Chapter 2
The Derivative
2.1
The Derivative and Slope of the Tangent Line
• The Slope of the Tangent Line • The Derivative of a Function • Cases When
the Derivative is Undefined • Differentiable Functions are Continuous Functions
The Slope of the Tangent Line
Mathematics advances whenever a solution to a long standing mathematical problem is
found. Such solutions are groundbreaking that they bring forth new ideas and techniques.
In the 18th century, Newton and Leibniz invented the theory of differential calculus in
solving the tangent line problem. In this section we describe the slope of a tangent line.
An example of a tangent line to the graph of a function f at a point P (c, f (c)) is shown
in Figure 1. In order to define the slope of a tangent line, we consider the slopes of secant
lines.
The moving power of
mathematical invention
is not reasoning but
imagination.
- A. De Morgan
y
f �c�
y
P
c
x
f �c��x�
Figure 1
Tangent line at point P (c, f (c)).
Let Q(c + ∆x, f (c + ∆x)) be another point on the graph of y = f (x) where ∆x �= 0,
see Figure 2. Since the slope of the line through the points (x1 , y1 ) and (x2 , y2 ) is given
by
y 2 − y1
x2 − x1
71
f �c�
Q
P
c
c��x
Figure 2
The secant line
through P (c, f (c)) and
Q(c + ∆x, f (c + ∆x)).
x
72
CHAPTER 2. THE DERIVATIVE
we obtain by substitution that the slope of the secant line joining P (c, f (c)) and Q(c +
∆x, f (c + ∆x)) is
msecant line
=
=
f (c + ∆x) − f (c)
(c + ∆x) − c
f (c + ∆x) − f (c)
.
∆x
Note, when Q(c + ∆x, f (c + ∆x)) is very near P (c, f (c)), the position of the secant
line is almost in the same position as the tangent line. In other words, when ∆x is close
to but not equal 0, the slope of the secant line joining P (c, f (c)) and Q(c + ∆x, f (c + ∆x))
is close to the slope of the tangent line at P (c, f (c)). This idea motivates the following
definition.
Definition 1 The Slope of the Tangent Line
The slope of the tangent line to the graph of y = f (x) at the point (c, f (c)) is
lim
∆x→0
f (c + ∆x) − f (c)
∆x
provided the limit exists.
The slope of the tangent line at (c, f (c)) is also called the slope of the graph of
y = f (x) at (c, f (c)).
Example 1 The Slope of the Graph of a Linear Function
Find the slope of the graph of f (x) = 4x − 2 at the point (3, 10).
Solution Using Definition 1 with c = 3, the slope of the graph at (3, 10) is
lim
∆x→0
f (3 + ∆x) − f (3)
∆x
=
=
=
lim
∆x→0
(4(3 + ∆x) − 2) − (4(3) − 2)
∆x
12 + 4∆x − 2 − 10
∆x
4∆x
lim
= 4.
∆x→0 ∆x
lim
∆x→0
Then the slope of the graph at (3, 10) is 4.
✷
Try This 1
Find the slope of the graph of g(x) = −3x + 12 at the point (2, 6).
Note, the slope of the graph of a linear function is the same at every point in the line.
For nonlinear functions, this is not the case, i.e., the slope of the tangent line may vary
from one point to another point. This is shown in the next example.
2.1. THE DERIVATIVE AND SLOPE OF THE TANGENT LINE
73
Example 2 The Slopes of the Graph of a Nonlinear Function
Find the slopes of the tangent lines to the graph of f (x) = x2 − 4 at the points (3, 5) and
(−1, −3).
y
Solution The slope of the graph at an arbitrary point (c, f (c)) is given by
lim
∆x→0
f (c + ∆x) − f (c)
∆x
=
=
=
=
=
lim
∆x→0
f (c + ∆x) − f (c)
∆x
=
lim
((c + ∆x)2 − 4) − (c2 − 4)
∆x
lim
c2 + 2c∆x + (∆x)2 − 4 − c2 + 4
∆x
∆x→0
∆x→0
�1
2c∆x + (∆x)2
lim
∆x→0
∆x
∆x(2c + ∆x)
lim
∆x→0
∆x
Figure 3
lim (2c + ∆x)
∆x→0
The slope of the graph of
f (x) = x2 − 4 at the point
(c, f (c)) is m = 2c.
2c
Then the slope of the tangent line at (c, f (c)) is m = 2c. Thus, at the point (3, 5) where
c = 3 the slope is m = 6. At (−1, −3), where c = −1 the slope is m = 2(−1) = −2. The
tangent lines are shown in the Figure 3.
✷
Try This 2
Find the slopes of the tangent lines to the graph of g(x) = 3x2 at the points (1, 3) and
(−2, 12).
Example 3 The Slope of a Tangent Line to a Rational Function
Find the slope of the tangent line to the graph of the given function at the indicated point.
�
�
2
2
g(x) = , 3,
.
x
3
Solution Using Definition 1, the slope m of the tangent line when x = 3 is
m
=
=
=
m
=
lim
∆x→0
g(3 + ∆x) − g(3)
∆x
2
2
−
3
+
∆x
3 =
lim
∆x→0
∆x
lim
∆x→0
−2∆x
=
3∆x(3 + ∆x)
2
− .
9
2
Hence, the slope of the tangent line is m = − .
9
3
lim
∆x→0
lim
6 − 2(3 + ∆x)
3∆x(3 + ∆x)
∆x→0
−2
3(3 + ∆x)
Substitute 0 into ∆x
✷
x
74
CHAPTER 2. THE DERIVATIVE
Try This 3
Find the slope of the tangent line to the graph of the given function at the indicated point.
�
�
1
1
1
a) g(x) = ,
,3
b) h(x) =
, (0, 1)
x
3
x+1
In the previous three examples, all the tangent lines had slopes. However, it is possible
that a tangent line may not have a slope1 . Such is the case when we have a vertical
tangent line.
Definition 2 Definition of a Vertical Tangent Line
If f (x) is continuous at x = c and either
lim
∆x→0
f (c + ∆x) − f (c)
f (c + ∆x) − f (c)
= ∞ or lim
= −∞
∆x→0
∆x
∆x
we say the graph of y = f (x) has a vertical tangent line at the point (c, f (c)).
In Figure 4, we see a vertical tangent line at the point (c, f (c)). Note, for ∆x �= 0, the
slope of the secant line joining the points (c, f (c)) and Q(c + ∆x, f (c + ∆x)) is positive.
As ∆x approaches 0, the position of the secant line is moving towards the vertical position
. Thus, the slope of the secant line is increasing without bound as ∆x approaches 0, i.e.,
lim
∆x→0
f (c + ∆x) − f (c)
= ∞.
∆x
y
f �c�
Q
P
c
Figure 4
1 It
A vertical tangent line at (c, f (c)).
is also possible that the tangent line may not exist at all.
x
2.1. THE DERIVATIVE AND SLOPE OF THE TANGENT LINE
75
y
Example 4 Verifying the Existence of a Vertical Tangent Line
3
Prove x = 1 is a vertical tangent line of the graph of
√
f (x) = 1 − 3 x − 1.
1
Solution
We apply Definition 2 with c = 1:
�
� �
�
√
√
1 − 3 1 + ∆x − 1 − 1 − 1 − 1
f (1 + ∆x) − f (1)
lim
=
lim
∆x→0
∆x→0
∆x
∆x
√
− 3 ∆x
−1
=
lim
= lim �
∆x→0
∆x→0 3 (∆x)2
∆x
=
Figure 5
The graph of
f (x) = 1 −
Note, f is continuous at x = 1 by Theorem 1.2, page 24. Thus, by Definition 2, x = 1 is
vertical tangent line.
Finally, in Figure 5 we see that the slope of a secant line joining P (1, 1) and Q(1 +
∆x, f (1 + ∆x)) is negative. This explains why the above limit is −∞, and not ∞.
✷
Try This 4
√
Show that the graph of f (x) = 3 x has a vertical tangent line at the point (0, 0) using
Definition 2.
The Derivative of a Function
We begin an important phase in studying of calculus. The limit that we used to define
the slope of a tangent line will now be used to define the derivative of a function. The
process of finding the derivative of a function is called differentiation.
Definition 3 The Derivative of a Function
The derivative of a function f is a function f � whose value at x is given by
∆x→0
provided the limit exists.
f (x + ∆x) − f (x)
∆x
x
1
−∞
f � (x) = lim
P
Q
(1)
√
3
x−1
has a vertical tangent
line at P (1, 1).
76
CHAPTER 2. THE DERIVATIVE
The domain of the derivative f � consists all x’s for which the limit in (1) exists. Recall,
f � (x) is the slope of the tangent line to the graph of f at the point (x, f (x)).
We read f � (x) as ‘f prime of x’. We say f is differentiable at x if f � (x) exists, and f
is differentiable on an open interval (a, b) if f � (x) exists for every x in (a, b). Other
notations for the derivative of y = f (x) are
dy
, read as “the derivative of y with respect to x”, or “dy dx”
dx
d
2.
[f (x)], read as “ddx of f (x)”, and
dx
dy
�y
3.
= lim
where �y = f (x + �x) − f (x).
�x→0 �x
dx
1.
Example 5 Using the Definition to find the Derivative
Find the derivative of f (x) = x3 using the limit process in Definition 3.
Problem-Solving Tips
In the limit process, simplify
the difference quotient ∆y/∆x
Solution
f � (x)
so that ∆x does not appear
in the denominator. This is
usually done by factoring when
polynomials are involved.
=
=
=
=
=
=
=
�
f (x)
=
lim
f (x + ∆x) − f (x)
∆x
lim
(x + ∆x)3 − x3
∆x
lim
x3 + 3x2 ∆x + 3x∆x2 + ∆x3 − x3
∆x
lim
3x2 ∆x + 3x∆x2 + ∆x3
∆x
lim
∆x(3x2 + 3x∆x + ∆x2 )
∆x
∆x→0
∆x→0
∆x→0
∆x→0
∆x→0
lim (3x2 + 3x∆x + ∆x2 )
∆x→0
3x2 + 3x(0) + (0)2
3x
Substitute ∆x = 0
2
Then the derivative is f � (x) = 3x2 .
✷
Try This 5
Use the limit process in Definition 3 to find the derivative of each function.
a) p(x) = 4x
b) r(x) = x2
2.1. THE DERIVATIVE AND SLOPE OF THE TANGENT LINE
77
Example 6 Derivatives and Tangent Lines
√
Given y = x, find the following:
dy
a)
dx
√
b) the slope of the graph of y = x at the point (4, 2), and
c) an equation of the tangent line at (4, 2).
Problem-Solving Tips
Solution
a) Applying Definition 3, we find
√
√
dy
x + �x − x
=
lim
�x→0
dx
�x
=
lim
�x→0
=
lim
�x→0
x + �x −
�x
�x
lim √
=
dy
dx
√
�x→0
√
x
To simplify a difference
quotient ∆y/∆x involving radicals, try rationalizing.
√
√
x + �x + x
·√
√
x + �x + x
(x + �x) − x
�√
√ �
x + �x + x
1
√
x + �x + x
y
2
Simplify
1
√ .
2 x
=
Rationalize
Substitute �x = 0
4
b) The slope m at (4, 2) is obtained by substituting x = 4 into the derivative.
Figure 6.
1
1
Slope = m = √ =
4
2 4
√
Since dy/dx = 1/(2 x),
the slope at (4, 2) is
√
m = 1/(2 4) = 1/4.
c) Using the slope-intercept equation of the tangent line, we find
y
=
mx + b
Slope-intercept equation
2
=
1
(4) + b
4
x = 4, y = 2, m =
1
=
b.
1
4
Thus, the tangent line is given by
y=
1
x + 1.
4
✷
Try This√6
If f (x) = 2x, find f � (x) and the slope-intercept equation of the tangent line to the graph
of f at (8, 4).
x
78
CHAPTER 2. THE DERIVATIVE
An alternative definition of the derivative is given by
f � (c) = lim
x→c
f (x) − f (c)
x−c
(2)
provided the limit exists. The verification that the alternate definition (2) agrees with
Definition 3 is left as an exercise at the end of the section. The one-sided limits
lim
x→c+
f (x) − f (c)
f (x) − f (c)
and lim
x−c
x−c
x→c−
are called the derivative from the right and derivative from the left at c, respectively, if they exist. Moreover, we say f (x) is differerentiable on a closed interval
[a, b] if f (x) is differentiable on (a, b), f (x) has a derivative from the right at a, and f (x)
has a derivative from the left at b.
Problem-Solving Tips
To simplify an improper
fraction, multiply and
divide by the least
common denominator.
Example 7 The Derivative of a Rational Function
Find the derivative f � (x) of f (x) =
y
Solution
Applying the alternative definition (2), we find
�
f (c)
1
x
1
1
using the alternative definition (2).
x
=
=
=
=
Figure 7
1
The graph of f (x) = .
x
=
f � (c)
y
=
1
1
−
f (x) − f (c)
x
c
lim
= lim
x→c
x→c x − c
x−c
1
1
−
x
c · xc
lim
x→c x − c
xc
lim
c−x
xc(x − c)
x→c
lim
−(x − c)
xc(x − c)
lim
−1
xc
x→c
x→c
−
1
c2
Multiply and divide by xc
Rewrite
Simplify
Substitute c into x
Rewriting, the derivative is
1
.
x2
�
See the graphs of y = f (x) and y = f (x) in Figures 7 and 8.
f � (x) = −
1
x
1
✷
Try This 7
Figure 8
The graph of f � (x) = −
Find the derivative of g(x) = −
2
using the alternative definition (2).
x
1
.
x2
Cases When the Derivative is Undefined
If the slopes of the graph abruptly change as the values of x passes through c, we say
the graph of y = f (x) has a ‘sharp turn’ or ‘cusp’ at (c, f (c)). In such a case , f � (c) is
undefined. For instance, see Figure 9.
Another case when f � (c) fails to exist is when the line x = c is a vertical tangent line
to the graph of y = f (x), see Figure 10.
2.1. THE DERIVATIVE AND SLOPE OF THE TANGENT LINE
79
Example 8 Derivatives Fail to Exist at Sharp Turns
Verify f (x) = |x − 2| + 3 is not differentiable at x = 2.
y
Solution
We evaluate the one-sided derivatives at x = 2 as follows:
lim
x→2+
f (x) − f (2)
x−2
=
lim
|x − 2| + 3 − 3
x−2
lim
|x − 2|
x−2
= lim
x−2
x→2+ x − 2
x→2+
=
x→2+
=
1
Since f (2) = 3
3
and similarly
x
2
lim
x→2−
f (x) − f (2)
x−2
=
=
lim
x→2−
−1.
|x − 2|
−(x − 2)
= lim
x−2
x−2
x→2−
Figure 9.
The graph of y = f (x) has a
‘sharp turn’ at point (2, 3) since
the left and right side derivatives are not the same at x = 2.
Since the one-sided derivatives are not equal, the derivative
f � (2) = lim
x→2
f (x) − f (2)
x−2
fails to exist.
In Figure 8, the line or graph to the left of point (2, 3) has slope −1, and the graph to
the right of (2, 3) has slope 1. Thus, the slopes change abruptly from −1 to 1 as x passes
through 2. Geometrically, for this reason f � (2) is undefined.
✷
Try This 8
a) Verify f (x) = 2|x − 1| is not differentiable at x = 1.
b) Verify g(x) = [[x + 2]] is not differentiable at x = 0.
g�x�
Example 9 Vertical Tangent Lines and Non-Differentiability
2
√
Verify g(x) = 2 5 x is not differentiable at x = 0.
Solution
Using the alternative definition of the derivative (2), we obtain
g � (0)
=
=
=
=
g � (0)
=
g(x) − g(0)
x−0
√
√
25x−250
lim
x→0
x−0
√
25x
lim
x→0
x
2
lim √
x→0 5 x4
∞.
2
lim
x→0
In particular, g � (0) does not exist. Since the limit is ∞, the graph of y = g(x) has a
vertical tangent line at the origin (0, 0), as seen in Figure 10.
✷
Figure 10
The derivative g � (0) does
not exist since x = 0
is a vertical tangent
line at (0,0).
x
80
CHAPTER 2. THE DERIVATIVE
Try This 9
Verify f (x) =
√
3
x + 1 is not differentiable at x = −1.
Differentiable Functions are Continuous Functions
If a function is continuous at a point, the function is not necessarily differentiable at the
point. Examples of such functions is seen in Figure 11, and the previous two examples.
However, if a function is differentiable at a point then it is continuous at the point.
Theorem 2.1 Differentiability Implies Continuity
If a function f (x) is differentiable at x = c, then f (x) is continuous at x = c
Proof
To prove f (x) is continuous at x = c, we must verify
lim f (x) = f (c)
x→c
or equivalently to show
lim f (c + h) = f (c).
h→0
Let h �= 0 and consider the identity
f (c + h) = f (c) +
y
f (c + h) − f (c)
·h
h
Applying limits and the definition of the derivative, we find
3
lim f (c + h)
2
h→0
=
1
�1
=
1
2
3
x
lim f (c + h)
h→0
=
lim f (c) + lim
h→0
h→0
f (c + h) − f (c)
· lim h
h→0
h
f (c) + f � (c) · 0
f (c).
