TEACH YOURSELF VISUALLY - CALCULUS

2Algebraic Methods to Calculate Limits chapter

LIMIT INVOLVING A TRIGONOMETRIC FUNCTION

Determine lim cos x .
2x
x"π

1 Begin with the original limit statement. lim cos x
2x
x"π

2 Substitute π for x; then simplify. = c2o$sππ

= 2-π1

LIMIT OF A QUOTIENT OF RATIONAL EXPRESSIONS

5 + 3 + 7
7 - x +
Determine lim 4 x2 .
x 6
x"3

x2

1 Begin with the original limit statement. 5 + 3 + 7
7 - x + x2
lim 4 6
x x2
x"3

2 Substitute ϱ for x; then simplify. = 5+ 0 +0
7- 0 +0

= 5
7

37

Indeterminate
and
Forms ! 3 0
3 0

Sometimes when using direct substitution to calculate a limit, you may encounter expressions such

as ! 3 or 00. These are known as indeterminate forms. When you encounter indeterminate forms,
3

appropriate algebraic methods must be used to alter the form of the expression the limit of which

you are attempting to calculate.

INDETERMINATE FORM INVOLVING TRIGONOMETRIC lim sin 3x
FUNCTION x
x"0

Here, x is being replaced with the number that x is approaching— =sin ^3 $ 0h
it’s what substitution is all about. Meanwhile, the colors show that 0
an appropriate number is being substituted for the x.
=si0n 0
Determine lim sin 3x .
x = 0
x"0 0

INDETERMINATE FORM INVOLVING RATIONAL FUNCTION lim x 2 +3x
7 -2x 2
Determine lim x 2 + 3x . x"3
7 - 2x2
x"3 = 32 +3 $3

7 -2 $ 32

= -33

INDETERMINATE FORM INVOLVING RECIPROCALS 1 +x 2
x 3 +1
1 2 lim
Determine lim x + x + 1.
3 x"3
x"3
x2 +x

x2+ x 31 + 2
3 +1
= 3

32 +3

= 0 +0
0

= 0
0

38

Dealing with 2chapter
Indeterminate Forms

When you encounter an indeterminate form, you can use a variety of algebraic techniques to
determine the limits. Among these techniques are factoring and reducing, dividing by the largest
power of the variable, using the common denominator, and rationalizing the denominator (or the
numerator).

Factor and Reduce

Using this technique, you factor the numerator and denominator, cancel like factors, and then use direct
substitution to evaluate the resulting expression.

LIMIT OF A RATIONAL FUNCTION lim x2 -5x +6
x2 +3x -10
Determine lim x2 - 5x + 6 . x"2
x2 + 3x - 10
x"3 = 22 -5 $ 2 +6
22 +3 $ 2 -10
1 Try direct substitution, 2 for x.
= 4 -10 +6
4 +6 -10

= 0
0

2 Since you ended up with an indeterminate form, return to the lim x2 -5x +6
original limit statement and then factor both the numerator and x2 +3x -10
the denominator, cancel the common factor, and then use direct x"2
substitution.
=lim ^x -3h^x -2h
x"2
^ x +5h^ x -2h

=lim x -3
x"2 x +5

= 2 -3
2 +5

= -1
7

39

Dealing with Indeterminate
Forms (continued)

LIMIT OF A RATIO OF TRIGONOMETRIC FUNCTIONS

Determine lim sin x + sin 2x .
sin x
x"0

1 Try direct substitution, 0 for x. lim s in x +sin 2x
sin x
x"0

=sin ^0h +sin 2 ^0h
sin ^0h

=0 +0
0

=00

2 Again, you ended up with an indeterminate form. Return to lim sin x +sin 2x
the original limit statement and substitute 2 sinx cosx for sin x
sin2x. Then factor, reduce, and use direct substitution. x"0

=lim sin x +2 sin x cos x
x"0 sin x

=lim sin x ^1 +2 cos xh
x"0 sin x

=lim ^1 +2 cos xh
x"0

=1 +2 cos^0h

=1 +2 $ 1

=3

Divide by Largest Power of the Variable

When the limit involves a rational function, you can divide all terms by the highest power of the variable
in the rational function—or you can multiply by the reciprocal of the highest powered term instead.

LIMIT OF A RATIONAL FUNCTION

Determine lim 5x 2 + 3x - 2
7x 2 - 4x + 6
x"3

40

2Algebraic Methods to Calculate Limits chapter

1 Try direct substitution, ϱ for x. lim 5x 2 +3x -2
7x 2 -4x +6
x"3

= 5 ^3h2 +3 ^3h -2
7 ^3h2 -4 ^3h +6

= 3
3

2 Return to the original limit statement and multiply the $^5x2 +3x -2h 1
x2
numerator and denominator by 1 , x2 being the highest lim ^7x 2 -4x +6h 1
x2
x"3

powered variable term. x2

=lim 5x 2 + 3x - 2
x"3 x2 x2 + x2
7x 2
x2 - 4x 2
x2 x2

=lim 5 + 3 - 2
x"3 7 - x + x2
4 6
x x2

= 5 + 3 - 2 2
7 - 3 3
4 6
3 + 3 2

= 57 +0 -0
-0 +0

= 5
7

41

Dealing with Indeterminate
Forms (continued)

LIMIT INVOLVING A RADICAL FUNCTION

Determine lim x2 - 2x
x"3 5x - 3

1 Try direct substitution, ϱ for x. lim x2 -2x
5x -3
x"3

32 -2 ^3h
= 5^3h -3

= 33

2 Since you got another indeterminate form, multiply the 1
numerator and the denominator by 1 .
x2 x2 -2x x2
5x -3 1
TIP $lim

Use the largest power of the variable x in x"3
whatever form it appears.
For x2 - 2x, use 1 . x2

x2 =lim x2 -2x
For 5x + x3, use 1 . x"3 x2

x3 5x -3

42 x2

x2 - 2x
x2 x2
=lim
x"3 5x - 3
x x

=lim 1 - 2
x"3 x

5 - 3
x

= 1 - 2
3

5 -33

= 1 -0
5 -0

= 1
5

2Algebraic Methods to Calculate Limits chapter

Use the Common Denominator
When an expression involves rational terms, you use the least common denominator of all rational terms.

