seven

The Definition of a Derivative
> Average Rate of Change (Section 1.3, pp. xx–xx) > Point–Slope Form of a Line (Section 0.8, pp. xx–xx)
> Secant Line (Section 1.3, pp. xx–xx) > Difference Quotient (Section 1.2, pp. xx–xx)
> Factoring (Section 0.3, pp. xx–xx)
OBJECTIVES 1 Find an equation of the tangent line to the graph of a function
2 Find the derivative of a function at a number c
3 Find the derivative of a function using the difference quotient
4 Find the instantaneous rate of change of a function
5 Find marginal cost and marginal revenue
The Tangent Problem
The geometry question that motivated the development of calculus was “What is the
slope of the tangent line to the graph of a function y  f (x) at a point P on its graph?”
See Figure 1.
We first need to define what we mean by a tangent line. In high school geometry, the
tangent line to a circle is defined as the line that intersects the graph in exactly one
point. Look at Figure 2. Notice that the tangent line just touches the graph of the circle.
y
x
P
Tangent line
to f at P
y = f (x)
FIGURE 1
P
FIGURE 2
The Definition of a Derivative 275
This definition, however, does not work in general. Look at Figure 3. The lines L1
and L2 only intersect the graph in one point P, but neither touches the graph at P.
Additionally, the tangent line LT shown in Figure 4 touches the graph of f at P, but also
intersects the graph elsewhere. So how should we define the tangent line to the graph
of f at a point P?
The tangent line LT to the graph of a function y  f (x) at a point P necessarily contains
the point P. To find an equation for LT using the point–slope form of the equation of a
line, it remains to find the slope mtan of the tangent line.
Suppose that the coordinates of the point P are (c, f (c)). Locate another point Q 
(x, f (x)) on the graph of f. The line containing P and Q is a secant line. (Refer to
Section 1.3.) The slope msec of the secant line is
Now look at Figure 5.
As we move along the graph of f from Q toward P, we obtain a succession of secant
lines. The closer we get to P, the closer the secant line is to the tangent line. The limiting
position of these secant lines is the tangent line. Therefore, the limiting value of the
slopes of these secant lines equals the slope of the tangent line. But, as we move from Q
toward P, the values of x get closer to c. Therefore,
The tangent line to the graph of a function y  f (x) at a point P  (c, f (c)) on its
graph is defined as the line containing the point P whose slope is
mtan  lim
x:c
msec  lim
x:c
f (x)  f (c)
x  c
msec 
f (x)  f (c)
x  c
y
x
L1
L2
P
FIGURE 3
y
x
LT
P
FIGURE 4
Lt
Q1 = (x1, f (x1))
Q = (x, f (x))
P = (c, f (c))
y
x
c
y = f (x)
FIGURE 5
mtan  lim (1)
x:c
f (x)  f (c)
x  c

provided that this limit exists.
If mtan exists, an equation of the tangent line is
y  f (c)  mtan(x  c) (2)

276 Chapter 4 The Derivative of a Function
EXAMPLE 1 Finding an Equation of the Tangent Line
Find an equation of the tangent line to the graph of at the point .
Graph the function and the tangent line.
The tangent line contains the point . The slope of the tangent line to the graph
of at is
An equation of the tangent line is
y  f (c)  mtan(x  c)
Figure 6 shows the graph of and the tangent line at 1, . ?
1
4 y  
x2
4
y 
1
2
x 
1
4
y 
1
4

1
2
(x  1)
 lim
x:1
x  1
4

1
2
mtan  lim
x:1
f (x)  f(1)
x  1
 lim
x:1
x2
4

1
4
x  1
 lim
x:1
(x  1) (x  1)
4(x  1)
1,
1
4 f (x)  
x2
4
1,
1
4 
1,
1
4 f (x)  
x2
4
SOLUTION
(1, )
y
x
1
2
1
2 1
4
1
1
1
2
1
4 y = x –
y = x2
4
FIGURE 6
f(c)  lim (3)
x:c
f(x)  f(c)
x  c

provided that this limit exists.
NOW WORK PROBLEM 3.
The limit in formula (1) has an important generalization: it is called the derivative of f at c.
The Derivative of a Function at a Number c.
Let y  f (x) denote a function f. If c is a number in the domain of f, the derivative
of f at c, denoted by f (c), read “f prime of c,” is defined as
1
The Definition of a Derivative 277
Steps For Finding the Derivative of a Function at c
STEP 1 Find f (c).
STEP 2 Subtract f (c) from f (x) to get f (x)  f (c) and form the quotient
STEP 3 Find the limit (if it exists) of the quotient found in Step 2 as x : c:
f (c)  lim
x:c
f (x)  f (c)
x  c
f (x)  f (c)
x  c

EXAMPLE 2 Finding the Derivative of a Function at a Number
Find the derivative of f (x)  2x2  5x at 2. That is, find f (2).
Step 1: f(2)  2(4)  5(2)2
Step 2:
Step 3: The derivative of f at 2 is
?
NOW WORK PROBLEM 13.
Example 2 provides a way of finding the derivative at 2 analytically. Graphing utilities
have built-in procedures to approximate the derivative of a function at any number c.
Consult your owner’s manual for the appropriate keystrokes.
f (2)  lim
x:2
f (x)  f(2)
x  2
 lim
x:2
(2x  1)(x  2)
x  2
 3
f (x)  f(2)
x  2