Thus, f (x) is continuous at x = c.
✷
Figure 11
f (x) = [[x]] is not
continuous at
x = 0, ±1, ±2, ...
and thus not differentiable
at the same values of x.
Theorem 2.1 implies that if a function is not continuous at x = c, then it not differentiable at x = c. For instance, the greatest integer function f (x) = [[x]] is not continuous at
integer values of x, see Figure 11. Then f � (x) is undefined if x is an integer.
2.1 Check-It Out
1. Find the slope of the graph of f (x) = x2 at the point (3, 9).
2. Find the derivative f (x) = 5x2 .
2.1. THE DERIVATIVE AND SLOPE OF THE TANGENT LINE
81
True or False. If false, explain or rewrite the statement to make it true.
(x + ∆x)2 − x2
∆x
√
√
3 4 + ∆x − 6
�
2. If f (x) = 3 x, then f (4) = lim
∆x→0
∆x
3. If f (x) = x2 and g(x) = x2 − 1, then f � (x) = g � (x).
1. If f (x) = x2 , then f � (x) = lim
∆x→0
4. The tangent line to the graph of f (x) = x2 at (0, 0) has slope 0.
5. At every point on the graph of f (x) = mx + b the slope of the tangent line is m.
6. If f � (c) exists, then lim f (x) = f (c).
x→c
7. The graph of f (x) = 1/x has a vertical tangent line.
8. The tangent line to the graph of f (x) = x3 at (2, 8) is y = 12x − 16.
√
9. If f (x) = x, then f � (0) = 0
10. If g(x) = |x|, then g � (x) = 1.
Exercises for Section 2.1
In Exercises 1-16, find the slope of the tangent line at the given point. Apply the limit process of Definition 1.
3x
, (−7, 1)
7
1.
f (x) = 10x − 4, (2, 16)
2.
f (x) = 4 +
4.
f (x) = 4 − x2 , (−3, −5)
5.
7.
g(t) =
√
f (x) = 4x − x2 , (2, 4)
8.
g(t) =
10.
g(t) =
√
13.
h(x) =
15.
−4
h(x) = √ , (4, −2)
x
t + 3, (1, 2)
4t − 3 − 1, (1, 0)
3
, (1, 3)
x
√
t − 5, (9, 2)
11.
g(t) = t(2 − t2 ), (−2, 4)
14.
h(x) =
16.
�
2
, 2,
7x
√
h(x) = 3 x − 5,
1
7
3.
f (x) = 2x2 + 3, (−2, 11)
6.
f (x) = 3x2 − 12x + 1, (2, −11)
9.
g(t) =
12.
�
√
4t + 1, (2, 3)
h(t) = t(t2 − 3), (2, 2)
(−3, −2)
In Exercises 17-20, estimate the slope of the tangent line at the indicated point (x, y).
17.
18.
19.
�x,y�
�x,y�
Figure for No. 17
20.
�x,y�
�x,y�
Figure for No. 18
Figure for No. 19
Figure for No. 20
In Exercises 21-34, find the derivative by applying the limit process in Definition 3.
21.
f (x) = x
22.
f (x) = 5x
23.
y = 3x − 4
25.
f (x) = 3x2 + 5
√
m(x) = x + 4
26.
f (x) = x2 − 6x
√
m(x) = 2 x − 1
27.
g(t) =
29.
33.
f (x) = 2x
3
30.
34.
f (x) = x
4
31.
1
t+4
√
f (x) = 3x
24.
28.
32.
2
x+8
3
2
g(t) =
t+1
√
f (x) = 4 2x
y=
82
CHAPTER 2. THE DERIVATIVE
In Exercises 35-40, find the derivative using the alternate definition in (2), page 78.
f (x) = 3x2 − x
35.
36.
f (x) = −x2 + 2x − 4
37.
f (x) =
4
x
√
√
8
39.
f (x) = 4 x
40.
f (x) = x + 9
x−4
In Exercises 41-46, prove each statement. Include a sketch of the graph of y = f (x).
38.
f (x) =
41. If f (x) = |x − 1| + 2, then f � (1) does not exist.
�
43. Prove f (0) = 0 if f (x) =
46. f (x) =
√
3
�
x2 + 1
42. f (x) = |1 − x2 | is not differentiable at x = 1.
if x ≥ 0
3
x +1
�
44. Show f (0) does not exist if f (x) =
if x < 0
x + 2 has a vertical tangent line at (−2, 0).
47. f (x) =
√
3
�
x2 + 1
if x ≥ 0
x+1
if x < 0
1 − x has a vertical tangent line.
In Exercises 47-52, determine where f (x) is not differentiable.
47. f (x) = |x − 3| + 2
f (x) = |4 − x2 |
48.
f �x�
49.
f �x�
6
5
4
3
√
3
f (x) =
x+2
f �x�
x
�3 �2 �1
2
1
3
�2
x
5
Figure for No. 47
50. f (x) =
f �x�
√
5
2
x
�2
Figure for No. 48
3−x+1
51.
f (x) =
f �x�
Figure for No. 49
1
(x − 1)2
52.
f (x) =
f �x�
15
2
8
x2 − 9
3
10
1
x
3
�1
�2
�4 �2
5
Figure for No. 50
2
4
x
�3
1
2
Figure for No. 51
x
Figure for No. 52
In Exercises 53-54, refer to graph of y = f (x). The points (2, 0), (1, 3), and (3, −2) lie on the given graph.
y
3
1 2 3
�2
Figure for Nos. 53-54
x
2.1. THE DERIVATIVE AND SLOPE OF THE TANGENT LINE
83
53. Identify or sketch the following.
b) f (1) − f (3)
a) f (1) and f (3)
c) y =
f (1) − f (3)
(x − 3) + f (3)
1−3
54. Which is greater?
a)
f (2) − f (1)
f (1) − f (0)
or
2−1
1−0
b)
f (1) − f (3)
f (2) − f (3)
or
1−3
2−3
55. Sketch the graph of y = f (x) in each of the following cases:
a) f � (x) is always positive
b)
f � (x) is always negative
56. Sketch the graph of y = g(x) in each of the following cases:
a) g � (x) is positive on (−∞, 0), and g � (x) is negative on (0, ∞)
b) g � (x) is positive on (0, ∞), and g � (x) is negative on (−∞, 0)
57. Use the given graph of y = g(x) to answer the following:
y
C
D
B
E
x
A
Figure for No. 57
a) Between which two consecutive points is the average rate of change of g(x) the
least?
b) Is the derivative of g(x) at C greater than or less than the average rate of change
of g(x) between B and C?
c) Sketch a tangent line to the graph of y = g(x) between A and B such that the
slope of the line is the same as the slope of the line segment AB.
f (c + ∆x) − f (c)
f (x) − f (c)
exists, use the ε-δ definition to prove that lim
x→c
∆x
x−c
exists and equals the former limit. This exercise shows that the alternative definition
of the derivative in (2) agrees with Definition 3.
58. If lim
∆x→0
Odd Ball Problems
59. Determine if f (x) = x|x| is differentiable at x = 0.
60. Determine if f (x) = sin |x| is differentiable at x = 0.
61. Find the values of a, b, c if the graph of f (x) = ax3 + bx + c passes through (−1, 0),
and y = x − 1 is the tangent to the graph at (1, 0).

1 − cos x

if x �= 0

x
�
62. Show f (0) exists if f (x) =


0
otherwise
63. Consider the function given below.

2

 3+x
f (x) =

 3 + x4
�
a) Prove f (0) exists. b)
if x is a rational number
otherwise
Prove f continuous but not differentiable at x = ±1.
84
CHAPTER 2. THE DERIVATIVE
2.2
Basic Differentiation Formulas
• Derivative of a Constant Function • The Power Rule • The Constant Multiple
Rule • Sum and Difference Rules • The Derivatives of the Sine and Cosine
Functions • Rate of Change
Derivative of a Constant Function
In this section, we develop rules for evaluating the derivative of several basic functions.
These rules are specially practical when we need to find the derivative without using the
limit process of Definition 3.
f �x�
Theorem 2.2 The Derivative of a Constant Function
f �x��c
For any constant c, we have
d
[c] = 0.
dx
x
Proof
Figure 1
Let f (x) = c be a constant function. Then the derivative is
The derivative or slope of
the constant function is 0:
d
[c]
dx
d
[c] = 0
dx
=
=
f � (c)
lim
f (c + ∆x) − f (c)
∆x
lim
c−c
∆x
∆x→0
for each real number c.
=
d
[c]
dx
=
∆x→0
Since f (c + ∆x) = c
lim 0 = 0
∆x→0
✷
In Figure 1, we see the graph of a constant function. Since the slope of a horizontal
line is 0, the derivative of a constant function is the zero function.
Example 1 When the Derivative is the Zero Function
The following is a list of constant functions and their derivatives.
Function
Derivative
a) y = 16
dy
=0
dx
b) a(t) = −32
a� (t) = 0
c) f (x) = π
d
[f (x)] = 0
dx
✷
Try This 1
Find the derivative of each function.
a) p(x) = 2 − π
b) a(t) = −9.8
2.2. BASIC DIFFERENTIATION FORMULAS
85
The Power Rule
The next theorem provides a rule for finding the derivative of a power function y = xn
such as y = x2 , f (x) = x3 , and s(t) = t1/3 .
Theorem 2.3 The Power Rule
If n is a real number, then
Proof
d n
[x ] = nxn−1 .
dx
Let n ≥ 2 be an integer. Applying the Binomial Theorem, we obtain
(x + ∆x)n = xn + nxn−1 ∆x +
n(n − 1) n−2
x
(∆x)2 + ... + (∆x)n .
2!
Then by Definition 3 in Section 2.1, we find
d n
[x ]
dx
=
=
=
=
lim
∆x→0
lim
∆x→0
lim
(x + ∆x)n − xn
∆x
�
xn + nxn−1 ∆x +
nxn−1 (∆x) +
=
∆x
n(n−1) n−2
x
2!
�
(∆x)2 + ... + (∆x)n − xn
(∆x)2 + ... + (∆x)n
∆x
∆x→0




n(n − 1) n−2


lim nxn−1 +
x
(∆x) + ... + (∆x)n−1 
∆x→0 
2!
�
��
�
∆x
d n
[x ]
dx
n(n−1) n−2
x
2!
nxn−1
is a factor of these terms
Substitute ∆x = 0
The proof when n = 1 is left as an exercise, see Exercise 21 in Section 2.1. The complete
proof which covers the case when n is an arbitrary real number is given in Chapter 5, see
Theorem 5.15.
✷
y
Example 2 Applying the Power Rule
f �x��x
The following is a list of power functions and their derivatives.
Function
f '�x��1
Derivative
a) g(t) = t
4
b) f (x) = x
c) y =
1
x5
d) y =
√
x
�
g (t) = 4 · t
4−1
= 4t
3
f � (x) = 1 · x1−1 = 1 · x0 = 1
� �
d
1
d � −5 �
5
=
x
= −5x−6 = − 6
dx x5
dx
x
d � 1/2 �
1
1
y� =
x
= x−1/2 = √
dx
2
2 x
Figure 2.
The graph of the
derivative of f (x) = x
is the horizontal
line y = 1.
x
86
CHAPTER 2. THE DERIVATIVE
In Example 2b), the derivative f � (x) = 1 is the slope of the linear function f (x) = x.
We see the graphs of f (x) = x and its derivative in Figure 2.
√
In Examples 2c) and d), we had to rewrite y = 1/x5 and y = x before applying the
Power Rule.
✷
Try This 2
Find the derivative of the function.
a) f (x) = x5
b) g(x) =
1
x2
1
c) y = √
x
d) p(t) = t
y
Example 3 The Slopes of a Graph
��2,4�
m��4
�0,0�
m�0
�1,1�
m�2
f �x��x2
Find the slope of the graph of f (x) = x2 at the given value of x.
x
a) x = −2
b) x = 0
c) x = 1
Solution Applying the Power Rule to f (x) = x2 , we obtain
f � (x) = 2x.
a) If x = −2, the slope of the graph is f � (−2) = 2(−2) = −4.
Figure 3.
The slope m of the
graph at (c, f (c)) is
m = f � (c).
b) If x = 0, the slope of the graph is f � (0) = 2(0) = 0.
c) If x = 1, the slope of the graph is f � (1) = 2(1) = 2.
In Figure 3, we see the tangent lines with their slopes.
✷
Try This 3
Find the slope of the graph of y = g(x) at the indicated value of x.
a) g(x) = x2 , x = 3
b) g(x) =
1
, x=1
x2
c) g(x) =
√
3
x, x = 8
y
Example 4 Finding an Equation of a Tangent Line
Find the slope-intercept form of the equation of the tangent line to the graph of f (x) = x4
at the point (2, 16), see Figure 3.
�2,16�
Solution To find the slope of the tangent line at the point (2, 16), we substitute x = 2
into the derivative f � (x) = 4x3 . Then the slope is
x
Figure 4
At the point (2, 16) on
the graph of f (x) = x4
the tangent line is
y = 32x − 48.
m = f � (2) = 4(2)3 = 32.
Next, substitute the slope m = 32 and the coordinates of (2, 16) into the slope-intercept
form of the equation of the line.
y
=
mx + b
16
=
32(2) + b
−48
=
b
Hence, the slope-intercept equation of the tangent line is
y = 32x − 48.
✷
2.2. BASIC DIFFERENTIATION FORMULAS
87
Try This 4
Find the slope-intercept equation of the tangent line to the graph of
f (x) = x3 at the point (2, 8).
The Constant Multiple Rule
Theorem 2.4 The Constant Multiple Rule
If f (x) is a differentiable function and c is any constant, then cf (x) is differentiable and
d
d
[cf (x)] = c [f (x)].
dx
dx
Proof
Applying Definition 3 in Section 2.1, we find
d
[cf (x)]
dx
=
=
=
d
[cf (x)]
dx
=
cf (x + ∆x) − cf (x)
∆x
�
�
f (x + ∆x) − f (x)
lim c
∆x→0
∆x
f (x + ∆x) − f (x)
c lim
∆x→0
∆x
d
c [f (x)]
dx
lim
∆x→0
This proves the theorem.
✷
By combining the Power Rule and the Constant Multiple Rule, we obtain the following
theorem.
Theorem 2.5 The Combination Rule
If c is any constant, then
d
[cxn ] = cnxn−1 .
dx
In using the combination rule, we may have to rewrite a function in the form y = cxn .
This way, the coefficient c and exponent n are explicit.
88
CHAPTER 2. THE DERIVATIVE
Example 5 Combining the Power and Constant Multiple Rules
Apply the Combination Rule to find the derivative of each function.
Function
Derivative
a) f (x) = 5x2
f � (x) =
b) y =
d � 2�
5x = 5 · 2 · x2−1 = 10x
dx
�
�
d 4 5
4
20 4
y� =
x = · 5x5−1 =
x
dx 3
3
3
4x5
3
s� (t)
7
c) s(t) = √
8 t
=
�
�
d 7 −1/2
7
t
= − t−3/2
dt 8
16
=
−
7
16t3/2
✷
Try This 5
Evaluate the derivative by using the Combination Rule.
a) f (x) = 3x4
b) g(x) =
4
x3
12
c) y = − √
x
Sum and Difference Rules
The following is another helpful technique for evaluating the derivative of a function.
Theorem 2.6 Sum and Difference Rules
If f (x) and g(x) are differentiable functions, then the sum (f + g)(x) and difference (f −
g)(x) are differentiable functions. In addition,
Proof
d
[f (x) + g(x)]
dx
=
f � (x) + g � (x)
Sum Rule
d
[f (x) − g(x)]
dx
=
f � (x) − g � (x).
Difference Rule
Applying the definition of the derivative, we find
d
[f (x) + g(x)]
dx
=
=
=
d
[f (x) + g(x)]
dx
=
lim
f (x + ∆x) + g(x + ∆x) − (f (x) + g(x))
∆x
lim
f (x + ∆x) − f (x) + g(x + ∆x) − g(x)
∆x
lim
f (x + ∆x) − f (x)
g(x + ∆x) − g(x)
+ lim
∆x→0
∆x
∆x
∆x→0
∆x→0
∆x→0
f � (x) + g � (x).
Applying the Sum Rule, we obtain
2.2. BASIC DIFFERENTIATION FORMULAS
d
[f (x) − g(x)]
dx
d
[f (x) − g(x)]
dx
=
d
[f (x) + (−1)g(x)]
dx
=
d
d
[f (x)] +
[(−1)g(x)] Sum Rule
dx
dx
=
d
d
[f (x)] + (−1) [g(x)] Constant Multiple Rule
dx
dx
=
f � (x) − g � (x)
89
✷
The Sum and Difference Rules extend to sums and differences of any finite number of
differentiable functions, by repeated applications of Theorem 2.6.
Example 6 Applying the Sum and Difference Rules
We list certain functions and evaluate their derivatives.
Function
Problem-Solving Tips
Before differentiating,
rewrite the function.