LIMIT INVOLVING RATIONAL EXPRESSIONS: EXAMPLE 1

Determine lim 1 + 1
x + 3
x"-3 x 3

1 Try direct substitution, –3 for x. 1 + 1
x 3
lim x +3

x " -3

= 1 + 13
-3 +3
-3

= 0
0

2 Since you encountered an indeterminate form, return to the lim 1 + 13
original limit. The least common denominator for all fractions is x +3
3x. Change both fractions in the numerator to this common x " -3 x
denominator.
1 b 3 l + 13 b x l
Note: Multiplying all terms in the numerator and denominator x 3 +3 x
by 3x gives you the same result that appears in Step 3 (see =lim
following page.) x " -3 x

=lim 3 + x
x " -3 3x 3x
+3
x

3 +x
3x
=lim x +3
x " -3

43

Dealing with Indeterminate 3 +x 1
Forms (continued) 3x x +3
$=lim
3 Invert and multiply, simplify the fraction, and then use direct x " -3
substitution, –3 for x.

=lim 1
x " -3 3x

=3^-1 3h

=-19

LIMIT INVOLVING RATIONAL EXPRESSIONS: EXAMPLE 2

Determine lim 1 + x 2 1
x 3 +
x"3

x2+ x

1 Try direct substitution, ϱ for x. 1 +x 2
x 3 +1
lim

x"3

x2 +x

= 1 + 2
3 3 +1

3

TIP 32 +3

= 0 +0
0
1 approaches 0, as do 1 and 1 3.
3 3+1 32 + 0
0
=

44

2Algebraic Methods to Calculate Limits chapter

2 Having encountered an indeterminate form, return lim 1 + x 2
to the original limit statement, writing all fractions x +1
in terms of the least common denominator of all x"3
of the terms: x(x + 1). x2 +x

Note: The goal is to eventually get rid of all of the =lim 1 +x 2
denominators in both the top and bottom of the x"3 x 3 +1
original fraction.
x ^x +1h

$ $1 x +1 + x 2 x
x +1 +1 x
=lim x
x"3 3

x ^x +1h

x +1 + x 2x
x ^x +1h ^x +1h
=lim
x"3 3

x ^x +1h

3x +1

=lim x ^x +1h
x"3 3

x ^x +1h

3 Invert and multiply, then simplify the resulting 3x +1 x ^x +1h
expression. x ^x +1h 3
$=lim
x"3

=lim 3x +1
x"3 3

=3^33h +1

=3

Therefore, limit does not exist.

45

Dealing with Indeterminate
Forms (continued)

Rationalize the Numerator (or Denominator)

In this technique, use the conjugate of a radical expression to calculate the limit.

RATIONALIZE THE NUMERATOR

Calculate lim xx--93.
x"9

1 Try direct substitution, 9 for x; an indeterminate form results. x -3
x -9
lim

x"9

= 9 -3
9 -9

= 0
0

2 Multiply the numerator and denominator by x +3, the x -3 x +3
conjugate of x - 3, simplify the resulting expression, and then x -9 x +3
use direct substitution. $lim

TIP x"9

` 1st term - numberj` 1st term + numberj x -9
just equals 1st term + ^numberh2. =lim

x " 9 ^x -9h` x +3j

=lim 1
x " 9 x +3

=1
9 +3

=3 1 = 1
+3 6

46

2Algebraic Methods to Calculate Limits chapter

RATIONALIZE THE DENOMINATOR lim x -2
Calculate lim x - 2 x " 2 x +3 - 5

x"2 x + 3 - 5

1 Direct substitution, 2 for x, leads to an indeterminate form.

= 2 -2
2 +3 - 5

= 0
0

2 Multiply the numerator and denominator of the $lim x -2 x +3 + 5
original expression by x +3 + 5 , the conjugate
of x + 3 - 5. Then simplify the result. x " 2 x +3 - 5 x +3 + 5

3 Use direct substitution in the resulting expression. =lim ^x -2h` x +3 + 5j
x"2
^x +3h -5

=lim ^x -2h` x +3 + 5j
x"2
^x -2h

=lim ` x +3 + 5j
x"2

= 2 +3 + 5
= 5+ 5
=2 5

47

Limits at Infinity:
Horizontal Asymptotes

This section discusses the behavior of the graph of a function as x approaches ± ϱ, in other words,
limits at infinity.

Definition of a Horizontal Asymptote

The line y = L is a horizontal asymptote f(x)
y = f(x)
of the graph of y = f(x) if either
y=L
lim f ^xh =L or lim f ^xh =L. Infrequently, x
x"3 x"3
f(x)
lim f ^xh and lim f ^xh are not the same y = f(x)
x"3 x"3
y=L
number; in which case there can be two x

different horizontal asymptotes (see the fourth f(x)
y = f(x)
example at the right.)

In the first figure, lim = L.
x"-3

In the second figure, lim = L.
x"!3

In the third figure, lim = L. y=L
x"-3 x

In the fourth figure, lim = M, but lim = L. f(x) y=L
x"-3 x"+3 y = f(x) x

y=M

48

2Algebraic Methods to Calculate Limits chapter

A FUNCTION WITH THE X-AXIS AS ITS HORIZONTAL ASYMPTOTE

Find the horizontal asymptote for the graph of f ^xh = x 3 1.
-

1 Set up the limit statement as x approaches ϱ and then evaluate the limit. lim x 3 1
-
x"3

= 3 1
3-

=0

2 At the right is the graph of f ^xh = x 3 1, with f(x)
- y = f(x)

its horizontal asymptote at y = 0. Note that the x
y=0
graph also has a vertical asymptote at x = 1, the
x=1
point at which the function is undefined, i.e., its

denominator equals 0.

