2 − 6x = 6x(2x − 1).
Hence f has stationary points x = 0 and x = 3 . Since
4
f (0) = 0
and
3
9
f
=
> 0,
4
4
we see that f has a local minimum at x = 3 . Although the second derivative
4
test tells us nothing about the nature of the critical point x = 0, we know, since
f has a local minimum at x = 3 , that f is decreasing on 0, 3
and increasing
4
4
on 3 , ∞ . Moreover, since 4x − 3 < 0 for all x < 0, it follows that f (x) < 0 for
4
all x < 0, and so f is also decreasing on (−∞, 0). Hence f has neither a local
maximum nor a local minimum at x = 0. Finally, since f (x) < 0 for 0 < x < 12
and f (x) > 0 for all other x, we see that the graph of f is concave downward
on the interval 0, 1 and concave upward on the intervals (−∞, 0) and 1 , ∞ .
2
2
See Figure 1.12.2.
62
CHAPTER 1. DERIVATIVES
Exercise 1.12.5.
Use the second-derivative test to find all local maximums
and minimums of
1
f (x) = x +
.
x
Exercise 1.12.6.
Find all local maximums and minimums of g(t) = 5t7 − 7t5.
Chapter 2
Integrals
2.1
Integrals
We now turn our attention to the other side of Zeno’s arrow paradox. In the
previous chapter we began with the problem of finding the velocity of an object
given a function which defined the position of the object at every instant of
time. We now suppose that we are given a function which specifies the velocity
v of an object, moving along a straight line, at every instant of time t, and we
wish to find the position x of the object at time t. There are two approaches
to finding x; we will investigate both, leading us to the fundamental theorem of
calculus.
First, from our earlier work we know that v is the derivative of x. That is,
dx = v.
(2.1.1)
dt
Hence to find x we need to find a function which has v for its derivative.
Definition 2.1.1. Given a function f defined on an open interval (a, b), we call
a function F an integral of f if F (x) = f (x) for all x in (a, b).
Example 2.1.1. If f (x) = 3x2, then F (x) = x3 is an integral of f on (−∞, ∞)
since F (x) = 3x2 for all x. However, note that F is not the only integral of f :
for other examples, both G(x) = x3 + 4 and H(x) = x3 + 15 are integrals of f as
well. Indeed, since the derivative of a constant is 0 the function L(x) = x3 + c
is an integral of f for any constant c.
In general, if F is an integral of f , then G(x) = F (x) + c is also an integral
of f for any constant c. Are there any other integrals of f ? That is, if we start
with both F and G being integrals of f , does it follow that G(x) − F (x) = c for
some constant c and for all x? To answer this question, first note that if we let
H(x) = G(x) − F (x), then
H (x) = G (x) − F (x) = f (x) − f (x) = 0
(2.1.2)
63
64
CHAPTER 2. INTEGRALS
for all x. Hence H is an integral of the constant function g(x) = 0 for all x. So
our question becomes: If H (x) = 0 for all x, does it follow that H(x) = c for
some constant c and all x? If it does, then
c = H(x) = F (x) − G(x),
(2.1.3)
and indeed F and G differ by only a constant. So suppose we are given H (x) = 0
for all x in an open interval (a, b). Then, for any two points u < v in (a, b), it
follows from the mean-value theorem that
H(v) − H(u) = H (d)(u − v)
(2.1.4)
for some d in (a, b). But then H (d) = 0, so H(v) − H(u) = 0, that is, H(u) =
H(v). Since this is true for any arbitrary points u and v in (a, b), it follows that
H must be constant on (a, b).
Theorem 2.1.1. If F (x) = G (x) for all x in (a, b), then there exists a constant
c such that G(x) = F (x) + c for all x in (a, b).
In particular, if F (x) = 0 for all x in (a, b), then F is constant on (a, b).
Example 2.1.2. Since
d
3 x2 + 4x = 3x + 4,
dx
2
any integral of f (x) = 3x + 4 must be of the form
3
F (x) =
x2 + 4x + c
2
for some constant c.
We denote an integral of a function f by
f (x)dx.
(2.1.5)
The motivation for this notation will be more evident once we discuss the fun-
damental theorem of calculus.
Example 2.1.3. Since
d (4x3 − sin(x)) = 12x2 − cos(x),
dx
it follows that
(12x2 − cos(x))dx = 4x3 − sin(x) + c,
where, as before, c is some constant.
2.1. INTEGRALS
65
From our rules for differentiation, it follows easily that
1
xndx =
xn+1 + c
(2.1.6)
n + 1
for every rational n = −1, and
sin(x)dx = − cos(x) + c,
(2.1.7)
cos(x)dx = sin(x) + c,
(2.1.8)
sec2(x)dx = tan(x) + c,
(2.1.9)
csc2(x)dx = − csc(x) + c,
(2.1.10)
sec(x) tan(x)dx = sec(x) + c,
(2.1.11)
and
csc(x) cot(x)dx = − csc(x) + c,
(2.1.12)
where in each case c represents an arbitrary constant. Note that differentiation
of the right-hand side of each of the above verifies these statements. Moreover,
if follows from our work with derivatives that for any functions f and g and any
constant k,
(f (x) + g(x))dx =
f (x)dx +
g(x)dx,
(2.1.13)
(f (x) − g(x))dx =
f (x)dx −
g(x)dx,
(2.1.14)
and
kf (x)dx = k
f (x)dx.
(2.1.15)
Example 2.1.4.
5
5x3 − 6x + 8 dx =
x4 − 3x2 + 8x + c.
4
Example 2.1.5.
(sin(x) − 4 cos(x))dx = − cos(x) − 4 sin(x) + c.
Example 2.1.6. Making an adjustment for the chain rule, we see that
1
sin(5x)dx = −
cos(5x) + c.
5
66
CHAPTER 2. INTEGRALS
y
40
30
20
3
y = 5 x
7x + 46
3
−
3
10
0
3
2
1
0
1
2
3
−
−
−
x
10
−
20
−
30
−
Figure 2.1.1: Parallel curves y = 5 x3 − 7x + c
3
Example 2.1.7. Suppose we wish to find the integral F (x) of f (x) = 5x2 − 7
for which F (1) = 10. Now
5
(5x2 − 7)dx =
x3 − 7x + c,
3
so
5
F (x) =
x3 − 7x + c
3
for some constant c. Now we want
5
10 = F (1) =
− 7 + c,
3
so we must have
5
46
c = 10 + 7 −
=
.
3