Derivative
f � (x)
a) f (x) = 8x3 − 12
d � 3�
d
8x −
[12]
dx
dx
=
24x2
=
dy
dx
3
9x
b) y = 2 +
x
4
=
=
�
�
d � −2 �
d 9 1
3x
+
x
Rewrite
dx
dx 4
−
6
9
+
Combination Rule, simplify
x3
4
✷
Try This 6
Find the derivative of each function.
a) f (x) = 7x6 − 32x + 1
b) g(x) =
4
8x2
+
− 5x
x7
3
The Derivatives of the Sine and Cosine Functions
We recall the following limits
lim
∆x→0
sin ∆x
= 1 and
∆x
lim
∆x→0
1 − cos ∆x
=0
∆x
(3)
For reference, see Theorem 1.7 and Example 8 in Section 1.3. The limits (3) are essential
in the proof of the next theorem.
90
CHAPTER 2. THE DERIVATIVE
Theorem 2.7 The Derivatives of the Sine and Cosine Functions
a)
d
[sin x] = cos x
dx
Proof
d
[cos x] = − sin x
dx
b)
Applying Definition 3 in Section 2.1 and the sum formula
sin(α + β) = sin α cos β + cos α sin β
we obtain
d
[sin x]
dx
=
=
=
d
[sin x]
dx
=
sin(x + ∆x) − sin x
∆x
sin x cos ∆x + cos x sin ∆x − sin x
lim
∆x→0
∆x
cos x sin ∆x − sin x (1 − cos ∆x)
lim
∆x→0
∆x
�
�
sin ∆x
1 − cos ∆x
lim cos x
− sin x
∆x→0
∆x
∆x
lim
∆x→0
If we fix the value of x in the limit process as ∆x approaches 0, then sin x and cos x are
constants. Since the limits in (3) exist, we rewrite the previous equation as follows:
d
[sin x]
dx
=
=
d
[sin x]
dx
=
sin ∆x
1 − cos ∆x
− sin x lim
∆x→0
∆x
∆x
cos x · 1 − sin x · 0
cos x lim
∆x→0
cos x
This proves part a) of the Theorem. The proof of part b) is assigned to Exercise 67 at
the end of the section.
✷
Example 7 Using the Derivatives of Sine and Cosine
Evaluate the derivative of each function.
Function
a) f (x) = 3 sin x
b) y =
4 cos x
5
c) y = 7x − cos x
Derivative
d
[sin x] Constant Multiple Rule
dx
f � (x)
=
3
f � (x)
=
3 cos x
dy
dx
dy
dx
=
4 d
[cos x] Constant Multiple Rule
5 dx
=
4
− sin x
5
=
=
dy
dx
Theorem 2.7
=
Theorem 2.7
d
d
[7x] −
[cos x] Difference Rule
dx
dx
7 − (− sin x)
Theorem 2.7
7 + sin x
✷
2.2. BASIC DIFFERENTIATION FORMULAS
91
Try This 7
Find the derivative of each function.
a) y = π cos x
b) y = 2 sin x − πx
c) y =
√
3
1
cos x + sin x
2
2
y
Example 8 Finding a Tangent Line to a Sine Function
4
Find an equation of the tangent line to the graph of
f �x��4sin x
2
f (x) = 4 sin x
x
Π
6
where x =
π
. See Figure 4.
6
Solution
Since the derivative is
�4
d
d
[4 sin x] = 4
[sin x] = 4 cos x
dx
dx
π
the slope of the tangent line where x =
is
6
√
√
π
3
m = 4 cos = 4 ·
= 2 3.
6
2
�π �
The point of tangency is
, 2 for
6
�π�
π
1
f
= 4 sin = 4 · = 2.
6
6
2
Figure 4
The tangent line at
(π/6, 2) to the graph
of f (x) = 4 sin x.
π
Then substitute the coordinates of the point of tangency, x =
and y = 2, and the slope
6
√
m = 2 3 into the slope-intercept form of the line.
y
=
2
=
2
=
b
=
mx + b
� √ ��π�
2 3
+b
6
√
π 3
+b
3
√
π 3
2−
3
y
Hence, the tangent line is given by
f �x��2cos x
1
√
√
π 3
y = 2 3x + 2 −
.
3
Π
4
Π
✷
�2
Try This 8
Find an equation of the tangent line to the graph of f (x) = 2 cos x
at the point where x = π/4, see Figure 5.
Figure 5
The tangent line at
the point where x = π/4.
x
92
CHAPTER 2. THE DERIVATIVE
Rate of Change
Let y = f (x) be a function. In an interval [x1 , x2 ], denote the change in x by
∆x = x2 − x1
and the change in y is
∆y = f (x2 ) − f (x1 ).
The average rate of change of y = f (x) in [x1 , x2 ] is defined as
f (x2 ) − f (x1 )
∆y
=
, x2 �= x1 .
∆x
x2 − x1
s
s�t1�
s�t2�
t
t1 t2
Figure 6.
The average velocity of
the position function
s = s(t) from t1 to t2 is
the slope of a secant line
through (t1 , s(t1 )) and
(t2 , s(t2 )).
(4)
The average rate of change of a function describes the change in y per unit of change in
x. The units of measurement of the average rate of change of a function include ft/sec,
dollars/item, students/semester, and kilobytes/sec, to name a few. The quotient in (4) is
also called the rate of change of y = f (x) as x changes from x1 to x2 .
Furthermore, suppose f (x) be differentiable. Then the derivative f � (x1 ) is the limit
of the average rate of change (4) as x2 approaches x1 , by the alternative definition of the
derivative in page 78. For this reason, the derivative f � (x1 ) is called the instantaneous
rate of change, or simply, rate of change of y = f (x) at x1 . For instance, f � (x1 ) may
describe the rate at which water is spilling over a levee in gallons/hr when x = x1 hr.
Many applications of rates of change can be found in engineering, economics, chemistry, meteorology, and physics. Some examples of rates of change include velocity, data
transmission speed, population growth rates, rates of how temperature rise or fall, job
employment rates, and earnings per share EPS, for instance.
Next, we describe the motion of an object moving in a straight line. In many instances,
the line is a horizontal or vertical number line. Let s = s(t) be the directed distance
between the object and the origin of the number line at time t. Such a function is called
a position function. This means that at time t the distance between the object and the
origin is |s(t)| units. In terms of position, the object lies to the right of (or above) the
origin if s(t) > 0, and the object is to the left of (or below) the origin if s(t) < 0. The
average velocity in [t1 , t2 ] is defined as
(Average velocity) =
s(t2 ) − s(t1 )
∆s
=
, t2 �= t1 .
∆t
t2 − t1
(5)
In Figure 6, the average velocity is interpreted as the slope of a secant line.
s�0��60
s�1��44
s�1.5��24
Example 9 Finding the Average Velocity
A ball is dropped from a height of 60 feet. If the height of the ball after t seconds is given
by
s(t) = −16t2 + 60 feet
find the average velocity in each indicated time interval.
ground
a) [1 sec, 1.1 sec]
b) t1 = 1 sec to t2 = 1.01 sec
Figure 7.
A number line is used
to model the path of a
free-falling ball. The
ground is designated as
the origin.
Solution We use a vertical number line with the origin at ground level to model the
path of the ball, see Figure 7.
2.2. BASIC DIFFERENTIATION FORMULAS
93
a) The average velocity from t1 = 1 sec to t2 = 1.1 sec is
s(1.1) − s(1)
∆s
40.64 − 44
=
=
= −33.6 ft/sec.
∆t
1.1 − 1
0.1
b) The average velocity from t1 = 1 sec to t2 = 1.01 sec is
s(1.01) − s(1)
∆s
43.6784 − 44
=
=
= −32.16 ft/sec.
∆t
1.01 − 1
0.01
The average velocities are negative since the ball is moving downward.
✷
Try This 9
A silver dollar is dropped from a height of 144 feet. If the height of the silver dollar after
t seconds is
s(t) = −16t2 + 144 feet
find the average velocity in each time interval.
a) [2.9 sec, 3 sec]
b)
t1 = 2.99 sec to t2 = 3 sec
s
The instantaneous velocity or velocity of an object at time t1 is denoted by v(t1 )
and defined as
v(t1 )
=
=
lim
t2 →t1
s� (t1 )
s(t2 ) − s(t1 )
t2 − t1
s�t1�
Limit of average velocities
Alternative definition of the derivative
where s(t) is a position function. The velocity v(t1 ) is the slope of the tangent line at
(t1 , s(t1 )) to the graph of s = s(t), see Figure 8. We call v = v(t) the velocity function,
and |v(t)| is the speed at time t.
If an object is moving along a vertical line and its motion is influenced only by gravity
(neglect air resistance for instance), we say the object is in a free-fall. In a free fall, the
position function is given by
1
s(t) = gt2 + v0 t + s0
2
where s0 is the initial height, v0 is the initial velocity, g is the acceleration due to gravity,
and the origin s = 0 corresponds to ground level. Near the surface of the earth, assume
g is constant and approximately −32 ft/sec2 or −9.8 m/sec2 . The initial velocity v0 is
positive if the object is thrown upward, and v0 < 0 if the object is thrown downward.
Example 10 Velocities of a Free-falling Object
A ball is thrown vertically upward with an initial velocity of v0 = 64 ft/sec from a height
of 5 ft from the ground. The position function of the ball is
s(t) = −16t2 + 64t + 5 feet.
a) Find the velocity of the ball at time t = 1 sec.
b) Find the speed and the velocity when the ball hits the ground.
t1
Figure 8.
The velocity at time t
is the slope of the
tangent line to the
graph of the position
function s = s(t) at t.
t
94
CHAPTER 2. THE DERIVATIVE
Solution
a) The velocity function is
s
v(t) = s� (t) =
At t = 1 sec, the velocity is
�
d �
−16t2 + 64t + 5 = −32t + 64 ft/sec
dt
v(1) = −32(1) + 64 = 32
1
Figure 9
8� 69
4
t
ft/sec.
b) To find the time when the ball hits the ground, solve s(t) = 0 for t > 0. Applying the
quadratic formula to
−16t2 + 64t + 5 = 0
we obtain
t=
Velocity, which is the slope of
the position function, is positive when the ball is moving
upward, and is negative when
moving downward.
−64 ±
�
√
√
642 − 4(−16)(5)
−64 ± 4416
8 ± 69
=
=
.
−32
−32
4
Since t > 0, the ball hits the ground at
√
8 + 69
t=
sec ≈ 4.077 sec.
4
Substitute the time of impact into v(t) = −32t + 64.
√ �
√ �
�
�
√
8 + 69
8 + 69
v
= −32
+ 64 = −8 69 ≈ −66.5 ft/sec
4
4
Hence, the ball hits the ground with a velocity of about −66.5 ft/sec, and the speed
is approximately 66.5 ft/sec.
✷
Try This 10
The height of a baseball from the ground is given by
s(t) = −16t2 + 32t + 48 feet
where time t is in seconds. Find the velocity when the baseball hits the ground.
2.2 Check-It Out
In Exercises 1-4, find the derivative of each function.
1.
f (x) = π 2
3.
y = 12x4
2.
g(x) = x
4.
h(x) = cos x
2
5. Let s(t) = −16t + 32 be the position function of a free-falling object. Find the
average velocity as t changes from 1 sec to 1.1 sec.
6. Let s(t) = −16t2 + 64 be the position function of a free-falling object. Find the
instantaneous velocity when t = 1 sec.
True or False. If false, explain or give a counterexample.
1. If f (x) = x4 − 5x3 , then f � (x) = 4x3 − 15x2 .
√
2. If f (x) = 2 sin x, then f � (π/3) = 3.
3. If f � (x) = 2x, then f (x) = x2 .
4. The tangent line to the curve y = 4 −
√
x at x = 1 is y =
1
x.
2
2.2. BASIC DIFFERENTIATION FORMULAS
95
2
at t = π/2 is 8/π 2 .
t
d
d
6. If f (x) is differentiable function, then
[xf (x)] = x f (x).
dx
dx
7. If g(t) = |t|, then g � (2) = 1 = g � (−2).
√
8. The slope of the tangent line to the graph of y = 1 − x at the point where x = 0 is −1/2.
5. The derivative of y = 2 sin t −
9. If the position function measured in feet of a free-falling ball is s(t) = −16t2 + 40t + 10 then the average velocity
from t1 = 1 sec to t2 = 2 sec is −8 ft/sec.
10. If a ball is thrown vertically upward with an initial velocity of 32 ft/sec from a height of 96 ft, then the ball
strikes the ground in 3 sec.
Exercises for Section 2.2
In Exercises 1-16, find the derivative of each function.
1.
f (x) = 2x3 − 5x + 4
2.
f (x) = −3x5 + πx − 3
3.
g(x) = x2 (x + 1)
4.
g(x) = (4x − 1)(2x + 3)
5.
y = 4 sin x
6.
g(t) = π cos t
y = x − 3 cos x
8.
√
m(t) = 2 t − sin t
9.
√
3
12.
7.
10.
13.
15.
k(x) = x1/3 − x2/5
11.
√
1
g(x) = 4x3/2 − 3 x − √
x
2x2 − 4x + 5
f (x) =
x3
14.
16.
1
t+ √
t
√
4
f (x) = −10x3/2 + √ + 8 x
x
4x2 + π
g(x) =
(2x)3
m(t) =
1
x
x4 − 3x3 + x2
f (x) =
2x2
h(x) = 1 +
In Exercises 17-32, find f � (c) for the indicated value of c.
17.
f (x) = x4 , c = 2
20.
f (x) = x2 − x +
23.
18.
f (x) = 7x, c = 0
19.
21.
4
f (x) = √ , c = 9
x
22.
f (x) = 3 + 4 cos x, c = 5π/6
24.
f (x) = 12 − 4 sin x, c = 2π/3
25.
f (x) = 2 sin x + 3 cos x, c = π/4
26.
f (x) = 12 cos x − 8 sin x, c = π/4
27.
f (x) = x4 (2x + 3), c = −1
28.
f (x) = (x + 3)(x2 − 1), c = 0
30.
f (x) =
31.
f (x) =
1
,c=2
x
1
1
− 2, c = 2
x
x
|x|
, c = −1
x
1
, c = −2
3x3
10
f (x) = √
, c = 32
5
x4
f (x) =
29.
f (x) =
√
3
32.
f (x) =
x3 + 1
, c = −2
x
x+
√
5
x, c = 1
In Exercises 33-40, find an equation of the tangent line to the given point.
33.
f (x) = x2 , (3, 9)
34.
f (x) = 1 − x2 , (2, −3)
35.
f (x) = x3 − x2 , (2, f (2))
36.
f (x) = 4x − 2x3 + 5, (1, f (1))
37.
√
f (x) = 10 x, (4, 20)
38.
f (x) =
39.
f (x) = 2 sin x, (π/6, 1)
40.
f (x) =
4
, (−8, −1/2)
x
4 cos x
, (2π/3, −2/3)
3
In Exercises 41-44, find the average rate of change of y = f (x) over the given interval [x1 , x2 ].
Then evaluate the rate of change of y = f (x) at x1 .
41.
43.
f (t) = 3t2 + 4, [5, 5.1]
�π π�
f (x) = cos x,
,
4 3
42.
44.
f (t) = t3 − t, [4, 4.01]
�π π�
f (x) = 4 sin x,
,
6 4
96
CHAPTER 2. THE DERIVATIVE
In Exercises 45-48, find the average velocity of a free-falling object with position function s(t) over a time interval
[t1 , t2 ]. Then evaluate the instantaneous velocity and speed at t1 .
45.
s(t) = −16t2 + 20t + 24 feet, [1 sec, 1.02 sec]
46.
s(t) = −16t2 + 30t + 4 feet, [0.49 sec, 0.50 sec]
47.
s(t) = −4.9t2 + 25t + 26.4 meters, [5.99 sec, 6 sec]
48.
s(t) = −4.9t2 + 10t + 2 meters, [0.25 sec, 0.26 sec]
Applied Problems
49. The position function of a free-falling ball is s(t) = −16t2 + 24t + 40 feet where time t is in seconds. When does
the ball strike the ground? Find the velocity when the ball hits the ground.
50. A baseball is dropped from a height of 512 feet. Find the velocity when the baseball strikes the ground.
51. A coin is dropped from the top of a cliff that is over a waterfall. If the coin splashes the water below in 2
seconds, find the speed at which the coin hits the water?
52. A small piece of rock is thrown straight downward from the top of a cliff that is 200 meters above ground level.
If the rock strikes the ground in 2 seconds, find the velocity when the rock strikes the ground.
53. A baseball is thrown vertically upward from a height of 5 feet and reaches a maximum height of 69 feet. Find
the velocity at which the ball was thrown.
54. A tennis ball is thrown straight up from a height of 1.8 meters. If the maximum height of the tennis ball is 21.4
meters, find the initial velocity.
Odd Ball Problems
55. A tangent line to the graph of f (x) = x2 passes through (0, −4). Find the point of tangency on the graph of
f (x) = x2 .
56. Find the point of tangency on the graph of f (x) = 1/x if the tangent line passes through (10, 0).
57. Find all lines tangent to the graph of y = 2x3 + x where the slope is equal to 7.
58. Find the equations of all the tangent lines to the graph of
y = x3 /3 − x2 + 2x + 4/3 that are parallel to the line y = 2x − 3.