Note: Substituting –ϱ for x would have given you
the same result of y = 0.

HORIZONTAL ASYMPTOTE OF A RATIONAL FUNCTION

Find the horizontal asymptote for the graph of f ^xh = 5x2 - 3x + 2 .
7x2 + 7x - 42

1 Set up the limit statement as x approaches ϱ. lim 5x2 - 3x + 2
7x2 + 7x - 42
x"3

49

Limits at Infinity: Horizontal
Asymptotes (continued)

2 If you were to use direct substitution, lim 5x2 -3x +2
7x2 +7x -42
an indeterminate form would result. x"3
Instead, divide all terms by x2, then
=lim 5x 2 - 3x + 2
simplify. x"3 x2 x2 x2
7x 2
x2 + 7x - 42
x2 x2

=lim 5 - 3 + 2
x"3 7 + x x2
7 42
x - x2

3 Now substitute ϱ for x and simplify = 5 - 3 + 2
the result. 7 + 3 - 32
7 42
3 32

= 5-0+0
7+0-0

= 57 ; therefore y = 5 is the horizontal asymptote.
7

Note: Substituting –ϱ for x would have given
you the same result of y = 75.

50

2Algebraic Methods to Calculate Limits chapter

4 At the right is the graph of f(x)

f ^xh = 5x2 - 3x + 2 with its horizontal
7x2 + 7x - 42
y = f(x)
asymptote at y =57, along with its two

vertical asymptotes at x = –3 and x = 2. y = 5
7

x

x = –3 x = 2

HORIZONTAL ASYMPTOTE OF A RATIONAL FUNCTION USING MULTIPLE LIMIT TECHNIQUES

Find the horizontal asymptote for the graph of f ^xh = 2x2 - x - 68.
x2 + 2x -

1 Set up the limit statement as x approaches ϱ. lim 2x2 - x - 6
x2 + 2x - 8
x"3

2 If you used direct substitution, you would encounter an lim 2x2 -x -6
indeterminate form. Next factor the numerator and x2 +2x -8
denominator and then reduce the resulting expression. x"3

=lim ^2x +3h^x -2h
x"3
^x +4h^ x -2h

=lim 2x +3
x"3 x +4

51

Limits at Infinity: Horizontal =lim 2x + 3
Asymptotes (continued) x"3 x x
x 4
3 If you were to directly substitute at this point, an indeterminate form x x
again would result. Instead, divide all terms by x, and then simplify.

+

=lim 2 + 3
x"3 1 + x
4
x

4 Now use direct substitution, ϱ for x. =2 + 33
Note: Substituting –ϱ for x would have given us the same result of y = 2. 1 4
+ 3

=12 +0
+0

=2

5 The graph of f ^xh = 2x2 - x - 6 is shown f(x)
x2 + 2x - 8

at the right with its horizontal asymptote at y = f(x)

y = 2.

Note the “hole” in the graph at x = 2. In Step y=2
2 (see preceding page), the factor x – 2 was
cancelled, so x =/ 2. x

2

x = –4

52

2Algebraic Methods to Calculate Limits chapter

FUNCTION WHOSE GRAPH HAS 2 HORIZONTAL ASYMPTOTES

Find the horizontal asymptote(s) for the graph of f ^xh = 3x + 2 .
x2+ 3 - 1

1 Set up the limit statement as x approaches ϱ. lim 3x +2
x " 3 x 2 +3 -1

2 Direct substitution would result in an indeterminate form. Divide lim 3x + 2
all terms by x and then simplify where possible. x x

x"3 x2 +3 -1
x

=lim 3 + 2
x"3 x

=lim x 2 +3 - 1
x"3 x2 x

3 + 2
x

1 + 3 - 1
x2 x

3 Last, substitute ϱ for and simplify. 3 + 2
3
=
3 1
1 + 32 - 3

= 3 +0
1 +0 -0

=3

So, y = 3 is a horizontal
asymptote.

53

Limits at Infinity: Horizontal f ^xh = 3x + 2
Asymptotes (continued) x2+ 3 - 1

4 Let’s take another look at the original function. f ^xh = positive number so the
positive number
a) x → +ϱ
horizontal asymptote is y = 3.
b) but as x → –ϱ
f ^xh = negative number so the
positive number

horizontal asymptote is y = –3.

5 At the right is the graph of f ^xh = 3x + 2 1 with its f(x)
x2+ 3 - y = f(x)

two horizontal asymptotes, y = 3, and y = –3. y=3
x

y = –3

54

3chapter

Introduction to
the Derivative

A major topic in the study of calculus is What Can Be Done With
differentiation, the process of finding a a Derivative?. . . . . . . . . . . . . . . . . . . . . 56
derivative. After introducing some common uses
of a derivative, this chapter shows how the formal Derivative as the Slope of
definition is used to compute derivatives. The a Tangent Line . . . . . . . . . . . . . . . . . . . 58
chapter then covers some alternate notations for the
derivative, and discusses a variety of applications Derivative by Definition. . . . . . . . . . . . . 60
using the derivative. Finally, this chapter concludes
with the relationship between differentiability and Find the Equation of a Line
continuity. Tangent to a Curve. . . . . . . . . . . . . . . . 67

Horizontal Tangents. . . . . . . . . . . . . . . . 68

Alternate Notations for a Derivative . . . 70

Derivative as a Rate of Change. . . . . . . 72

Differentiability and Continuity . . . . . . 74

What Can Be Done f(x)
With a Derivative? T

This section introduces some common uses for a derivative.