59. Determine the values of the constants a, b, and c so that the graph of y = ax2 + bx + c will pass through the
point (−1, 6) and be tangent to the line 2x + y − 1 = 0 at the point (0, 1)
60. The graphs of y = x2 + bx + c and y = ax − x2 are tangent to each other at P (1, 0) for certain constant values
a, b, and c. That is, the tangent lines at P are the same. Determine the values of these constants. What is the
equation of their common tangent line?
61. Normal Line A normal line to the graph of a function y = f (x) at (a, f (a)) is perpendicular to the the tangent
line at (a, f (a)). Find the slope-intercept equation of the normal line to the graph of f (x) = x2 at (1, 1).
√
√
62. A normal line to the graph of f (x) = x contains the point (1, 0). Find the point on the graph of f (x) = x
that the normal line passes through.
63. At what point or points on the graph of y = x2 /4 do the normal lines pass through (4, −1)?
64. Find the slope-intercept equation of the normal line to the graph of y = (x − 1)2 that is perpendicular to the
line y = 2x.
65. Fitting a Curve to the Tangent Lines The graph of a quadratic function is tangent to the lines at y =
−2x − 1, y = 0, and y = −2x + 1 when x = −1, 0, 1, respectively. Find an equation of the quadratic function.
66. Find a third degree polynomial with tangent lines y = 5x + 2, y = 2x, and y = 5x − 2 when x = −1, 0, 1,
respectively.
d
67. Verify the differentiation rule
[cos x] = − sin x. Here, x is in radians. Hint: Expand cos(x + ∆x).
dx
68. Why Use Radians? Let f (x) = sin x where x is in degrees. The derivative of f is
f � (x) =
sin(x + ∆x) − sin x
d
[sin x] = lim
.
∆x→0
dx
∆x
a) Use a calculator to approximate the f � (0◦ ). Compare your answer with π/180.
b) Approximate f � (30◦ ). Compare your answer with cos(π/6)π/180.
2.3. PRODUCT AND QUOTIENT RULES
2.3
97
Product and Quotient Rules
• Product Rule • Quotient Rule • Derivatives of the Remaining Trigonometric
Functions • Higher Order Derivatives
Product Rule
In the previous section, we discussed the derivative of the sum and difference of two
functions, namely,
d
d
d
[f (x) + g(x)] =
[f (x)] +
[g(x)]
dx
dx
dx
Sum Rule
and
d
d
d
[f (x) − g(x)] =
[f (x)] −
[g(x)] .
Difference Rule
dx
dx
dx
In the present section, we develop rules for finding the derivative of the product and
quotient of two functions.
Theorem 2.8 Product Rule
If f (x) and g(x) are two differentiable functions of x, then their product f (x)g(x) is a
differentiable function and
d
[f (x)g(x)] = f (x)g � (x) + f � (x)g(x).
dx
Proof
In the proof, we add and subtract f (x + ∆x)g(x) as follows:
d
[f (x)g(x)]
dx
=
=
=
=
d
[f (x)g(x)]
dx
=
lim
∆x→0
f (x + ∆x)g(x + ∆x) − f (x)g(x)
∆x
f (x + ∆x)g(x + ∆x) − f (x + ∆x)g(x) + f (x + ∆x)g(x) − f (x)g(x)
∆x
�
�
g(x + ∆x) − g(x)
f (x + ∆x) − f (x)
lim f (x + ∆x)
+
g(x)
∆x→0
∆x
∆x
lim
∆x→0
lim f (x + ∆x) lim
∆x→0
∆x→0
g(x + ∆x) − g(x)
f (x + ∆x) − f (x)
+ lim
lim g(x)
∆x→0
∆x→0
∆x
∆x
f (x)g � (x) + f � (x)g(x)
This completes the proof of the Product Rule.
✷
The Product Rule may be rewritten either as
d
[f (x)g(x)] = f � (x)g(x) + f (x)g � (x)
dx
or
d
[f (x)g(x)] = f (x)g � (x) + g(x)f � (x).
dx
In words, the derivative of a product f (x)g(x) is the first function times the derivative of
the second, plus the second function times the derivative of the first.
The Product Rule can be generalized to include the product of any finite number of
functions. For three functions, we have
d
[f (x)g(x)h(x)] = f � (x)g(x)h(x) + f (x)g � (x)h(x) + f (x)g(x)h� (x).
dx
98
CHAPTER 2. THE DERIVATIVE
To illustrate, we evaluate the derivative of y = x3 sin x cos x.
dy
dx
=
d � 3�
d
d
x sin x cos x + x3
[sin x] cos x + x3 sin x
[cos x]
dx
dx
dx
=
3x2 sin x cos x + x3 cos x cos x + x3 sin x(− sin x)
=
3x2 sin x cos x + x3 cos2 x − x3 sin2 x
Example 1 Using the Product Rule
Find the derivative of
y = (3x + 5)(x2 − 4x)
using the Product Rule.
Solution Let f (x) and g(x) be the factors of the product. We list their derivatives below:
f (x) = 3x + 5
f � (x) = 3
g(x) = x2 − 4x
g � (x) = 2x − 4
By the Product Rule, the derivative is
dy
dx
dy
dx
=
f (x)g � (x) + f � (x)g(x)
=
(3x + 5)(2x − 4) + 3(x2 − 4x)
=
(6x2 − 2x − 20) + (3x2 − 12x)
=
9x2 − 14x − 20.
✷
Try This 1
Find the derivative of
p(x) = (4x2 + x)(3x + 1)
using the Product Rule.
In Example 1 we may evaluate the derivative without using the Product Rule. If we
multiply the factors, we obtain
y = (3x + 5)(x2 − 4x) = 3x3 − 7x2 − 20x.
Then apply the Sum and Difference Rules:
�
dy
d � 3
=
3x − 7x2 − 20x = 9x2 − 14x − 20.
dx
dx
We obtain the same result as in Example 1. However, circumventing the Product Rule is
not always possible.
2.3. PRODUCT AND QUOTIENT RULES
99
Example 2 Using the Product Rule
Find the derivative of h(x) = x2 cos x.
Solution
We write the factors as f (x) and g(x), and include their derivatives.
f (x) = x2
f � (x) = 2x
g(x) = cos x
g � (x) = − sin x
Applying the Product Rule, we obtain
h� (x)
=
f (x)g � (x) + f � (x)g(x)
=
x2 · (− sin x) + 2x · cos x
=
−x2 sin x + 2x cos x.
✷
Try This 2
Find the derivative of each function.
a) y = x sin x
b) y = sin x cos x
To find the derivative of a product of functions where one factor is a constant, applying
the Constant Multiple Rule may simplify the procedure.
Example 3 Using the Product and Constant Multiple Rules
�
�
Find the derivative of y = π x2 + x − 3 cos x.
Solution
Applying the Constant Multiple Rule, we obtain
dy
dx
=
=
=
�
�
d �� 2
x + x − 3 cos x
dx
��
�
�
π x2 + x − 3 (− sin x) + (2x + 1) cos x
π
�
�
π (2x + 1) cos x − (x2 + x − 3) sin x
Product Rule
Rewrite
Try This 3
Find the derivative of h(x) = 40x3 sin x by the two methods indicated below.
�
d � 3
a) Constant Multiple and Product Rules: h� (x) = 40
x sin x
dx
�
d �
b) Product Rule: h� (x) =
(40x3 ) sin x
dx
Verify that both derivatives are equal.
✷
100
CHAPTER 2. THE DERIVATIVE
Quotient Rule
We discuss a rule for evaluating the derivative of a quotient of two functions.
Theorem 2.9 Quotient Rule
If f (x) and g(x) are two differentiable functions of x, then the quotient f (x)/g(x) is a
differentiable function and
�
�
g(x)f � (x) − f (x)g � (x)
d f (x)
=
provided g(x) �= 0.
dx g(x)
[g(x)]2
Proof
In the proof, we add and subtract f (x)g(x) as highlighted below.
�
�
d f (x)
dx g(x)
=
=
=
=
f (x + ∆x)
f (x)
−
g(x + ∆x)
g(x)
lim
∆x→0
∆x
lim
g(x)f (x + ∆x) − f (x)g(x + ∆x)
∆xg(x)g(x + ∆x)
lim
g(x)f (x + ∆x)−g(x)f (x) − f (x)g(x + ∆x)+f (x)g(x)
∆xg(x)g(x + ∆x)
lim
g(x) (f (x + ∆x) − f (x)) − f (x) (g(x + ∆x) − g(x))
∆xg(x)g(x + ∆x)
∆x→0
∆x→0
∆x→0
g(x)
=
lim
∆x→0
f (x + ∆x) − f (x)
g(x + ∆x) − g(x)
− f (x)
∆x
∆x
g(x)g(x + ∆x)
lim g(x) lim
=
∆x→0
∆x→0
f (x + ∆x) − f (x)
g(x + ∆x) − g(x)
− lim f (x) lim
∆x→0
∆x→0
∆x
∆x
lim (g(x)g(x + ∆x))
∆x→0
�
d f (x)
dx g(x)
�
=
g(x)f � (x) − f (x)g � (x)
[g(x)]2
This proves the Quotient Rule.
✷
In words, the derivative of a quotient f /g is the denominator times the derivative of the
numerator minus the numerator times the derivative of the denominator, and all divided
by the square of the denominator.
Example 4 Using the Quotient Rule
Apply the Quotient Rule to find the derivative of
y=
2x − 3
.
4 + x2
Solution Let f (x) and g(x) be the numerator and denominator of the function. We list
their derivatives as follows:
f (x) = 2x − 3
f � (x) = 2
g(x) = 4 + x2
g � (x) = 2x
2.3. PRODUCT AND QUOTIENT RULES
101
Using the Quotient Rule, we obtain the following.
�
�
g(x)f � (x) − f (x)g � (x)
d f (x)
=
dx g(x)
[g(x)]2
�
�
d f (x)
dx g(x)
=
(4 + x2 )(2) − (2x − 3)(2x)
(4 + x2 )2
=
(8 + 2x2 ) − (4x2 − 6x)
(4 + x2 )2
=
−2x2 + 6x + 8
(4 + x2 )2
=
−2(x2 − 3x − 4)
(4 + x2 )2
Factoring the numerator, we find that the derivative is
−2(x − 4)(x + 1)
dy
=
.
dx
(4 + x2 )2
✷
Try This 4
Find the derivative of y =
2x + 1
.
3x − 4
For more clarity, when using the Quotient Rule use parentheses ( ) or brackets [ ]
to separate the factors as needed, and to group the terms that are subtracted from one
another. Also, we usually find it convenient to rewrite a function before differentiating.
Example 5 Rewrite, Differentiate, and Simplify
Find the derivative of f (x) =
4x/3
.
2 + 1/x
Solution Rewrite f (x) as follows:
f (x)
4x
3
=
1
2+
x
4x2
6x + 3
=
·
3x
3x
provided x �= 0
Rewrite
Applying the Quotient Rule, we obtain
�
f (x)
f � (x)
(6x + 3)
=
d � 2�
d
4x − (4x2 )
[6x + 3]
dx
dx
2
(6x + 3)
=
(6x + 3)(8x) − (4x2 )(6)
24x2 + 24x
=
2
(6x + 3)
(6x + 3)2
=
24x(x + 1)
9 (2x + 1)2
=
=
8x(x + 1)
3 (2x + 1)2
provided x �= −
1
2
Simplify
✷
102
CHAPTER 2. THE DERIVATIVE
Try This 5
Find the derivative of y =
x/3
.
5 + x/2
In differentiating a quotient f (x)/g(x), we do not have to use the Quotient Rule all
the time. For instance, if a quotient can be written as a constant times a function, the
Constant Multiple Rule may be more appropriate to apply than the Quotient rule. See
the next example.
Example 6 Using the Constant Multiple Rule on a Quotient
Function
y=
x3
4
f (x) =
g(x) =
Rewrite
2
3x5
x − 9x2
8x
y=
1 3
x
4
Simplify
dy
3
= x2
dx
4
dy
3x2
=
dx
4
2 −5
x
3
f � (x) = −
1
(1 − 9x)
8
g � (x) =
f (x) =
g(x) =
Differentiate
10 −6
x
3
1
(−9)
8
f � (x) = −
10
3x6
g � (x) = −
9
8
✷
Try This 6
Rewrite each function and find the derivative by the Constant Multiple Rule.
a) y =
3x8
16
b) y =
1
4x2
c) y =
8x + 5
4
Next we prove the Power Rule when the power n is a negative integer.
Example 7 Proving the Power Rule for Negative Exponents
If n is a negative integer, prove the Power Rule formula
d n
[x ] = nxn−1 .
dx
The above formula has been verified when n is a positive integer,
see Theorem 2.3 in page 85 and Exercise A.2 in Section 2.1.
Solution Let k = −n. Then
d n
[x ]
dx
=
�
�
� �
d
1
d
1
=
dx x−n
dx xk
xk
=
d
d � k�
[1] − 1 ·
x
dx
dx
(xk )2
Since k is a positive integer, we have
d � k�
x = kxk−1 .
dx
Quotient Rule
2.3. PRODUCT AND QUOTIENT RULES
103
Continuing with the above calculation, we obtain
d n
[x ]
dx
=
xk · 0 − kxk−1
−kxk−1
=
(xk )2
x2k
=
−kx−k−1
=
nxn−1 .
Simplify
Since n = −k
This completes the proof.
✷
The proof of the Power Rule when the exponent n =
implicit differentiation, see Exercise 45 in Section 3.5.
p
q
is a rational number uses
Try This 7
If n is an integer, prove the formula
� �
d
1
n
= − n+1 .
dx xn
x
Derivatives of the Remaining Trigonometric Functions
In Section 2.2, we obtained the derivative of y = sin x and y = cos x. Next, we derive
the derivatives of the remaining four trigonometric functions by applying the Quotient
Rule.
Theorem 2.10 Derivatives of the other Trigonometric Functions
a)
d
[tan x] = sec2 x
dx
b)
d
[sec x] = sec x tan x
dx
Proof
c)
d)
d
[cot x] = − csc2 x
dx
d
[csc x] = − csc x cot x
dx
We prove only part b). By the Quotient Rule, we find
�
�
d
d
1
[sec x] =
dx
dx cos x
cos x
=
d
d
[1] − 1 ·
[cos x]
dx
dx
2
cos x
=
(cos x) · 0 − 1(− sin x)
sin x
=
cos2 x
cos2 x
=
1 sin x
= sec x tan x
cos x cos x
The remaining identities are proved similarly. The student is asked to prove these in
Exercise 65.
✷
104
CHAPTER 2. THE DERIVATIVE
Example 8 More on Derivatives of Trigonometric Functions
We list certain functions and evaluate their derivatives.
Function
Derivative
a) f (x) = 2 cos x + 3 csc x
x
b) y =
tan x
d
d
[cos x] + 3 [csc x]
dx
dx
f � (x)
=
2
f � (x)
=
−2 sin x − 3 csc x cot x
tan x
dy
dx
=
dy
dx
=
d
d
[x] − x
[tan x]
dx
dx
tan2 x
tan x − x sec2 x
tan2 x
✷
Try This 8
Find the derivative of y =
2 tan x
.
1 − tan x
Example 9 Using Trigonometric Identities in Finding Derivatives
Differentiate both forms of the function
y=
1 − sin x
= sec x − tan x
cos x
and show the derivatives are equivalent.
Solution
Applying the Quotient Rule, we obtain
�
�
dy
d 1 − sin x
=
dx
dx
cos x
d
d
cos x
[1 − sin x] − (1 − sin x)
[cos x]
dx
dx
=
cos2 x
cos x(− cos x) − (1 − sin x)(− sin x)
=
cos2 x
2
− cos x + sin x − sin2 x
=
cos2 x
dy
sin x − 1
=
.
Since sin2 x + cos2 x = 1
dx
cos2 x
Applying the Difference Rule to the second form, we find
dy
dx
=
=
dy
dx
=
d
[sec x − tan x]
dx
sec x tan x − sec2 x
sec x(tan x − sec x)
Finally, the two derivatives are equal since
�
�
�
�
sin x − 1
1
sin x − 1
1
sin x
1
=
=
−
= sec x(tan x − sec x).
cos2 x
cos x
cos x
cos x cos x
cos x
✷
2.3. PRODUCT AND QUOTIENT RULES
105
Try This 9
1 − cos x
Differentiate both forms of the function y =
= csc x − cot x.
sin x
Then show that the derivatives are equivalent.
Higher Order Derivatives
In Section 2.2, the position function s(t) of an object moving in a straight line was discussed. Also, we defined the average velocity in [t, t + ∆t] as
(average velocity) =
s(t + ∆t) − s(t)
∆s
=
.
∆t
∆t
and the velocity function v(t) as a derivative
v(t) = lim
∆t→0
∆s
= s� (t).
∆t
Next, we discuss acceleration. The average acceleration in [t, t + ∆t] is the average rate
of change of the velocity function in [t, t + ∆t]:
v(t + ∆t) − v(t)
∆v
=
.
∆t
∆t
The acceleration function is the derivative of the velocity function:
(average acceleration) =
a(t) = lim
∆t→0
∆v
= v � (t).
∆t
Note, the acceleration function a(t) is obtained by differentiating the position function s(t)
twice. We say a(t) is the second derivative of s(t), and write
a(t) = v � (t) = s�� (t) =
d2
[s(t)].
dt2
Example 10 Average Acceleration and Acceleration
The position function of an object moving along a horizontal line is
s(t) = 15t2 − t3 + 6
with distance s given in meters, and time t in seconds.
a) Find the average acceleration in the interval [2 sec, 5 sec].
b) Find the acceleration when t = 4 sec and t = 6 sec.
Solution
a) Since the velocity function is
v(t) = s� (t) =
d
[15t2 − t3 + 6] = 30t − 3t2
dt
we obtain
v(5)
=
30(5) − 3(5)2 = 75 m/sec
v(2)
=
30(2) − 3(2)2 = 48 m/sec.
Then the average acceleration in [2 sec, 5 sec] is
∆v
∆t
=
=
v(5) − v(2)
75 m/sec − 48m/sec
=
5 sec − 2 sec
3 sec
m/sec
m
9
= 9
.
sec
sec2
On average, we say the velocity is increasing at 9 m/sec per sec during the time
interval [2 sec, 5 sec].
106
CHAPTER 2. THE DERIVATIVE
b) To find the acceleration function a(t), evaluate v � (t).
a(t) = v � (t) =
d
[30t − 3t2 ] = 30 − 6t
dt
The acceleration when t = 4 sec and t = 6 sec, respectively, are given by
a(4)
=
30 − 6(4) = 6 m/sec2
a(6)
=
30 − 6(6) = −6 m/sec2 .
✷
Try This 10
The position function of a free-falling object is s(t) = −16t2 − 16t + 96 in feet where t
is in seconds.
a) Find the average acceleration in the interval [2 sec, 3 sec].
b) Find the acceleration when t = 2 sec.
The second derivative of a function is the derivative of the derivative of the function.
Moreover, the third derivative is the derivative of the second derivative. A higher order
derivative is the derivative of a derivative, i.e., the (n + 1)st derivative is the derivative of
the nth derivative. The higher order derivatives are denoted as follows.
First Derivative
y�
f � (x)
dy
dx
d
[f (x)]
dx
Second Derivative
y ��
f �� (x)
d2 y
dx2
d2
[f (x)]
dx2
Third Derivative
y (3)
f (3) (x)
d3 y
dx3
d3
[f (x)]
dx3
y (n)
f (n) (x)
dn y
dxn
dn
[f (x)]
dxn
..
.
nth Derivative
Example 11 Higher Order Derivatives
If f (x) =
Solution
1
, find the fifth derivative f (5) (x).
x
We evaluate the higher order derivatives as follows:
f (x) = x−1
f � (x) = −x−2
f �� (x) = 2x−3
f ��� (x) = −6x−4
f (4) (x) = 24x−5
f (5) (x) = −120x−6
Then the fifth derivative is f (5) (x) = −
120
.
x6
Try This 11
√
Evaluate the third derivative f (3) (x) of f (x) = 3 x.
✷
2.3. PRODUCT AND QUOTIENT RULES
107
2.3 Check-It Out
1. Find the derivative of f (x) = x4 sin x.
3. If y =
2. Evaluate the derivative of y =
x2
.
1 + 3x
5
dy
, find
by applying the Combination Rule.
6x3
dx
4. The position of a free-falling object is s(t) = −16t2 + 64t + 100 feet where t is in seconds.
Find the acceleration when t = 1 sec.
True or False. If false, write a counter-example. Assume f (x), g(x), h(x) are differentiable functions.
1. If r(x) = f (x)g(x), then r� (x) = f � (x)g � (x).
2. If y = (x2 + x + 2)(x + 3), then
dy
= (x2 + x + 2) + (2x + 1)(x + 3).
dx
3. If f (x) = xg(x), then f �� (x) = 0.
4. If R(x) = f (x)g(x) and f (x) is a constant function, then R� (x) = f (x)g � (x).
5. If p(x) =
f (x)
f � (x)
, then p� (x) = �
.
g(x)
g (x)
6. The derivative of y =
2x
8
is y � =
.
x+4
(x + 4)2
7. It is possible to use the Constant Multiple Rule to find the derivative of y =
c
xn
where c, n are constants.
8. If s(t) is a position function of an object moving along a straight path, then the acceleration function is s�� (t).
−1
9. If s(t) = t+1
is a position function of an object moving in a straight line, then the acceleration is undefined
when t = 1.
�
�
�
�
f (x)
d
d f (x)/h(x)
10.
=
dx g(x)h(x)
dx
g(x)
Exercises for Section 2.3
In Exercises 1-26, find the derivative. When applicable, a and c denote real constants.
1.
y = (2 − x)(2x + 1)
2.
y = (2x3 − 1)(x − 4)
3.
f (x) = (x − a)(x2 + ax − 1)
4.
f (x) = (x2 − ax)(x2 + ax − 1)
5.
y = x tan x
9.
f (x) = (2x − 5)2
6.
y = x2 tan x
10.
y = (2x + 1)3
x2 + c 2
x2 − c 2
7.
y = x2 sec x
8.
p(x) = sin x cos x
11.
f (x) =
x
x−1
12.
f (x) =
3 + 2x
3 − 2x
15.
m(x) =
cos x
x
16.
m(x) =
x
sin x
−6x2 + x + 2
3x2
13.
g(x) =
x2
x3 + a3
14.
g(x) =
17.
c(x) =
x2 + 1
x2 + 5
18.
r(x) =
x3 − 1
x+1
19.
g(x) =
x2 + 5x
x
20.
f (x) =
21.
f (x) =
3x2
4
22.
f (x) =
4x3
5
23.
f (x) =
2
3x2
24.
f (x) = −
25.
f (x) =
2 sin x
x
−
5
3
26.
f (x) =
tan x
π
+
2
3
3
2x4
In Exercises 27-38, evaluate the derivative y � at the given value of x.
27.
y=
4 − x2
, x = −1
2x + 3
28.
y=
x3 − 4x + 2
,x=0
(x − 2)2
29.
√
y = 2 x(2 − 3x), x = 4
108
30.
33.
36.
CHAPTER 2. THE DERIVATIVE
x+1
1
y= √ ,x=
4
x
√
x
y= 2
,x=4
x −1
y = (x + 1) cot x, x = π/2
31.
y = (2x2 − 5x + 1)(3x − 2), x = 1
34.
y=
37.
g(x) = |x| − |x + 2|, x = −1
sin x
, x = π/2
x
32.
y = (x − a)(x2 − a2 ), x = a
35.
y = x cos x, x = π
38.
g(x) = |x − 3| − |x|, x = 1
In Exercises 39-46, find an equation of the tangent line to the graph of the function. The x-coordinate of the point of
tangency is indicated.
x2
, x = −2
x2 + 1
39.
y=
1
,x=1
1 + x2
40.
y=
x−1
, x = −1
x+3
41.
y=
43.
y = tan x, x = π/6
44.
y = sec x, x = π/3
45.
y = csc x, x = π/4
3
,x=1
x(x + 2)
42.
y=
46.
y = 3 cot x, x = 2π/3
In Exercises 47-50, find p� (1) given f (1) = 1, g(1) = 3, f � (1) = a, g � (1) = b.
47.
p(x) = f (x)g(x)
48.
p(x) =
f (x)
g(x)
49.
p(x) =
g(x)
f (x)
50.
p(x) = 2g(x)f (x)
In Exercises 51-52, find F � (x) using the following methods: a) Perform the indicated multiplication and differentiate
term by term, b) Product Rule, and c) Show that the answers to a) and b) are equal.
51.
F (x) = x2 (a − x) where a is a constant
52.
F (x) = (3x − 1)(x2 + 4x − 3)
53. The position function of an object traveling in a straight line is s(t) = t3 − 6t2 + 9t with distance s(t) given in
meters, and time t in seconds.
a) Find the acceleration function.
b) When is the rate of change of the velocity function equal to zero?
54. An ant is crawling along a horizontal line and the position function of the ant is s(t) =
with distance s(t) given in inches, and time t in seconds.
a) Find the average acceleration in [2 sec, 2.1 sec]
t4
3t3
23t2
15t
−
+
−
8
2
4
2
b) Find the acceleration when t = 2 sec.
55. A ball is dropped from an unknown height and hits the ground at 60 mph. Find the unknown height.
56. A pebble is thrown downward with a speed of 8 ft/sec. If the pebble strikes the ground at 88 ft/sec, find the
the height at which the pebble was thrown.
In Exercises 57-62, evaluate the indicated higher order derivative.
57.
f (x) = x3 , f (3) (x)
60.
f (x) =
1
,
x
f (n) (x), n ≥ 1
58.
f (x) = x7 , f (7) (x)
59.
f (x) =
61.
f (x) = 5 sin x, f (27) (x)
62.
f (x) = 2 cos x, f (14) (x)
1
,
x2
f (3) (x)
Miscellaneous Problems
63. The unit price p in the sale of x items is p = 1 − 0.002x dollars.
a) Find the revenue R(x) from the sale of x items.
b) Find the marginal revenue R� (x).
64. The length and width of a rectangle are given as functions of time t:
L(t) = 5 + 1.05t2 and W (t) = 3 + 1.04t
Find the rate of change of the area of the rectangle with respect to t
65. Prove the differentiation rules listed below.
a)
d
[tan x] = sec2 x
dx
b)
d
[cot x] = − csc2 x
dx
c)
d
[csc x] = − csc x cot x
dx
2.4. THE CHAIN RULE AND ITS APPLICATIONS
2.4
109
The Chain Rule and Its Applications
• The Chain Rule • The General Power Rule • Using Several Differentiation
Rules • Applying the Chain Rule to Trigonometric Functions
The Chain Rule
In this section, we discuss how to evaluate the derivative of a composite function (f ◦g)(x).
The appropriate rule needed is called the Chain Rule. For instance, the following can be
evaluated by applying the chain rule:
�
�
d ��
d �
1 + x3 and
sin(x2 ) .
dx
dx
Theorem 2.11 Chain Rule
If f (x) and g(x) are differentiable functions, then
d
[(f ◦ g)(x)] = f � (g(x)) · g � (x).
dx
Proof
If F (x) = f (g(x)), then
F � (x)
=
=
F � (x)
=
lim
F (x + ∆x) − F (x)
∆x
lim
f (g(x + ∆x)) − f (g(x))
∆x
lim
f (g(x) + ∆g) − f (g(x))
∆x
∆x→0
∆x→0
∆x→0
where
∆g = g(x + ∆x) − g(x).
Fix x and define the following function:


 f (g(x) + w) − f (g(x)) − f � (g(x))
w
G(w) =


0
if
w �= 0
if
w=0
Since f � (g(x)) exists, G(w) is continuous at w = 0. For all w, we have
�
�
f (g(x) + w) − f (g(x)) = f � (g(x)) + G(w) w.
If w = ∆g, we obtain
F � (x)
=
=
F � (x)
=
lim
∆x→0
lim
∆x→0
[f � (g(x)) + G(∆g)] ∆g
∆x
�
f � (g(x)) + G(∆g)
f � (g(x))g � (x)
�
lim
∆x→0
∆g
∆x
Since lim G(∆g) = G(0) = 0.
∆x→0
This proves the Chain Rule.
✷
In a composite function f (g(x)), we usually call g(x) the inner function and f (x) the
outer function. The Chain Rule says that the derivative of a composite function is the
derivative of the outer function evaluated at the inner function, which is f � (g(x)), times
the derivative g � (x) of the outer function.
110
CHAPTER 2. THE DERIVATIVE
Example 1 Illustrating Composite Functions
√
If f (x) = 10 x and g(x) = 3x2 − 4x, evaluate (f ◦ g)(x) and (f ◦ g)� (x).
Solution
We find
(f ◦ g)(x)
=
f (g(x)) = f (3x2 − 4x)
=
10
Before applying the chain rule, we note
�
3x2 − 4x.
5
f � (x) = √ , g � (x) = 6x − 4.
x
Applying the chain rule, we obtain
(f ◦ g)� (x)
=
f � (g(x))g � (x)
=
f � (3x2 − 4x) · (6x − 4)
=
√
5
3x2
− 4x
· (6x − 4)
✷
Try This 1
If f (x) = 3x4 and g(x) = 5x − 7, evaluate (f ◦ g)(x) and (f ◦ g)� (x).
Example 2 Illustrating Composite Functions
Find two functions f (u) and g(x) such that f (g(x)) is equal to the given function.
y = f (g(x))
y = f (u)
u = g(x)
a) y = (3x + 1)4
b) y = tan x2
c) y = 3 cos2 5x
y = u4
y = tan u
y = 3 cos2 u
u = 3x + 1
u = x2
u = 5x
Try This 2
Find f (u) and g(x) such that y = f (g(x)).
a)
y = (5x − 2)3
b)
y=
√
1 − x2
Example 3 Illustrating the Chain Rule
Apply the Chain Rule to evaluate the derivative.
a)
�
d �
sin(x2 )
dx
b)
d � 2 �
cos x
dx
c)
y = tan(2πx)
2.4. THE CHAIN RULE AND ITS APPLICATIONS
111
Solution
a) We write f (g(x)) = sin(x2 ) where
i) f (u) = sin u, g(x) = x2 , and
ii) f � (u) = cos u, g � (x) = 2x.
Using the Chain Rule, we obtain
�
d �
sin(x2 )
= f � (g(x)) · g � (x)
dx
�
�
=
cos(x2 ) 2x
=
Chain Rule
2x cos(x2 ).
Simplify
2
b) Let f (g(x)) = cos x where
i) f (u) = u2 , g(x) = cos x, and
ii) f � (u) = 2u, g � (x) = − sin x.
Applying the Chain Rule, we find
d � 2 �
cos x
= f � (g(x))g � (x)
dx
= (2 cos x)(− sin x)
=
− sin(2x).
Double-angle identity
✷
Try This 3
Find the derivative by using the Chain Rule.
a)
y = sec 2x
b)
y = (5x − 2)3
c)
y=
√
1 + x3
The General Power Rule
Several functions are of the form y = (u(x))n where n is a constant, for instance see
Example 3b). The General Power Rule, a special case of the Chain Rule, tells us how
to obtain the derivative of y = (u(x))n .
Theorem 2.12 General Power Rule
Let y = [u(x)]n where u(x) is a differentiable function of x, and n is a constant. Then y
is a differentiable function of x, and
dy
d
= n[u(x)]n−1 [u(x)]
dx
dx
or equivalently
Proof
d n
[u ] = nun−1 u� .
dx
If f (x) = xn and g(x) = u(x), then
f (g(x)) = [u(x)]n .
Note,
f � (x) = nxn−1 and f � (g(x)) = f � (u(x)) = n[u(x)]n−1 .
Using the Chain Rule, we get
dy
dx
=
f � (g(x))g � (x) = n[u(x)]n−1 u� (x).
✷
112
CHAPTER 2. THE DERIVATIVE
Example 4 Using the General Power Rule
Evaluate the derivative of f (x) = (3x2 + 3 cos x)4 .
Solution
we find
Write f (x) = [u(x)]4 with u(x) = 3x2 + 3 cos x. By the General Power Rule,
f � (x)
d
[3x2 + 3 cos x]
dx
=
4[3x2 + 3 cos x]3
=
4[3x2 + 3 cos x]3 (6x − 3 sin x)
=
324[x2 + cos x]3 (2x − sin x) .
General Power Rule
Differentiate
Factor
✷
Try This 4
Find the derivative f � (x).
a)
f (x) = (sin x + cos x)2
f (x) = 4(1 − cos x)3
b)
Example 5 Applying the General Power Rule
Find the derivative of y =
Solution
2
.
(8x2 − x3 + 5)4
Rewrite the function as
y = 2(8x2 − x3 + 5)−4 .
Using the General Power Rule, we obtain
dy
dx
=
=
dy
dx
=
d
[8x2 − x3 + 5]
dx
�
�
−8[8x2 − x3 + 5]−5 16x − 3x2
Differentiate
−8[8x2 − x3 + 5]−5 ·
−
8x(16 − 3x)
.
(8x2 − x3 + 5)5
Factor and simplify
✷
Try This 5
Find the derivative y � .
a)
y=
3
(x2 − 1)2
b)
y= √
4
1 + x2
Example 6 Using the General Power Rule With Radicals
√
Determine the points on the graph of f (x) = 12 3 x2 − 4 where the tangent line is vertical.
�
Then determine the points where f (x) = 0.
2.4. THE CHAIN RULE AND ITS APPLICATIONS
Solution
113
We evaluate f � (x) as follows:
�
f (x)
=
y
�
d � 2
12
(x − 4)1/3
dx
2
=
4(x − 4)
=
�
3
−2/3
Constant Multiple Rule
4
d 2
[x − 4] General Power Rule
dx
8x
.
(x2 − 4)2
�3
lim �
8x
=∞
(x2 − 4)2
and
lim �
x→−2 3
8x
= −∞
(x2 − 4)2
Try This 6
√
Repeat Example 6 but f (x) = 3 x2 − x − 6.
Using Several Differentiation Rules
In order to evaluate the derivative of a function, we may have to apply several differentiation rules. The order in which the rules are to be applied usually matters. In such
situations, the students’ algebraic skills become tested and sharpened.
Example 7 Combining the Differentiation Rules
Find the derivative of y = (x2 − 4)3 (x2 − 1).
dy
dx
First, we apply the Product Rule.
=
=
=
=
dy
dx
=
3
�20
Since f (±2) = 0, we have vertical tangent lines at (±2,√0), see Figure 6.