FIND SLOPE OF A TANGENT LINE
You can use a derivative to find the slope of a
line tangent to the graph of a function at a given
point P.

y = f(x)
x

MAXIMUM AND MINIMUM ON GRAPH OF A f(x) y = f(x)
FUNCTION
Rel. Max.
You can also use a derivative to find points on the inc.
graph of a function where the relative maximum
and relative minimum occur. dec.

Intervals on the graph that are increasing or inc.
decreasing can also be found.
Rel. Min. x

56

3Introduction to the Derivative chapter

ANALYZE RATES OF CHANGE V =sphere 4 πr3
3
You can use a derivative to analyze rates of change. Given the
dV 2 dr
dt = 4πr dt
formula for the volume of a sphere, you can use the derivative to

relate the rate of change of the volume, dV , to the rate of change of
dt
its radius, dr .
dt

ANALYZE MOTION ON A OBJECT s(t) = t3 – 5t2 + 7t – 15
v(t) = 3t2 – 10t + 7
Given a function, s(t), that describes the position of an object, a(t) = 6t – 10
you can use derivatives to find both the velocity function, v(t),
and the acceleration function, a(t).

OPTIMIZE WORD PROBLEMS x 10 x
10 – 2x
Let’s say that equal squares are cut from each 6 – 2x x x 6
corner of a rectangular sheet of metal which is x x
10 inches by 6 inches. After removing the
squares at each corner, the “flaps” are folded up x x
to create a box with no top.

You can use a derivative to find the size of each
square to be removed so that the resulting box
has the maximum volume.

x 6 – 2x
10 – 2x

Volume Box = x (10 – 2x)(6 – 2x)

57

Derivative as the Slope f(x)
of a Tangent Line f(x) = x2 – 3x

Find the slope of the line tangent to the graph of 2P x
f(x) = x2 – 3x at the point with x-coordinate 2. T f(2 +Dx) – f(2)
1 Select a point P that is ∆x units to the right of
Dx
point T.
2 The coordinates of point T are _2, f ^2hi. slope of TP = y2 --xy11
x2
The coordinates of point P are ( 2 + ∆x, f(2 + ∆x) )
= f ^2 +Dxh -f ^2h
3 Find the slope of the secant line TP. ^2 +Dxh -2

4 As ∆x → 0, the slope of the secant line TP gets The slope of the tangent line at
closer and closer to the line tangent at x = 2.
point T =lim f ^2 +Dxh -f ^2h
Dx " 0 ^2 +Dxh -2 .

58

3Introduction to the Derivative chapter

5 Find f(2 + ∆x) and f(2) using f(x) = x2 – 3x, =lim $9^2 +Dxh2 -3^2 +DxhC -722 -3 2A
then simplify. Dx " 0
2 +Dx -2

$=lim 4 +4 Dx +^Dxh2 -6 -3Dx -4 +6
Dx " 0 Dx

=lim Dx +^Dxh2
Dx " 0 Dx

6 Factor out ∆x, cancel like factors, then substitute 0 for ∆x. =lim D x ^1 +Dxh
Dx " 0 Dx

=lim ^1 +Dxh
Dx " 0

=1 +0

=1

7The last expression, 1, above on the right, is the slope of Therefore, the slope of line
the line tangent to the graph of f(x) = x2 – 3x at the point tangent at x = 2 is 1.
where x = 2.
f l^2h =lim f ^2 +Dxh -f ^2h
The limit process used above to find the slope of the line Dx " 0 Dx
tangent at x = 2 is called the derivative of f(x) at x = 2,
denoted as f'(2) (read “f prime of 2”).

59

Derivative
by Definition

As you can see from the previous example, when you replace the ∆x with an h and the 2 with an
arbitrary number c, you end with a formal definition of the derivative of f(x) at x = c.

FIRST FORM OF THE DEFINITION f l^ch =lim f ^c +hh -f ^ch
This is the first of the two most common definition forms. h"0 h
The derivative of a function f at a number c, denoted by
f '(c), is given by the statement to the right. f(x) (c + h, f(c + h))

As h → 0, point P gets closer and closer to point T. P y = f(x)
Note: The process of finding a derivative is called
differentiation.

(c, f(c)) f (c + h) – f(c)

T

slope of TP = f ^c +hh -f ^ch h
^c +hh -c

= f ^c +hh -f ^ch
h

slope of line tangent at point T = lim f ^c +hh -f ^ch =f l^ch x
h c c+h
h"0

60

3Introduction to the Derivative chapter

Determine the Derivative of a Specific Function at a Specific Number

USING THE FIRST FORM OF THE DEFINITION
Using the definition above, find the derivative of f(x) = x2 – 5x + 3 at x = 2; that is, find f '(2).

1 Set up the limit statement from the definition. f l^2h =lim f ^2 +hh -f ^2h
h"0 h

2 Find f(2+h) and f(2) by using =lim $9^2 +hh2 -5^2 +hh +3C -722 -5 2 +3A
f(x) = x2 – 5x + 3. h"0
h

3 Expand the numerator and simplify the =lim 4 +4h +h2 -10 -5h +3 -4 +10 -3
resulting expression. h"0 h

Therefore, f '(2) = –1 -h +h 2
h
Note: –1 is actually the slope of the line =lim
tangent to the graph of f(x) = x2 – 5x + 3 at h"0
the point with x-coordinate 2.
=lim h^-1 +hh
h"0 h

=lim ^-1 +hh
h"0

=-1

61

Derivative by
Definition (continued)

SECOND FORM OF THE DEFINITION f(x) (x, f(x))

The derivative of a function f at a number c, denoted by P y = f(x)
f '(c) is given by:

f l^ch =lim f ^xh -f ^ch (c, f(c)) f(x) – f(c)
x"c x -c
T

As x → c, the point P gets closer and closer to point T. x–c

slope of TP = f ^xh -f ^ch
x -c
x
Slope of line tangent at point T cx

=lim f ^xh -f ^ch =f l^ch
x"c x -c

Find the Derivative of a Specific Function at a Specific Number

USING THE SECOND FORM OF THE DEFINITION
Using the definition above, find the derivative of f(x) = x2 – 5x + 3 at x = 2, that is, find f '(2).