Moreover,
if f � (x) = 0 then x = 0 and f (0) = −12 3 4. Hence, f � (x) = 0 at the point
√
3
(0, −12 4). See Figure 6.
✷
Solution
1
Simplify
At x = ±2, we have infinite limits for f � (x).
x→2 3
�1
�
�
d � 2
d � 2
x −1 +
(x − 4)3 · (x2 − 1) Product Rule
dx
dx
� 2
�
2
3
(x − 4) · (2x) + 3(x − 4)2 (2x) · (x2 − 1) General Power Rule
(x2 − 4)3 ·
�
�
2x(x2 − 4)2 (x2 − 4) + 3(x2 − 1)
Factor
2x(x2 − 4)2 (4x2 − 7)
2x(x − 2)2 (x + 2)2 (4x2 − 7)
Factor
✷
Try This 7
Find the derivative of y = (3x − 1)4 (x + 4)3 .
Figure 6
The derivative f � (x) is
undefined at (±2, 0),√and
f � (x) = 0 at (0, −12 3 4).
x
114
CHAPTER 2. THE DERIVATIVE
Example 8 Evaluating a Derivative With a Radical
Find the derivative of g(x) = √
Solution
4x
.
1 + x2
We apply the Quotient Rule.
g � (x)
√
=
=
g � (x)
1 + x2
√
�
d
d ��
1 + x2
[4x] − 4x
dx √
dx
[ 1 + x 2 ]2
Quotient Rule
1 + x2 · 4 − 4x · 12 (1 + x2 )−1/2 (2x)
General Power Rule
1 + x2
=
4(1 + x2 ) − 4x2
(1 + x2 )3/2
=
4
(1 + x2 )3/2
√
1+x2
Multiply by √ 2
1+x
✷
Try This 8
Find f � (x) if f (x) =
√
1 + sin x
.
cos x
Example 9 Combining the Differentiation Rules
�
x
Find the derivative of h(x) = 3 2
.
x +1
Solution
Rewrite the given function as
h(x) =
�
x
x2 + 1
�1/3
.
Then apply the General Power and Quotient Rules as follows:
h� (x)
h� (x)
=
1
3
�
x
x2 + 1
�−2/3
=
1
3
�
x
x2 + 1
�−2/3 �
=
1
3
�
x2 + 1
x
�2/3
=
General Power Rule
(x2 + 1) · 1 − x · 2x
(x2 + 1)2
1 − x2
(x2 + 1)2
1 − x2
+ 1)4/3
3x2/3 (x2
�
�
d
x
dx x2 + 1
�
Quotient Rule
Rewrite Negative Exponent
Simplify
✷
Try This 9
�
Find h (x) if h(x) =
�
1−x
.
1+x
2.4. THE CHAIN RULE AND ITS APPLICATIONS
115
Applying the Chain Rule to Trigonometric Functions
Let u(x) be a differentiable function. We write the Chain Rule versions of the derivatives
of the six trigonometric functions.
d
du
[sin u] = cos u ·
dx
dx
d
du
[cos u] = − sin u ·
dx
dx
d
du
2
[tan u] = sec u ·
dx
dx
d
du
[csc u] = − csc u cot u ·
dx
dx
d
du
[sec u] = sec u tan u ·
dx
dx
d
du
2
[cot u] = − csc u ·
dx
dx
Example 10 Trigonometric Functions and the Chain Rule
We evaluate the derivative of certain trigonometric functions.
d
d
[cos(2x)] = − sin(2x)
[2x] = − sin(2x) · 2 = −2 sin(2x)
dx
dx
√
√ �
√ d �√ �
√
cos( x)
d �
1
√
b)
sin( x) = cos( x)
x = cos( x) · √ =
dx
dx
2 x
2 x
a)
c)
�
d �
d � 4�
tan(3x4 ) = sec2 (3x4 )
3x = sec2 (3x4 ) · 12x3 = 12x3 sec2 (3x4 )
dx
dx
✷
Try This 10
Evaluate the derivatives:
a)
�
d �
cos(x2 )
dx
b)
d
[sec(2x + 1)]
dx
c)
√ �
d �
cot( 3 x)
dx
Example 11 Applying the Chain Rule Twice
Find the derivative of y = csc2 (5x).
Solution
y
dy
dx
=
=
=
dy
dx
=
(csc(5x))2
Rewrite
d
2 csc(5x) ·
[csc(5x)]
General Power Rule
dx
�
�
d
2 csc(5x) − csc(5x) cot(5x) ·
[5x]
Chain Rule
dx
−10 csc2 (5x) cot(5x)
Simplify
✷
Try This 11
Find the derivative of f (x) = 4 sin2 (3x).
116
CHAPTER 2. THE DERIVATIVE
Summary of Differentiation Rules
Let u, v, and f be differentiable functions of x, and let c be a constant.
Basic Differerentiation
Rules
Constant Multiple Rule
d
[cu] = cu�
dx
Sum and Difference Rules
d
[u ± v] = u� ± v �
dx
Product Rule
Quotient Rule
d �u�
vu� − uv �
=
dx v
v2
d
[uv] = uv � + u� v
dx
Constant Rule
d
[c] = 0
dx
Chain Rule
Power Rule
d n
d
[x ] = nxn−1 ,
[x] = 1
dx
dx
Chain Rule
d
[f (u)] = f � (u)u�
dx
General Power Rule
d n
[u ] = nun−1 u�
dx
Derivatives of
d
[sin u] = cos(u)u�
dx
d
[cos u] = − sin(u)u�
dx
Trigonometric Functions
d
[tan u] = sec2 (u)u�
dx
d
[cot u] = − csc2 (u)u�
dx
d
[sec u] = sec(u) tan(u)u�
dx
d
[csc u] = − csc(u) cot(u)u�
dx
2.4 Check-It Out
1. If f (x) = 2x3 and g(x) = 4x + 1, then evaluate (f ◦ g)� (x).
2. Find f (x) and g(x) such that (f ◦ g)(x) = 10 csc(12x).
3. Evaluate
�
d �
3(1 − x2 )4
dx
4.
Evaluate
�
d �
sin(x2 )
dx
True or False. If false, rewrite the statement so that it becomes true.
√
1 + x2 , then f (x) =
1.
If f (g(x)) =
3.
�
�
d
1
1
=
dx (1 + x)2
2(1 + x)
5.
7.
9.
�
d �
5(π 2 + x2 )4 = 20(2x)3
dx
�
d � 2
cos (2x) = −4 sin 2x cos 2x
dx
If p(x) =
√
3
x2 − 4, then p� (0) = 0.
√
x and g(x) = 1 + x2 .
2.
4.
6.
8.
10.
�
d �
(1 + tan x)3 = 3(1 + tan x)2
dx
�
d ��
x
1 − x2 = − √
dx
1 − x2
�
d �
7(sec x + tan x)2 = 14 sec x(sec x + tan x)2
dx
1
, then f (f (x)) = x.
x
√
�
x
2
�
If s(x) =
, then s (1) =
.
x+1
8
If f (x) =
2.4. THE CHAIN RULE AND ITS APPLICATIONS
117
Exercises for Section 2.4
In Exercises 1-8, evaluate the function (f ◦ g)(x) and the derivative (f ◦ g)� (x).
1.
f (x) = 2x3 , g(x) = 4x − 5
4.
f (x) =
7.
f (x) = tan x, g(x) = x3 − 10
√
3
x, g(x) = 9 − x
√
2.
f (x) = 4x + 1, g(x) = 3x2
3.
f (x) =
5.
f (x) = 2 sin x, g(x) = 12x
6.
f (x) = −3 cos x, g(x) = 8x
8.
f (x) =
x, g(x) = 1 − 2x
csc x
, g(x) = x2
2
In Exercises 9-10, find two functions f (u) and g(x) such that f (g(x)) is equal to the given function. There are several
possible answers. See Example 2
9.
10.
y = f (g(x))
y = f (u)
u = g(x)
a) y = (3x + 1)4
y=
u=
b) y = tan(x2 )
y=
u=
c) y = 3 cos2 (5x)
y=
u=
y = f (g(x))
y = f (u)
u = g(x)
a) y =
√
2x + 1
y=
u=
�
y=
u=
c) y = (2 sin x + 1)2
y=
u=
b) y = sin
�
x
x+1
In Exercises 11-32, find the derivative with respect to x of each given function.
11.
f (x) = (3x + 1)5
12.
f (x) = (3 − x2 )4
13.
g(x) =
√
14.
g(x) =
√
3
15.
φ(x) = (2x2 + x − 7)−4/3
16.
φ(x) =
1
(x2 + 4)2/3
17.
y = tan(2x3 )
18.
g(x) = 2 cot(x3 )
19.
y = sin2 (3x)
20.
g(x) = cos2 (4x)
21.
y = 3 sec2 x + 2 tan2 x
22.
y = cot2 (3x) − 2 csc2 (3x)
23.
y=
24.
y=
1 2 1
x − cos(3x)
4
6
25.
26.
w(x) =
27.
c(x) = cot3 (sin x)
28.
r(x) = 4 tan (cos 2x)
29.
f (x) = √
30.
φ(x) =
√
31.
�
x�
y = tan2 cos
2
32.
x3 + 1
√
2x − 1 3 x + 1
1
1
sin(2x) − x2
4
2
�
�
x
m(x) = sec
x+1
7x + 4
1
csc
2
�
2x
x+1
x2
x−3
� � x ��
y = 2 tan sin
4
�
118
CHAPTER 2. THE DERIVATIVE
In Exercises 33-38, determine the points on the graph of y = f (x) where the tangent lines are vertical lines, if any.
Also, find the points where f � (x) = 0.
33.
f (x) = 4(x3 − 8)1/3
34.
36.
�
f (x) = 3 3 (x2 − 4)2
37.
1
2(8 − x3 )1/3
√
f (x) = 3 sin x
f (x) =
35.
f (x) = 3(x2 + x − 20)1/5
38.
f (x) = √
5
4
cos x
In Exercises 39-44, find the slope-intercept form of the tangent line to the graph of the function at point P .
√
39.
f (x) = x2 − 5, P (3, 2)
40.
f (x) = (3 − x)7 , P (2, 1)
41.
g(x) = 2 sin 3x, P (π/18, 1)
42.
g(x) = 16 cos3 x, P (π/3, 2)
43.
y = 12 sin2 x, P (π/6, 3)
44.
y = 12 cos(2x), P (π/6, 6)
In Exercises 45-48, evaluate the derivative of the function given g(a) = 4, g � (a) = 6, f (4) = 5, and f � (4) = 3.
45.
47.
If p(x) = (f ◦ g)(x), find p� (a) = (f ◦ g)� (a)
�
If r(x) = g(x), evaluate r� (a)
46.
If q(x) = f (2x), evaluate q � (2)
48.
if s(x) = x2 f (x), find s� (4)
Odd Ball Problems
49. Let f and g be differentiable functions that are inverse functions of each other. Apply the Chain Rule to prove
1
f � (g(x)) = �
, if g � (x) �= 0.
g (x)
50. Let f and g be differentiable functions that are inverse functions of each other. If f � (g(x)) =
second derivative g �� (x).
1
,
x3
evaluate the
51. If f (x) = x7 + x and g(f (x)) = x, then evaluate f (1) and g � (2).
52. If f (x) = 1 − x3 − x5 and y = f −1 (x) is the inverse function, then evaluate f (1) and (f −1 )� (−1).
Applied Problems
√
53. A rectangle inscribed in a semicircle y = 1 − x2 , see Figure for 53 . The base of the rectangle lies on the
x-axis, and the y-axis divides the rectangle into two equal smaller rectangles. Express the area of the rectangle
as a function A(w) where 2w is the length of the base. Then find A� (w).
y
y
�w
�w
Figure for No. 53
w
w
x
x
Figure for No. 54
54. An isosceles triangle is inscribed in a circle of radius 1, see Figure for 53 . The base of the triangle is 2w. Write
the area A(w) of the triangle as a function w. Then find A� (w).
√
√
55. Let P (x, 2 x), x ≥ 0, be a point on the graph of y = 2 x. If d(x) is the distance between P and the point
�
(11, 0), find d (x).
56. Let P (x, 1/x), x < 0, be a point on the graph of f (x) = 1/x. Denote by d(x) the distance between P and the
point (1, 1). Find d� (x).
2.5. IMPLICIT DIFFERENTIATION
2.5
119
Implicit Differentiation
• Implicitly Defined Functions • Implicit Differentiation • Finding the Second
Derivative Using Implicit Differentiation
Implicitly Defined Functions
The following functions are written in explicit form:
y = x3 − 4x2 + 5, p = sin t, and R(x) =
2x
.
x−5
In general, if an equation expresses a dependent variable in terms of an independent
variable, we say the function is given in explicit form. However, if a function is given by
an equation which is not in explicit form, the function is said to be in implicit form. In
this section, we discuss the process of finding the derivative of a function in implicit form.
For example, the equation below
xy − 1 = x2
expresses y implicitly as a function of x. We can easily solve the above equation for y and
thereby express the function in explicit form. Then we evaluate dy/dx as shown below.
Implicit Form
xy − 1 = x2
Explicit Form
y=
Derivative
x2 + 1
= x + x−1
x
dy
1
= 1 − x−2 = 1 − 2
dx
x
In general, it is not easy to solve for y in terms of x such as in
x3 + y 3 = 6xy.
In fact, it is not necessary to solve for y in order to find dy/dx. We will use the method
of implicit differentiation to find dy/dx. If a student wishes to see the derivative by
skipping some explanations, see Example 3.
Before we proceed further we recall a few elementary facts about derivatives. If y is
dy
a function of x, the derivative of y 3 is 3y 2 dx
by the General Power Rule. Moreover, the
dy
dy
derivative of sin(y) is cos(y) dx . Also, by the product rule the derivative of xy is x dx
+ y.
Note, y is the dependent variable, x is the independent variable, and we differentiate with
respect to x. Consequently, in particular the derivative of y n is
d n
dy
[y ] = ny n−1
dx
dx
and the derivative of xn is
d n
[x ] = nxn−1
dx
where n is a constant.
Example 1 The Derivative with Respect to x
Evaluate the derivative. Assume y is a differentiable function of x.
d
[x] = 1
dx
d
dy
b)
[y] =
dx
dx
d
dy
c)
[3x5 + y 3 ] = 15x4 + 3y 2
dx
dx
a)
Power Rule
Equivalent Notations
Power Rule, Chain Rule
120
CHAPTER 2. THE DERIVATIVE
d)
d
[3x2 y 2 ]
dx
d � 2�
d � 2� 2
y +
3x · y
dx
dx
=
3x2
=
3x2 · 2y
=
6x2 y
dy
+ 6xy 2
dx
=
�
�
dy
12(1 + y 2 )3 · 2y
dx
=
24y(1 + y 2 )3
Product Rule
dy
+ 6x · y 2
dx
Chain Rule
e)
d
[3(1 + y 2 )4 ]
dx
Chain Rule
dy
dx
✷
Try This 1
Let y be a differentiable function of x. Evaluate the derivative.
a)
d
[3y]
dx
b)
d � 3�
2y
dx
c)
d
[xy]
dx
�
d ��
3
1 + y3
dx
d)
Implicit Differentiation
We describe below general guidelines for finding dy/dx from an equation that implicitly
defines y as a function of x. In this section2 , unless otherwise noted we assume y is a
differentiable function of x.
The Method of Implicit Differentiation
1. Differentiate both sides of the equation with respect to x, treating y as a differentiable
function of x.
2. Collect all terms with dy/dx on one side of the equation.
3. Factor out dy/dx.
4. Solve for dy/dx.
Example 2 Differentiating Implicitly
Find dy/dx if x2 + 2y − y 3 = 1.
Solution
1. Differentiate both sides of the equation with respect to x.
d 2
[x + 2y − y 3 ]
dx
dy
dy
2x + 2
− 3y 2
dx
dx
=
d
[1]
dx
=
0
Chain Rule
2 Some equations do not implicitly define y as a differentiable function of x. For instance,
the graph of x2 + y 2 = 0 consists of exactly one point, namely, (0, 0); consequently, y is not a
differentiable function of x.
2.5. IMPLICIT DIFFERENTIATION
121
2. Collect all terms with dy/dx on one side of the equation.
2
4. Solve for dy/dx:
Find the derivative
a)
=
−2x
�
dy �
2 − 3y 2 = −2x
dx
dy
2x
= 2
dx
3y − 2
3. Factor dy/dx:
Try This 2
dy
dy
− 3y 2
dx
dx
✷
dy
.
dx
y3 − x − y = 1
b)
xy 2 = y + 2
Example 3 Using Implicit Differentiation to find the Slope
Find the slope of the graph of x3 + y 3 = 6xy at the point (3, 3).
The graph is called the folium of Descartes, see Figure 1.
Solution
y
First, we determine dy/dx.
�
d � 3
x + y3
dx
dy
3x2 + 3y 2
dx
dy
3x2 + 3y 2
dx
dy
dy
3y 2
− 6x
dx
dx
�
dy � 2
3y − 6x
dx
dy
dx
dy
dx
=
=
=
d
[xy]
dx
�
�
dy
6 x
+y
dx
dy
6x
+ 6y
dx
6
3
Sum and Product Rules
3
=
6y − 3x2
Collect all dy/dx
=
6y − 3x2
Factor dy/dx
Figure 1
The folium of Descartes
at the point (3, 3) has
slope dy/dx = −1.
2
=
=
6y − 3x
3y 2 − 6x
2y − x2
y 2 − 2x
x
Solve for dy/dx
Simplify
Next, substitute the coordinates of (3, 3) into the derivative.
Then the slope of the graph at (3, 3) is
y
2(3) − 32
dy
= 2
= −1.
dx
3 − 2(3)
1
✷
�1
Try This 3
Find the slope of the tangent line to the graph of y 2 + x2 = x − 2xy − 1 at the point
(2, −1), see Figure 2.
�2,�1�
2
Figure 2
The point (2, −1)
on the graph of
y 2 + x2 = x − 2xy − 1.
x
122
CHAPTER 2. THE DERIVATIVE
Example 4 Finding a Tangent Line
Find an equation of the tangent line to the graph of
4 sin x + 2 cos(x + y) = 1
�π π�
at the point
,
.
6 2
y
Solution
Differentiate both sides of the equation with respect to x.
d
d
[sin x] + 2
[cos(x + y)]
dx
dx �
�
dy
4 cos x − 2 sin(x + y) · 1 +
dx
4
Π
2
x
Π
6
Figure 3
The tangent line to
4 sin x + 2 cos(x + y) = 1
at the point (π/6, π/2).
=
d
[1]
dx
=
0
Chain Rule
4 cos x − 2 sin(x + y) − 2 sin(x + y)
dy
dx
=
0
−2 sin(x + y)
dy
dx
=
−4 cos x + 2 sin(x + y)
dy
dx
=
2 cos x
−1
sin(x + y)
Simplify
At the point (π/6, π/2), the slope m of the tangent line is
m
=
2 cos(π/6)
2 cos(π/6)
−1=
−1
sin(π/6 + π/2)
sin(2π/3)
=
√
2( 3/2)
√
− 1 = 1.
3/2
Using the slope-intercept form of the tangent line at (π/6, π/2), we find
y
=
mx + b
π
2
=
1·
π
3
=
b.
Hence, the tangent line is y = x +
π
+b
6
Note m = 1
π
as shown in Figure 3.
3
y
Try This 4
Find an equation of the tangent line at (0, 0) to the graph of
1
sin(x2 − y 2 ) + sin y = x
1
x
Figure 4
The graph of
sin(x2 − y 2 ) + sin y − x = 0
around (0, 0).
as seen in Figure 4.
✷
2.5. IMPLICIT DIFFERENTIATION
123
Finding the Second Derivative Using Implicit Differentiation
d2 y
The next example involves finding the second derivative dx
2 of a function that is defined
implicitly. The idea is that after finding the first derivative we apply implicit differentiation
to find the derivative of the derivative, i.e.,
� �
d dy
d2 y
=
.
dx dx
dx2
Example 5 Finding the Second Derivative Implicitly
Find the second derivative
Solution
d2 y
given x2 + y 2 = 1.
dx2
We determine the first derivative
dy
dx
�
d � 2
x + y2
dx
dy
2x + 2y
dx
dy
dx
as follows:
=
d
[1]
dx
=
0
=
−
(6)
x
y
Then evaluate the derivative of the derivative.
d2 y
dx2
dy
� �
y·1−x
d x
dx
−
=−
dx y
y2
=
�
x
y−x −
y
−
y2
=
−
=
�
Quotient Rule
Since
y 2 + x2
1
= − 3
y3
y
dy
−x
=
dx
y
Since x2 + y 2 = 1
✷
d2 y
is to differentiate both sides (6):
dx2
�
�
d
d
dy
d
[2x] + 2
y
=
[0]
dx
dx
dx
dx
Another way to find
� 2
�
d y
dy dy
2+2 y 2 +
·
dx
dx dx
=
0
d2 y
x2
+ 2
2
dx
y
=
0
d2 y
dx2
=
−
1
y
d2 y
dx2
=
−
1
y3
1+y
Divide by 2, let
�
1+
x2
y2
�
=
dy
x
=−
dx
y
−(x2 + y 2 )
y3
Since x2 + y 2 = 1
Note, the above answer agrees with the answer in the previous example.
✷
Try This 5
Find the second derivative
a)
x2 − y 2 = 1
d2 y
.
dx2
b)
y3 + y = x
124
CHAPTER 2. THE DERIVATIVE
2.5 Check-It Out
1. Evaluate the derivative. Assume y is a differentiable function of x.
a)
d
[2y + 3x]
dx
b)
�
d � 2
4y + x2
dx
�
d �
(x + y)12
dx
c)
2. Find the slope of the tangent line to the graph of 2y + 3x = y 2 at the point (0, 2).
3. Find
d2 y
if y 3 + y − x = 0.
dx2
True or False. If false, explain or show an example that shows it is false.
1.
3.
5.
7.
d
[y] = 1
dx
d � 4�
dy
3y = 12y 3
dx
dx
d �y�
y
=− 2
dx x
x
d2 � 2 �
y = 2[yy �� + (y � )2 ]
dx2
d
dx
d
dx
d
dx
d
dx
2.
4.
6.
8.
�
�
dy
x2 = 2x
dx
dy
[xy] = y + x
dx
� �
x
y − xy �
=
y
y2
[cos y] = − sin(y)y �
d � 2 2�
sin y = 4y sin(y 2 ) cos(y 2 )y �
dx
10. The slope of the graph of xy 2 + y = 3 at (2, 1) is −1/5.
9.
Exercises for Section 2.5
In Exercises 1-14, find dy/dx by implicit differentiation.
1.
x2 + y 2 = 4
2.
y 2 − x2 = 1
3.
4.
5.
x2 + 4xy + y 3 = 4
√
√
x+ y =9
6.
7.
(x + 3y)2 + 3xy = 6
8.
5x − y 2 + xy − x2 − 2 = 0
√
√
4 y − 6 x = 12
√
xy 2 + xy = 10
9.
sec(x + y) − tan(x + y) = 1
10.
cot(x + y) − csc(x + y) = 1
11.
x2 = sin(y 2 ) + sin(2y)
12.
x2 + cos(y 2 ) = cos(2y)
13.
sin−1 (x + y) = 2y Hint: sin(2y)
14.
cos−1 (xy) = x − y
In Exercises 15-20, find d2 y/dx2 .
15.
x2 + y 2 = 4
16.
y 2 − x2 = 1
17.
3y 2 = x3
18.
y 3 = x2
19.
1 3
y − y = x2
3
20.
y 2 = x3 − y
In Exercises 21-26, find dy/dx and evaluate the derivative at the given point.
21.
2x3 + y 3 = 5y, (1, 2)
23.
x=
25.
tan(y/4) = x − 1, (2, π)
y2 − 1
, (0, 1)
y2 + 1
22.
y 5 − x2 = −2x, (1, −1)
24.
x=
26.
cos(2πy) = 2y + x, (−2, 1/2)
8y
, (0, 0)
y2 + 1
2.5. IMPLICIT DIFFERENTIATION
125
In Exercises 27-34, find an equation of the tangent line to the graph of the equation at the
given point.
√
27.
(y − 1)2 = 4(x − 1), (2, −1)
28.
x2 + y 2 = 1, ( 3/2, −1/2)
√
√
29.
x2 − y 2 = 4, (−3, 5)
30.
25x2 + 9y 2 = 225, (3 3/2, 5/2)
31.
x3 + y 3 + 3y = 5, (1, 1)
32.
x3 − y 3 = 6y + 1, (2, 1)
33.
x2 y 3 = 4y − 3, (1, 1)
34.
x3 y 2 + 2 = 3y, (1, 1)
Normal and Tangent Lines
35. Find an equation of the normal line to the graph of y 2 = x3 − 2x at the point (2, 2).
36. Find an equation of the line through the point (1, 2) and normal to the graph of
y 2 = 4x.
37. A line tangent to the curve y 2 = 4x passes through the point (−2, 0). Find the point
or points of tangency.
2
3
38. At a certain
√ point P on the graph of y = x + x the normal line is parallel to
y = −x/ 2. Find the coordinates of P .
�0,c�
Odd Ball Problems
y
39. Let 0 < a < c. Show that the x- and y-intercepts of the tangent line to the graph
√
√
√
√
√
of x + y = c at the point (a, b) are a + ab and b + ab, respectively. Then
show the sum of the x- and y-intercepts is c. See Figure for 39.
40. Show that the tangent line to the ellipse
x2
y2
+
=1
a2
b2
at the point (x0 , y0 ) is
y=−
b2 x 0
b2
x+ .
2
a y0
y0
41. Find the points on the ellipse
x2
y2
+
=1
25
16
whose tangent line has y-intercept (0, 5). Hint: Use Exercise 40
42. Show that the tangent line to the hyperbola
x2
y2
− 2 =1
2
a
b
at the point (x0 , y0 ) is
y=
b2 x 0
b2
x
−
.
a 2 y0
y0
43. Find the points on the graph x3 + y 3 − 9xy = 0 (Folium of Descartes) that has a
horizontal or vertical tangent line. Use a graphing utility to sketch the graph.
44. Find the points on the graph x2 − xy + y 2 = 3 (Ellipse) where the tangent line is
horizontal or vertical. Include a sketch of the graph.
45. Verify the Power Rule when n = p/q is a rational number:
d n
[x ] = nxn−1 .
dx
Hint: Apply implicit differentiation to y q = xp
�c,0�
x
Figure for 39
√
√
√
The curve x + y = c.
126
2.6
CHAPTER 2. THE DERIVATIVE
Related Rates
• Computing a Related Rate from a Given Equation • Setting up an Equation
and Finding a Related Rate
Computing a Related Rate from a Given Equation
In the previous section, we discussed implicit differentiation which essentially is an application of the chain rule. In this section, we apply implicit differentiation to find rates of
change. In general, we may be given an equation involving variables such as x, y, and z.
Or, we must write such an equation. Then we use implicit differentiation to find the rate
of change dx/dt of one variable in terms of the rates of change of other variables.
Example 1 Related Rates and a Right Triangle
Let r, x, and y be differentiable functions of t such that
r 2 = x2 + y 2
as in Figure 1. Find
Solution
y
dr
dt
when r = 5, y = 3,
dy
dt
= 3, and
= 2.
Differentiate the given equation with respect to t:
d � 2�
r
dt
r
=
�
d � 2
x + y2
dt
2r
dr
dt
=
2x
r
dr
dt
=
x
x
Figure 1
A right triangle with
sides x and y, and
hypotenuse r that are
functions of time t.
dx
dt
dx
dy
+ 2y
dt
dt
Chain Rule
dx
dy
+y
dt
dt
Simplify
(7)
If r = 5 and y = 3, we find
52
=
x2 + 3 2
4
=
x.
For x > 0
Substituting into equation (7) we obtain
5
dr
dt
=
4(3) + 3(2)
dr
dt
=
18
.
5
Substitute known values
Solve for dr/dt
✷
Try This 1
Let A, h, and r be differentiable functions of t satisfying
y
A = r2 + 2rh.
Find dA/dt given r = 1, h = 2, dr/dt = 1/2, and dh/dt = 1/4.
Θ
50
Figure 2
A right triangle with
side y and opposite angle θ.
Example 2 Related Rates and an Angle
Let y and θ be differentiable functions of t satisfying
y = 50 tan θ.
Find the rate of change
dθ
dt
when
dy
dt
= 14 and θ = π/4. See Figure 2.
2.6. RELATED RATES
127
Solution Differentiate both sides of the given equation with respect to t, substitute the
known values, and solve for dθ/dt.
Then we obtain
dy
dt
=
50 sec2 (θ)
14
=
50 sec2
14
=
50
dθ
7
=
.
dt
50
dθ
dt
� π � dθ
4
Chain Rule
Since
dt
dy
π
= 14, θ =
dt
4
�√ �2 dθ
2
dt
✷
Try This 2
Let x and θ be differentiable functions of t, and let
10
x = 10 cos θ.
Θ
Find dx/dt if θ = π/6 and dθ/dt = 1/2. See Figure 3.
x
Figure 3
Setting up an Equation and Finding a Related Rate
In the previous two examples, we solved for a rate of change from a given equation. For the
remaining examples in this section, we set-up an equation and solve for a rate of change.
Example 3 The Surface Area of an Expanding Cube
Suppose the edges of a cube are increasing at a rate of 2 in./min. Determine the rate of
change of the surface area of the cube when an edge is 5 inches.
Solution
Let x be the length of an edge. If S is the surface area, then
S = 6x2
as seen in Figure 4. Note, dx
= 2 inches/minute.
dt
Applying the Chain Rule, we obtain
dS
dt
dx
dt
=
12x
=
�
�
in.
12 (5 in.) 2
min
=
120
in.2
min
Thus, the surface area is increasing at 120 in.2 /min when x = 5 inches.
✷
Try This 3
The edges of a cube without out a top are decreasing at the rate of 4 inches per minute.
How fast is the surface area of the cube changing when the surface area is 20 square inches?
Figure 4
The surface of a cube
is S = 6x2 where x is
the length of an edge.
128
CHAPTER 2. THE DERIVATIVE
Strategy For Solving Related Rate Problems
1.
Make sure you understand the problem. List the variables and make a sketch. The
variables may describe length, area, volume, or an angle, among others.
2.
Write an equation involving the variables from Step 1. Identify the unknown rate of
change to be determined and the given values of the variables.
3.
Differentiate the equation in Step 2 by using the Chain Rule. Usually, differentiate
with respect to t.
4.
Solve for the unknown rate of change in the differential equation from Step 3. We
may substitute the known values from Step 2 into the equation.
Example 4 Rate of Change of the Volume of a Cone
Concrete is being poured into the shape of a right circular cone at a rate of 1 ft3 /min.
How fast is the height increasing when the height is 2 ft? Assume the diameter of the base
is always equal to the height.
Solution Let V , r, and h represent the volume, radius, and height of the cone, respectively. Since 2r = h, the volume V satisfies
V
π 2
r h
3
π 3
h
12
=
=
Figure 6
The volume of a cone
with height h and radius r
π
is V = r2 h.
3
Note r2 =
� �2
h
.
2
Use the Chain Rule and differentiate with respect to t.
�
�
dV
π
dh
=
3h2
dt
12
dt
=
πh2 dh
4 dt
1
=
π
dh
dt
=
1
ft/min
π
dV
dt
dh
dt
Substitute
dV
= 1, h = 2
dt
Solve for dh/dt
Hence, when h = 2 ft, the height is increasing at a rate of
dh
1
dh
= ≈ 0.318 ft/min or
≈ 3.8 in./min.
dt
π
dt
✷
Figure 7
Water is added at a rate of
1 ft3 /hour. How fast is the
height increasing?
Try This 4
A circular cylindrical tank is being filled with water at a rate of 1 ft3 /hr. How fast is the
water level rising? Assume the radius of the tank is 2 ft, see Figure 7.
2.6. RELATED RATES
129
Example 5 Rate of Change of a Distance
2
A particle is moving along the graph
√ of y = x . Suppose the abscissa x of the particle’s
coordinates (x, y) is increasing at 10 units/minute. How fast is the distance between the
particle and the origin changing when the particle is at point (3, 9)? See Figure 8.
Solution
y
If (x, x2 ) are the coordinates of the particle’s position, then
�x,x2�
√
dx
= 10.
dt
The distance D between the origin and (x, x2 ) is
�
D =
(x − 0)2 + (x2 − 0)2
�
=
x2 + x4 .
x
Differentiating D with respect to t, we find
�1/2 �
d
d �� 2
[D] =
x + x4
dt
dt
�
�
1
dx
3 dx
√
=
2x
+ 4x
dt
dt
2 x2 + x4
=
x + 2x3 dx
√
.
x2 + x4 dt
Figure 8
The coordinates√(x, x2 )
satisfy dx/dt = 10.
Simplify
If x = 3, then
d
[D]
dt
=
3 + 54 √
√
10
9 + 81
Note
√
dx
= 10
dt
=
√
57 10
√
= 19
90
Simplify
y
�0,3�
Thus, the distance from the origin is changing at 19 units/min when x = 3.
D
✷
Try This 5
A particle is moving along the x-axis towards the right at a rate of 5 ft/sec. How fast is
the distance between the particle and the point (0, 3) increasing when the particle is at
(4, 0)? See Figure 9.
�x,0�
Figure 9
If dx/dt = 5 and
D is the length of
the diagonal, find
dD/dt when x = 4.
Example 6 Finding the Rate of Change of an Angle
A kite is flying at a constant height of 72 ft and is moving horizontally farther away at
a rate of 12 ft/sec. How fast is the angle θ between the√kite’s string and the horizontal
changing when the horizontal distance x of the kite is 72 3 ft as seen in Figure 10?
Solution
Θ
Applying right triangle trigonometry, we obtain
x
72
tan θ =
.
x
Differentiating with respect to t, we obtain
sec2 (θ)
dθ
dt
=
−
72 dx
·
.
x2 dt
72
(8)
Figure 10
In a right triangle,
opposite
72
tan θ =
=
.
adjacent
x
x
130
CHAPTER 2. THE DERIVATIVE
√
If we substitute x = 72 3 into tan θ =
72
,
x
then
tan θ
=
72
1
√ = √
72 3
3
θ
=
π
.