1 Set up the limit statement from the second form of the f l^2h =lim f ^xh -f ^2h
derivative definition. x"2 x -2

62

3Introduction to the Derivative chapter

2 Find f(x) and f(2) using f(x) = x2 – 5x + 3, and then =lim $7 x2 -5x +3A -722 -5 2 +3A
simplify. x"2 x -2

=lim x2 -5x +3 -4 +10 -3
x"2 x -2

=lim x2 -5x +6
x"2 x -2

3 Direct substitution leads to the indeterminate form 0/0. Instead, =lim ` x -2j^x -3h
factor the numerator, cancel the common factor, and then use x"2
direct substitution. x -2

Therefore, f '(2) = –1 =lim ^x -3h
x"2
Note: This is the same result as finding f'(2) using the first form of
the definition of the derivative. =2 -3

=-1

DERIVATIVE OF A SPECIFIC POLYNOMIAL FUNCTION f l^ch =lim f ^x +hh -f ^xh
For f(x) = 3x2 – 12x + 9, find f '(x), the derivative at any point. h"0 h

1 Set up the limit process.

2 Find f(x + h) and f(x) using f(x) = lim 93^x +hh2 -12 ^x +hh +9C -73x2 -12x +9A
3x2 – 12x + 9, expand the
h"0 h
numerator, and then simplify.
3x 2 +6xh +3h 2 -12x -12h +9 -3x 2 +12x -9
h
=lim
h"0

6xh +3h 2 -12h
h
=lim
h"0

63

Derivative by =lim h^6x +3h -12h
Definition (continued) h"0 h

3 Factor out h in the numerator, cancel like factors, and then =lim ^6x +3h -12h
finish up with direct substitution. h"0

Therefore f'(x) = 6x – 12 =6x +3 $ 0 -12

DERIVATIVE OF A RADICAL FUNCTION =6x -12
Find f' (x) for f ^xh = x + 3
1 Begin with the limit statement from the derivative f l^xh =lim f ^x +hh -f ^xh
h"0 h
definition.

2 Using f ^xh = x + 3, find f(x + h) and f(x). =lim x +h +3 - x +3
h"0 h
3 Direct substitution leads to an
indeterminate form. In this case, x +h +3 - x +3 x +h +3 + x +3
multiply the numerator and h x +h +3 + x +3
denominator by x +h +3 + x +3, $=lim
the conjugate of the numerator. h"0

64

3Introduction to the Derivative chapter

4 Expand the numerator and simplify it, but leave the =lim ^x +h +3h -^x +3h
denominator as is. h " 0 h ` x +h +3 + x +3j

TIP =lim h` x +h +3 -x -3
h"0 x+h +3 + x +3j
The product,
` 1st term - 2nd termj` 1st term + 2nd termj, =lim h
is just 1st term – 2nd term.
h " 0 h` x +h +3 + x +3j

=lim 1

h " 0 x +h +3 + x +3

5 Last, use direct substitution, 0 for h. = 1
x +0 +3 +
x +3

=1
x +3 + x +3

=1
2 x +3

Therefore, f l^xh = 1
2 x +3

TIP

Definition:
If the derivative of a function (or the derivative
at a number) can be found, the function is said
to be differentiable.

65

Derivative by
Definition (continued)

DERIVATIVE OF A GEOMETRIC FORMULA 4
3
The volume of a sphere with radius r, is given by: V = πr 3

Find V'(r). V l ^r h =lim V ^r +hh -V ^r h
1 Set up the limit process. h"0 h

2 Find V(r + h) and V(r) using V = 4 πr 3 4 π ^r +hh3 - 4 πr 3
3 3 3
=lim
h"0 h

3 Factor out the 4 π, expand the numerator, and 4 ^r +hh3 -r 3
3 3 h
$= π lim

then simplify. h"0

$=4 r 3 +3r 2 h +3rh 2 +h 3 -r 3
3 h
π lim

h"0

4$= 3r 2 h +3rh 2 +h 3
3 h
π lim

h"0

4 Factor out h in the numerator, simplify and then use $= 4 π lim h_3r 2 +3rh +h2i
direct substitution, 0 for h. 3
h"0 h
66
$= 4 π lim _3r 2 +3rh +h2i
3
h"0

$= 4 2 0 +^0h2k
3 π a3r +3r

= 4 π $ 3r 2
3

=4πr 2

Therefore, V' (r) = 4πr2

Find the Equation of a 3chapter
Line Tangent to a Curve

Let’s say you want to find an equation of the line tangent to the graph of f(x) = x3 – 6x2 + 9x – 13 at
the point with x-coordinate 2.

To find the equation of a line, two things are needed: a slope and a point on the line.

1 To find the slope of the tangent line, use f'(x), found in the f(x) = x3 – 6x2 + 9x – 13
“Derivative of a Cubic Polynomial” example earlier in this f '(x) = 3x2 – 12x + 9
chapter

2 Substitute 2 for x, in f '(x), the derivative. f '(2) = 3 • (2)2 – 12 • 2 + 9
f'(2) = 12 – 24 + 9
f'(2) = –3
–3 is the slope of the line

tangent at x = 2

3 Next find the y-coordinate of the point with x-coordinate 2. f(x) = x3 – 6x2 + 9x – 13
Substitute 2 for x in the original function f(x).
f(2) = 8 – 24 + 18 – 13
The point is (2, –11) f(2) = –11

4 Last, find the equation of the line having slope –3 and containing y – y1 = m(x – x1)
point (2, –11). y – (–11) = –3(x – 2)

Therefore, y = –3x – 5 is the equation of the line tangent to the y + 11 = –3x + 6
graph of f(x) = x3 – 6x2 + 9x – 13 at the point with x-coordinate 2.
y = –3x – 5

67

Horizontal
Tangents

For many problems in calculus, you need to locate the horizontal tangent to a curve. An example
of such a problem is finding the maximum and minimum values or a function. The slope (that is,
derivative) at the point of tangency will be zero.