6
Since the kite moves away horizontally at 12 ft/sec, we write
equation (8), we find
sec2
�
� π � dθ
6 dt
2
√
3
�2
dx
= 12. Substituting into
dt
√ dx
Note x = 72 3,
= 12
dt
=
−
72
√
(12)
(72 3)2
dθ
dt
=
−
1
18
dθ
dt
=
−
1
radians/sec
24
Thus, the angle θ is decreasing at a rate of 1/24 radians/sec.
Figure 11
The volume V = 4πr3 /3
of a sphere of radius r
satisfies dV /dt = 5 ft3 /hr.
✷
Try This 6
An compressor pumps air into a spherical balloon at a rate of 5 ft3 /hr. How fast is the
radius increasing when the diameter is 1 ft? See Figure 11.
For reference we list commonly used formulas from geometry and trigonometry.
Formulas from Geometry
1. Area of a triangle A =
1
bh
2
2. Area of a circle A = πr2
θ
Area of a sector A = r2
2
3. Area of an ellipse A = πab where
4. Volume of a sphere V =
4 3
πr
3
Circumference C = 2πr,
Length of an arc s = rθ
x2
y2
+ 2 =1
2
a
b
Surface area A = 4πr2
πr2
5. Volume of a right circular cone V =
h
3
√
2
2
Lateral surface area S = πr r + h
Ah
6. Volume of a cone with base area A and height h: V =
3
√
Surface area S = πr r2 + h2
2.6. RELATED RATES
131
Trigonometric Formulas
1. cos2 α + sin2 α = 1, tan2 α + 1 = sec2 α, cot2 α + 1 = csc2 α
2. sin
π
1
π
= = cos
6
2
3
3. sin θ =
sin
√
π
3
π
=
= cos
3
2
6
opposite
adjacent
opposite
, cos θ =
, tan θ =
hypotenuse
hypotenuse
adjacent
4. sin(α ± β) = sin α cos β ± cos α sin β
cos(α ± β) = cos α cos β ∓ sin α sin β
tan α ± tan β
tan(α ± β) =
1 ∓ tan α tan β
5. sin 2α = 2 sin α cos α, cos 2α = cos2 α − sin2 α
�
�
α
1 − cos α
α
1 + cos α
=±
cos = ±
2
2
2
2
α
1 − cos α
sin α
tan =
=
2
sin α
1 + cos α
6. sin
2.6 Check-It Out
Solve the related rates problems.
dx
ds
1
dc
dr
= 3, find
when x = .
2. If c = 2πr, find
when
= 3.
dt
dt
4
dt
dt
dθ
1
dx
π
3. If x = 4 cos θ and
= , find
when θ = .
dt
3
dt
6
4. The base of a right triangle is 4 inches. If the height is increasing at a rate of 2 in./min, how fast is the area of
the triangle increasing?
1. If s = 4x2 and
True or False. If false, explain or show an example that shows it is false.
dV
= 4.
dt
2.
If A = L · W , then
dv
= 3t2 .
dt
4.
If A = πr2 and
1.
If V = 4h, then
3.
If v = t3 , then
5. If s = 2πrh,
dA
dW
dL
=L
+
W.
dt
dt
dt
dr
dA
= 4, then
= 8π.
dt
dt
dr
dh
ds
= 2, and
= 3, then
= 2π(3r + 2h).
dt
dt
dt
In numbers 6-7, assume the radius of a circle is increasing at 0.5 cm/sec.
6. The circumference is increasing at π cm/sec.
7. The area of the circle is increasing at π cm2 /sec.
In numbers 8-10, suppose a right triangle has the same base and height. Assume the base and height are each
decreasing at 0.2 ft/sec.
8. The area of the triangle is decreasing at a rate of 1 ft2 /sec when the base is 5 ft.
√
9. The hypotenuse is decreasing at a rate of 2 cm/sec.
√
10. The perimeter is decreasing at a rate of (2 + 2)/5 cm/sec.
132
CHAPTER 2. THE DERIVATIVE
Exercises for Section 2.6
In Exercises 1-8, solve the related rates equation.
1. Let A = x2 and
3
2. Let V = x and
3. Let V =
4.
5.
6.
7.
8.
4
πr3
3
dx
dt
dx
dt
and
= 2. Find
=
dV
dt
2
.
9
dA
dt
when x = 3.
Evaluate
dV
dt
= π. Evaluate
when V = 27.
dr
dt
when V = 36π.
Let S = xy, dx
= 2, and dy
= 3. Find dS
when x = 10 and y = 5.
dt
dt
dt
dh
dA
Suppose A = 12 bh, db
=
2,
and
=
3.
Find
if A = 10 and b = 4.
dt
dt
dt
dy
Let T = xy + 4xz, dx
= 2, dt = −1, and dz
= 3. Evaluate dT
if x = y = z =
dt
dt
dt
1 2
dA
dθ
π
dr
Let A = 2 r θ, dt = 5π, and dt = 4 . Find dt when A = π and r = 2.
Suppose V = πr2 h, dV
= 2π, and dh
= 2. Find dr
when V = 2π and h = 8.
dt
dt
dt
1.
Applied Problems
9. The sides of a square are increasing at 2 in./min. How fast is the area changing when the side is exactly 4 in.
10. The area of a circle is increasing at a rate of 12π ft2 /hr. How fast is the radius increasing when the area of the
circle is 9π square feet?
11. The radius of a sphere is increasing at 2 ft/hr. How fast is the volume changing when the volume is exactly
32π/3 cubic feet?
12. If the radius of a sphere is increasing at 0.5 in./min, how fast is the surface area increasing when the surface
area is 16π square inches?
13. If the edges of a cube are increasing at 2 inches per minute, how fast is the volume of the cube changing when
every edge is 6 inches?
14. Suppose the edges of a cube are increasing at 5 inches per hour. Determine the rate of change of the surface
area when the surface area is 24 square inches. Assume the cube has six faces.
15. A swimming pool has the shape of a rectangular box. Water is pumped into the pool at a rate of 12 cu. ft/min.
If the length and width of the pool are 25 ft and 16 ft, respectively, how fast is the water level rising?
16. A block of ice is in the shape of a rectangular box with a square top. At a certain instant, every edge of the
square is 10 inches, and every edge is decreasing at 1 inch/hour. Moreover, the height of the ice is 6 inches, and
the height is decreasing at 1.5 inches/hour. Find the rate at which the ice is melting at that instant.
17. A 6-ft ladder is leaning against a wall. If the base of the ladder is sliding away from the wall at 2 ft/min, how
fast is the angle between the ladder and the ground changing when the base is 3 ft from the wall? How fast
is the top of the ladder falling at the same moment? At the same instant, how fast is the area of the triangle
formed by the ladder, wall, and the ground changing?
18. A right triangle in the first quadrant is formed by the coordinate axes and the line y = mx + 1 where m < 0. If
the slope m is decreasing at a rate of 5 units/hour, how fast is the angle of the triangle formed by the hypotenuse
and the y-axis changing when m = −1?
19. The shape of a container is a right circular cylinder where the radius is equal to the height of the cylinder. The
radius and height are each increasing at 1.5 ft/hr. How fast is the volume of the container changing when the
height is 4 ft?
20. The radius and height of a right circular cylinder are increasing at 0.5 ft/hr and 4 ft/hr, respectively. How fast
is the volume changing when the radius is 2 ft and height is 3 ft?
21. Sand is being added into a right circular cone at a rate of 5 cubic feet per minute Assume that the height of
the cone is always twice the radius. How fast is the height of the cone increasing when the cone is 3 feet high?
22. A water tank is in the shape of an inverted cone with diameter and altitude, 2 meters and 3 meters, respectively.
Water leaks from the tank at the rate of 1/2 cubic meter per hour. At what depth is the decrease of the depth
one meter per hour?
23. At a certain instant, ship A is 100 miles north of ship B. Ship A sails south at a rate of 25 miles per hour and
ship B is due east at a rate of 20 miles per hour. Find the rate of change of the distance between the two ships
after three hours.
2.6. RELATED RATES
133
24. Two ships S and T leave port O at the same time along routes making an angle of 60 degrees with each other.
Ship S sails at the rate of 20 miles per hour and ship T at 30 miles per hour. How fast are they separating from
each other after one hour?
25. A particle moves along the parabola y = x(x − 1) away from the point (1, 0) in the first quadrant. Find the rate
of change of its distance from the x-axis when it is at the point (3, 6), if the abscissa of its position increases at
the rate of 3 units/sec.
2
2
26. A particle moves along the ellipse x2 + y8 = 1. Find the rate of change of its distance from the x-axis when it
is at the point (1, 2) and its abscissa is increasing at the rate of two units per second.
27. A boy flies a kite steadily at a height of 36 ft above the ground. If the kite moves horizontally at the rate of 5
ft/sec, how fast is the string being let out when 39 ft of the string is out.
28. A reconnaissance plane flies parallel to the ground at a height of 2 miles and at the rate of 4 miles per minute. If
the plane passes directly over a specific point on the ground, what is the rate of change of the distance between
the plane and the specific point after half a minute later?
29. The volume of a sphere is changing at a constant rate. Show that its surface area changes at a rate inversely as
its radius.
30. Assume that a rain drop is a perfect sphere. If the rain drop accumulates moisture at a rate proportional to its
surface area , show that the radius increases at a constant rate.
31. A 5-ft tall girl is walking away from a light post at 2 ft/sec. If the light post is 18 ft high, how fast is her shadow
lengthening?
32. How fast is the tip of the shadow in Exercise 31 moving?
33. A video camera installed at the center of a circular track is focused on an athelete running around the track. At
a certain instant, the camera is turning at a rate of 1 degree per second. If the radius of the track is 36 yards,
how fast is the athlete running at that instant?
34. According to Boyle’s Law for the expansion of gas, P V = C, where P is the number of pounds per square unit
of pressure, V is number of cubic units of volume, and C is a constant. At a certain instant, the pressure of a
gas is 3600 pounds per sq. ft. and the volume is 6 cu. ft. If the volume is increasing at the rate of 4 cu.ft./min,
find the rate of change of the pressure at that instant.
35. An expanding water ripple is formed when a stone is dropped into a pond. If the radius of the ripple increases
at a rate of 3 in./sec, find the radius when the area of the ripple is increasing at 24π in.2 /sec.
36. From a dock which is 8 feet above sea level, a man is pulling a boat by means of a rope. Assume the rope is
tied to a lock in the boat and the lock is always positioned at sea level. If he pulls the rope at a rate of 1.5
ft/sec, how fast is the boat approaching the dock when it is 6 ft from the dock?
37. One side of a triangle is increasing at 15 cm/sec and another side is decreasing at 9 cm/sec. Their included angle
is decreasing at a rate of 2 radians per second. At the instant when the sides have lengths 12 cm and 18 cm,
respectively, the included angle measures 60◦ . Determine if the area of the triangle is increasing or decreasing.
38. A lamp at the top of a 50 ft post illuminates a flagpole which is 20 ft from the post. If a boy climbs the flagpole
at the rate of 4 ft/min, how fast does his shadow move along the ground when he is 10 ft from the ground?
39. A spherical ball with a 20 cm diameter is coated uniformly with a layer of ice. If the ice melts at a rate of 30
cm3 /min, how fast is the thickness of the ice decreasing when it is 5 cm thick?
40. In Exercise 39, how fast is the exposed surface area of the ice changing?
Chapter 2 Multiple Choice Test
Choose the best answer.
1. The slope of every line parallel to the x-axis is
A. 0
B. −1
C. 1
D. ∞
2. The line through the origin that forms a 45◦ angle with the x-axis passes through
A. (1, 0)
B. (0, 1)
C. (−1, 1)
D. (−1, −1)
3. The slope of the line x + 2y + 4 = 0 is
1
1
A. 2
B.
C. −
2
2
D. −2
134
CHAPTER 2. THE DERIVATIVE
4. The line tangent to the graph of y = x3 at x = 1 is
A. y + 1 = 3(x − 1)
B. y − 1 = −3(x − 1)
C. y − 1 = 3(x − 1)
D. y + 1 = 3(x − 1)
1
�
5. If f (x) = √ , then f (2) is equal to
x
√
√
√
1
2
2
A. √
B. 2
C. −
D.
8
8
2
6. The points (2, 5) and (3, 6) are on the graph of y = f (x). Which of the following is true?
A. 3f (2) = 4 − f (3)
B. 2f (2) = 4 + f (3)
C. 2f (2) − f (3) = 7
D. 2 + f (2) = 3 − f (3)
7. The line joining the points (0, 0) and (1, −2) passes through
A. (−2, 1)
B. (−1, 2)
C. (1, 2)
D. (2, 1)
8. If f � (x) = g � (x) for all real numbers x and c is a constant, then
A. f (x) − g(x) = 0
B. f (x) + g(x) = c
C. f (x) − g(x) = c
2
D. f (x) · g(x) = c
9. If the velocity of a particle moving in a straight line is v(t) = 24 − t then the acceleration at t = 3 is
A. 6
B. −6
C. 9
D. −9
10. If f (x) = x|x|, then f � (0) is
A. ∞
B. 0
C. 1
��
11. If f (x) = x|x|, then f (0)
A. does not exist
D. −1
C. is 1
D. is −1
dy
12. If xy = 4 defines y as function of x, then
at (−4, −1) is
dx
1
1
A. −4
B. 4
C.
D. −
4
4
13. The position of a particle moving along the x-axis at time t is x(t) = t2 − 4t + 3, −∞ < t < ∞. The open
interval where the particle is moving to the left is
A. (1, 3)
B. (−2, 2)
C. (2, ∞)
D. (0, 2)
B. is 0
14. For the particle in Exercise 13, its velocity is zero at t equals
A. 0
B. 1
C. 2
D. 3
1
15. The asymptotes of the graph of f (x) =
are the lines
x−1
A. y = x and y = −x
B. y = 0 and x = 0
C. x = 1 and y = 0
D. x = 0 and y = 1
4
16. If f (x) =
x + 3x − 1
, then f � (1) equals
x
A. 2
B. 3
C. 4
x4
3
17. If f (x) =
+ x − 1, then f �� (−1) equals
2
A. 12
B. 0
C. 6
D. 7
D. −6
18. The horizontal tangents of the graph of y = sin x, 0 ≤ x ≤ 2π are at
A. x = 0 and x = π
B. x = π and x = 2π
C. x =
19. The function f (x) = tan x has discontinuities at
A. a finite number of points
B. an inifinite number of points
2
π
3π
and x =
2
2
C. No points
D. No points
D. cannot be determined
2
20. If f (x) = sec (x) and g(x) = tan (x), then
A. f � (x) + g � (x) = 0
B. f � (x) − g � (x) = 0
2
2
C. [f � (x)] + [g � (x)] = 1
D. f � (x) · g � (x) = 1
2.6. RELATED RATES
135
Investigation Projects
Derivatives, Rates of Change, and Applications
1. The size of a population of bacteria is
P (t) = 107 +
90, 000t
10 + 5t2
where t is in minutes. Find the rate of change of P (t) when t = 2 minutes. Is the
population increasing or decreasing?
2. Flu Season
A health department estimates that the number of people who will be sick with flu
at time t days from the outset of a flu season is
N (t) = 90t2 − t3 , 0 ≤ t ≤ 90.
a) How fast will the flu be spreading when t = 30 days?
1
b) How many people are anticipated to be sick on the 31st day, i.e., find N (31) −
N (30)? Compare your answer with the answer to part a).
x
c) Will the flu spread at a rate of 1500 people per day?
3. Controlling the Cost
The cost of manufacturing x units of a certain item is
2
C(x) = 1800 + 3.2x − 0.0002x dollars.
a) How fast is the cost increasing when 2000 items are manufactured?
Figure for 5
Three circles tangent
to each other with
radii y < x < 1.
b) Find the cost of the 2001st unit, i.e., find C(2001) − C(2000). Compare your
answer with the answer to part a).
c) Interpret C � (x).
y
4. Marginal Revenue
As an incentive, a seller will lower the unit price p if a buyer will purchase a large
number x of items. Suppose a relationship between price and quantity is
p = 3x + 200 − 0.02x2 .
�1, 1�
a) Find the revenue R(x) from the sale of x units.
b) Find R� (100).
x
c) Find the revenue from the sale of the 101st unit? Compare your answer with
the answer to part b).
5. Related Rates on Three Circles
Three circles of radii 1, x, and y are tangent to each other as shown in Figure 5.
The smallest circle has radius y.
Figure for 6
Osculating circle to
y = x2 at (1, 1)
a) If y < x < 1, show that the radii satisfy
1
1
√ =1+ √ .
y
x
y
b) If x is increasing at a rate of 27 cm/min, how fast is y changing when x = 0.25
cm?
12
6. Osculating Circle
Find an equation of the circle
8
2
2
(x − h) + (y − k) = r
2
that is tangent to the parabola y = x2 at P (1, 1), and the second derivatives of the
circle and parabola are equal at P . See Figure 6.
7. Multiple Common Tangent Lines
Show that the graphs of y = x2 + 4 and y 2 = x − 4 have three tangent lines in
common. Find the equations of the lines.
2
Figure for No. 7
Common Tangent Lines
x