Find Points on a Curve at Which Tangent Line is Horizontal

Find the coordinates of each point on the graph of f(x) = x3 – 6x2 + 9x – 13 at which the tangent line is
horizontal.

1 The slope of the tangent line is given by f '(x). f(x) = x3 – 6x2 + 9x – 13
f ' (x) = 3x2 – 12x + 9

2 The slope of the horizontal tangent is 0. Set f'(x) = 0 0 = 3x2 – 12x + 9
and solve for x. 0 = 3(x2 – 4x + 3)
0 = 3(x – 1)(x – 3)
3 Find the y-coordinates for the points with so x = 1 or x = 3
x-coordinates 1 and 3.
These are the x-coordinates of
68 the points at which the tangent
line is horizontal.

f(x) = x3 – 6x2 + 9x – 13
f(1) = 13 – 6 • 12 + 9 • 1 – 13
f(1) = –9
f(3) = 33 – 6 • 32 + 9 • 3 – 13
f(3) = –13

The points on the graph of
f(x) = x3 – 6x2 + 9x – 13
at which the tangent lines are
horizontal are (1, –9) and (3, –13)

3Introduction to the Derivative chapter

4 At the right is the graph of f(x) = x3 – 6x2 + 9x – 13 f(x) f(x) = x3 – 6x2 + 9x – 13
with horizontal red lines at (1, –9) and (3, –13).
13 x
Note: In Chapter 8 you will study the larger topic of
relative extrema — the maximum and minimum
values of a function. In this example, –9 is a relative
maximum for f(x), while –13 is a relative minimum.

(1, –9)
(3, –13)

69

Alternate Notations
for a Derivative

There are many ways to indicate finding the derivative of the given function y = f(x). Listed below
in the left column are some directions you may encounter when doing a calculus problem. In the
right column is the notation you would use as you write out your solution (as well a “pronunciation
guide”).

• Find the derivative of f(x). f '(x) read “f prime of x”
• Find the derivative of f(x).
• For y = f(x), find the derivative of y. d/dx f (x) read “dee dee x of f(x)
• For y = f(x) (that is, y is a function of x), find the
y' read “y prime”
derivative of y.
dy/dx read “dee y dee x”

or “the derivative of y
with respect to x”

Earlier in this section, for the function f(x) = 3x2 – 12x + 9, it was For f(x) = 3x2 – 12x + 9
determined that f'(x) = 6x – 12. You can write this fact in many
ways. f '(x) = 6x – 12

or

d f ^xh =6x -12
dx

For y = 3x2 – 12x + 9

y' = 6x – 12

or

dy =6x -12
dx

70

3Introduction to the Derivative chapter

Some notations for the derivatives mentioned below are given in the right column.

• Second Derivative ym, f m ^ xh, d2 y , d2 8 f ^ xhB
dx 2 dx 2

• Third Derivative yn, f n ^ x h, d3 y , d3 8 f ^xhB
dx 3 dx 3

• nth Derivative y(n) , f (n)^ xh, dmy , dn 8 f ^ xhB
dxm dx n

for n $ 4

71

Derivative as a
Rate of Change

The derivative notation dy (or in other problems dx , ddyt , dV , ddhy ) can also be interpreted as a rate
dx dz dr

of change. A few examples of derivative as a rate of change are air being pumped into a balloon,

and the velocity and acceleration of a moving object.

BALLOON PROBLEM

Air is being pumped into a spherical balloon at the rate of 8π cubic inches per minute. Find the rate of
change of the radius at the instant the radius is 2 inches.

1 The information you are given is that r = 2 and dV =8π.
dt

2 Start with the formula for the volume of a sphere with radius r. V ^r h = 4 πr 3
3

3 You determined V'(r) in a previous example in this chapter, labeled as V'(r) = 4πr2
“Derivative of a Geometric Formula.”
dV = 4πr 2
4 Use an alternate notation for V'(r). dr

72

3Introduction to the Derivative chapter

5 Multiply both sides of the equation by the term dr. dV = 4πr2dr

6 Divide both sides of the equation by the term dt (dt on the left and dt dV =4πr 2 dr
on the right. dt dt

7 Using the given information, substitute 8π for dV and 2 for r, and 8π =4π $ 2 2 $ dr
dt dt
dr
then solve the resulting equation for dt . $ dr
dt
1 8π =16π
2
When the radius is 2 inches, the radius is changing at the rate of 8π = dr
16π dt
inches per minute.
1 dr
2 = dt

73

Differentiability
and Continuity

This chapter concludes with some comments on the relationship between differentiability and
continuity.

When a Function Fails to Be Differentiable

The graph of a function can reveal points at which the function fails to be differentiable. This can occur
at points at which the graph has a sharp turn, a vertical tangent, a “jump,” or a “hole.”

A GRAPH WITH A SHARP TURN f(x)
To the right is the graph of f ^xh = 1 -x2 , and a comment f(x) = |1 – x2|
about its differentiability at x = ±1.
x
A GRAPH WITH A VERTICAL TANGENT LINE
–1 1
1
f(x) is not differentiable at
To the right is the graph of f ^xh =x 3 , and a comment about its x = –1 and x = 1, since the
differentiability at x = 0.
slopes left and right of
each of these numbers are

not equal.

f(x)

1

f(x) = x3

0x

f(x) is not differentiable
at x = 0, because f has a
vertical tangent at x = 0.

74

3Introduction to the Derivative chapter

A GRAPH WITH A “JUMP” f(x) f(x) = 2 if x Ͼ 2
2 if x > 2 1 if x Յ 2

To the right is the graph of f ^xh*-1 if x # 2 and a comment 2
about its differentiability at x = 2. 1
x
2

f(x) is not differentiable at
x = 2 because it is not

continuous at x = 2 (there
is a “jump” in the graph

at x = 2).

A GRAPH WITH A “HOLE” f(x)

x 2 -2x -3 f(x) = x2 – 2x – 3
x -3 x –3
To the right is the graph of f ^xh = and a comment

about its differentiability at x = 3.

Note: The function at the right is not defined for x = 3, because it x
would result in a zero denominator. 3

f(x) is not
differentiable at x = 3

because there is a
“hole” in the graph.

75

Differentiability and
Continuity (continued)

Relationship between Differentiability and Continuity

If the function f is differentiable at x = c (that is in other words, f'(c) exists there), then f is continuous at
x = c.
Below are some graphs that we can analyze differentiability and continuity at a given point.

c f(x)
y = f(x)
y = f(x)
f(x) is differentiable at x
x = c and f(x) is c

continuous at x = c. f(x) is not differentiable at
x = c, but f(x) is continuous

at x = c.

f(x) f(x)
y = f(x)
y = f(x)
x cx
c
f(x) is not differentiable f(x) is not differentiable at
at x = c, but f(x) is x = c, and f(x) is not
continuous at x = c. continuous at x = c.

76

4chapter

Derivatives
by Rule

In Chapter 3 you used the limit definition to find Derivatives of Constant, Power,
derivatives. In this chapter you will start to make and Constant Multiple . . . . . . . . . . . . 78
use of many rules of differentiation, which enable
you to find derivatives without using the time- Derivatives of Sum, Difference,
consuming limit definition for derivatives. Polynomial, and Product . . . . . . . . . . . 80

Chapter 3 also introduces L’Hôpital’s Rule — The General Power Rule . . . . . . . . . . . . 84
another technique to help you find limits that are
one of the indeterminate forms. The Quotient Rule . . . . . . . . . . . . . . . . . 86

Rolle’s Theorem and the Mean
Value Theorem . . . . . . . . . . . . . . . . . . . 89

Limits: Indeterminate Forms and
L’Hôpital’s Rule . . . . . . . . . . . . . . . . . . 93

Derivatives of Constant, Power,
and Constant Multiple

This section introduces the Constant, Power, and Constant Multiple rules for finding derivatives
and then includes an application of finding the equation of a line tangent to a curve.

The Constant, Power, and Constant Multiple Rules

THE CONSTANT RULE d ^ch =0
dx
If c is a constant, then d ^ch =0.
dx d
dx `5π 3j =0

THE POWER RULE d _ x 5i =5x 4
dx
If n is any rational number, then

$d _ x ni =n x n -1 . If y = x, then

dx

yl= d ` xj
dx

= d a x 1 k = 1 x- 1 = 1
dx 2 2 2 2

x

If f ^xh = 1 , then
x3

f l ^ x h = d c 1 m = d _ x -3i =-3x -4 = -3
dx x3 dx x4

78

4Derivatives by Rule chapter

THE CONSTANT MULTIPLE RULE

If c is a real number, and f(x) is a differentiable _5x$ $d3i = 5 d _ x 3i = 5 3x2 =15x2
dx
function, then d 8c $ f ^ xhB = c $ ; d _ f ^ xhiE. dx
dx dx

Just move the constant in front of the variable If y= 4 x, then
function. Next, multiply the constant by
the function’s derivative. dy = d ` 4 xj= 4 $ d ` x j
dx dx dx

$ $ $=4d a x - 1 k = 4 1 x - 1 = 2
dx 2 2 2

x

$ $dJ3 6 N d J 3 1 N 6 d f 1 p
KK x2 OO =6 dx KK x2 OO= dx
dx P P 2
L L
x3

$ $ $=6 d a x - 2 k = 6 -2 x- 5
dx 3 3 3

= -4 = -4 or -4
3 x5 3 x2
5
x3

79

Derivatives of Sum, Difference,
Polynomial, and Product

This section introduces the rules for finding the derivatives of sums and differences of functions,
the derivative of a polynomial function, and the derivative of the product of two functions.

The Sum/Difference Rule d 8 f ^ xh ! g ^ xhB = d f ^xh ! d g^xh
dx dx dx
If f and g are both differentiable functions, then:
d 8 f ^ xh ! g ^ xhB =f l ^ xh ! gl ^ xh
Note: A function is differentiable if its derivative dx
can be found.
This can also be written as:

Or, using a popular shorthand notation, this could be written as: d _ f ! gi =f l! gl
dx

d _ x 3+ 6x 2 i
dx

= d _ x 3i + d _6x 2 i
dx dx

=3x2 +12x

FIND THE DERIVATIVE OF A POLYNOMIAL FUNCTION If f(x) = anxn + an – 1xn – 1 +...+a2x2 + a1x + a0,
then f '(x) = n • anxn – 1 + (n – 1) •
Using repeated applications of the first 4 an – 1xn – 2 +...+ 2 • a2x + a1
differentiation rules—Constant, Constant
Multiple, Sum/Difference, and Power—you If f(x) = 5x3 – 6x2 + 9x – 13,
can now find the derivative of any Then
polynomial function.
f '(x) = 5 • 3x2 – 6 • 2x + 9
= 15x2 – 12x + 9

80

4Derivatives by Rule chapter

The Product Rule g ^ xhB =SSRSSHdddxer. fof^1xsth D2nd V 6 4d4er.7of 244nd 8 E1st
g^xh W +; d f ^xh
If f and g are both differentiable $ $ $d8f ^ xh W dx g ^ xhE
functions, then: WW
dx

TX

This can also be written as: d 8 f ^xh $ g ^ xhB =f l ^ xh $ g ^ xh +gl ^xh $ f ^ xh
dx

Using shorthand d _ f $ gi =f l$ g +gl$ f
notation, it can also be dx
written as:
If h ^xh =_3x2 -5xi^4x +7h,

4d4er.7of 24nd4 8
$ $6 4 4d4er.7of 41st 44 8 6 6 44 71st 44 8
then hl ^ x h =; d _3x 2 -5xiE H2nd +; d ^4x +7hE _3x2 -5xi
dx ^4x +7h dx

=^6x -5h^4x +7h +4 _3x2 -5xi
=24x2 +42x -20x -35 +12x2 -20x
hl^xh =36x2 +2x -35

If you had first found the product of the two functions, h(x) = (3x2 – 5x)(4x + 7)
you would have: h(x) = 12x3 + x2 –35x,
then find the derivative
h'(x) = 36x2 + 2x – 35

81

Derivatives of Sum, Difference, h ^xh = x_3x2 - 6xi
Polynomial, and Product (continued)

ANOTHER PRODUCT RULE EXAMPLE
Find the derivative of h ^xh = x_3x2 -6xi.
1 Start with the original function, rewriting x as a power.

= x 1 _3x 2 - 6x i
2

2 Find h'(x) using the Product Rule. h ^ x h =x 1 _3x 2 -6x i
2

$ $Hder. of 1st D1st
hl ^ x h =; 1 x- 1 E 6 44 72nd 44 8 Hder. of 2nd 1
2 2 _3x 2 -6xi +^6x -6h
ax 2k

3 Next, expand the terms on the right and simplify. = 3 x 3 - 3x 1 + 6x 3 - 6x 1
2 2 2 2 2

= 15 x 3 - 9x 1
2 2 2

4 To finish up, factor 1 x 1 out of each term, and simplify to = 1 1 ^15x - 18h
2 2 2
x2

complete the process.

hl^xh = x ^15x -18h
2

82

4Derivatives by Rule chapter

ANOTHER LOOK AT THE PREVIOUS PROBLEM
Find the derivative of h ^xh = x_3x2 - 6xi, without the Product Rule.

1 Start with original function, rewriting x as a power; h ^xh = x_3x2 - 6xi
then distribute.
1

= x 2 _3x2 - 6xi

53

h ^xh = 3x 2 - 6x 2

2 Now find h'(x) by finding the derivative of each term; $ $hl^xh = 3 5 x 3 - 6 3 x 1
then simplify. 2 2 2 2

= 15 x 3 - 18 x 3
2 2 2 2

3 As in the previous problem, factor 1 x 1 out of each =125 x 3 -128 x 1
2 2 2 2

term, and simplify to complete the process. = 1 x 1 -18h
2
2 ^15x

= x ^15x -18h
2

hl^xh = x ^15x -18h
2

83

The General
Power Rule

The General Power Rule gives you a means to find the derivative of the power of any function. It is
a special case of the Chain Rule, which will be introduced in Chapter 5.

General Power Rule 64$ 4474448 H^ hn
n -1 der. of inside
If f is a differentiable function and n is a rational
number, then: inside function function

n n -1 d
dx
^ xhB 8 f ^xhB
$ $d8f = n f ^xh

dx

This can also be written as: $ ^ hn -1 der. of inside
Using shorthand notation, you can also write:
n inside function Hffulnc^tixonh
6 4 44 7 4 44 8
n n -1

^ xhB 8 f ^xhB
d 8 f = n

$ $dx

$ $d 7 f n A =n f n -1 fl

dx

6 44 7f n 4 8 ?n Hf n -1 ?f l
^2x -3h10 =10 ^2x -3h9 2
$ $d

dx

=20 ^2x -3h9

84

4Derivatives by Rule chapter

It is usually helpful to write a radical expression as a If f ^xh =3 4x +5, then
power before attempting to compute its derivative.
f l ^ x h = d 3 4x +5
After writing the radical expression in the dx
denominator as a power, move it up to the
numerator to avoid having to use the General = d ^ 4x +5h 1
Power Rule. dx 3

?n 6 44$^ hn -17 44 8 der. of
=13^4x inside+5h-
2 Cinside
3 4

= 4

2

3^4x +5h 3

f l^xh = 4

2
3 3 4x +5

d J x 5 N = d 5 1
dx KK 2 +3 OO dx +3i 2
L P _ x 2

$= d =5 -1
dx
_ x2 +3i 2 G

$=5 d =_ x 2 - 1 G
dx 2
+3i

R ^ hn -1 V
S inside Wder. of
S ?n 6 8 W
S– 447 44 Cinside W
S 3 ^2xh
1 - 2
2
3i
$ $=5 _ x 2 +

W
SS WW
TX

85

The Quotient
Rule

Like the Product Rule, the Quotient Rule involves putting pieces in the right places in the right
formula and then simplifying the resulting expression. As in the Product Rule, your Algebra skills
will be put to the test.

Statement of the Quotient Rule $ $6 4d4er.7of t4o4p 8
Dbottom 6de44r. o7f bo4t4tom8 Dtop
If f and g are differentiable d g^xh d g^xh
functions, and g(x) ≠ 0, then: ; dx f ^ xhE - ; dx g ^ xhE

d e f ^xh o = 9g ^xhC2
dx g ^xh

\
^ h2

bottom

$ $d

dx
This can also be written as: e f ^xh o = f l ^ xh g ^xh -gl^xh f ^xh
Or in shorthand notation, it can g ^xh
be written as: 2

86 8 g ^ xhB

d d f n = f l$ g -gl$ f
dx g 2

_gi

If h ^xh =35xx -+24, then

$ $hl ; d ^3x -2hE ^5x +4h -; d ^5x +4hE ^3x -2h
= dx dx
^ xh
^5x +4h2

= 3 $ ^5x +4h -5 $ ^3x -2h
^5x
+4h2

=15x +12 -15x +10
^5x +4h2

hl^xh =^5x2+24h2