(2x2  5x)  (2)
x  2

2x2  5x  2
x  2

(2x  1)(x  2)
x  2
2
SOLUTION
EXAMPLE 3 Finding the Derivative of a Function Using a Graphing Utility
Use a graphing utility to find the derivative of f (x)  2x2  5x at 2. That is, find f (2).
Figure 7 shows the solution using a TI-83 graphing calculator.
So f (2)  3. ?
NOW WORK PROBLEM 45.
SOLUTION
FIGURE 7
The steps for finding the derivative of a function are listed below:
278 Chapter 4 The Derivative of a Function
EXAMPLE 4 Finding the Derivative of a Function at c
Find the derivative of f (x)  x2 at c. That is, find f (c).
Since f (c)  c2, we have
The derivative of f at c is
?
As Example 4 illustrates, the derivative of f (x)  x2 exists and equals 2c for any number
c. In other words, the derivative is itself a function and, using x for the independent
variable, we can write f (x)  2x. The function f  is called the derivative function of f
or the derivative of f. We also say that f is differentiable. The instruction “differentiate
f ’’ means “find the derivative of f ’’.
It is usually easier to find the derivative function by using another form. We derive
this alternate form as follows:
Formula (3) for the derivative of f at c is
Formula (3)
Let h  x  c. Then x  c  h and

Since h  x  c, then, as x : c, it follows that h: 0. As a result,
f (c)  (4)
Now replace c by x in (4). This gives us the following formula for finding the derivative
of f at any number x.
f(c  h)  f(c)
h
lim
h:0
f(x)  f(c)
x  c
lim
x:c
f (c  h)  f (c)
h
f (x)  f (c)
x  c
f (c)  lim
x:c
f (x)  f (c)
x  c
f (c)  lim
x:c
f (x)  f (c)
x  c
 lim
x:c
(x  c)(x  c)
x  c
 2c
f (x)  f (c)
x  c

x2  c2
x  c

(x  c)(x  c)
x  c
SOLUTION
EXAMPLE 5 Using the Difference Quotient to Find a Derivative
(a) Use formula (5) to find the derivative of f (x)  x2  2x.
(b) Find f (0), f (1), f (3).
Formula for the Derivative of a Function y  f (x) at x
(5)
That is, the derivative of the function f is the limit of its difference quotient.
f (x)  lim
h:0
f (x  h)  f (x)
h

3
The Definition of a Derivative 279
(a) First, we find the difference quotient of f (x)  x2  2x.
Simplify.
Factor out h.
Cancel the h’s.
The derivative of f is the limit of the difference quotient as h: 0. That is,
(b) Since,
f (x)  2x  2
we have
f (0)  2  0  2  2
f (1)  2(1)  2  0
f (3)  2(3)  2  8 ?
NOW WORK PROBLEM 29.
Instantaneous Rate of Change
In Chapter 1 we defined the average rate of change of a function f from c to x as
The limit as x approaches c of the average rate of change of f, based on formula (3), is
the derivative of f at c. As a result, we call the derivative of f at c the instantaneous rate
of change of f with respect to x at c. That is,
y
x

f (x)  f (c)
x  c
f(x)  lim
h:0
f(x  h)  f(x)
h
 lim
h:0
(2x  h  2)  2x  2
 2x  h  2

h(2x  h  2)
h

2xh  h2  2h
h

x2  2xh  h2  2x  2h  x2  2x
h
f (x  h)  f (x)
h

[(x  h)2  2(x  h)]  [x2  2x]
h
SOLUTION
 (6) Instantaneous rate of
change of f with respect to x at c  f (c)  lim
x:c
f (x)  f (c)
x  c

EXAMPLE 6 Finding the Instantaneous Rate of Change
During a month-long advertising campaign, the total sales S of a magazine were given
by the fraction
S(x)  5x2  100x  10,000
where x represents the number of days of the campaign, 0  x  30.
4
280 Chapter 4 The Derivative of a Function
(a) What is the average rate of change of sales from x  10 to x  20 days?
(b) What is the instantaneous rate of change of sales when x  10 days?
(a) Since S(10)  11,500 and S(20)  14,000, the average rate of change of sales from
x  10 to x  20 is
(b) The instantaneous rate of change of sales when x  10 is the derivative of S at 10.
S(10) 

  (x  30)  5  40  200
The instantaneous rate of change of S at 10 is 200 magazines per day. ?
We interpret the results of Example 6 as follows: The fact that the average rate of sales
from x  10 to x  20 is S/x  250 magazines per day indicates that on the 10th day
of the campaign, we can expect to average 250 magazines per day of additional sales if
we continue the campaign for 10 more days. The fact that S(10)  200 magazines per
day indicates that on the 10th day of the campaign, one more day of advertising will
result in additional sales of approximately 200 magazines per day.
NOW WORK PROBLEM 39.
Application to Economics: Marginal Analysis
Economics is one of the many fields in which calculus has been used to great advantage.
Economists have a special name for the application of derivatives to problems in
economics—it is marginal analysis. Whenever the term marginal appears in a discussion,
involving cost functions or revenue functions, it signals the presence of derivatives
in the background.
5 lim
x:10
(x  30)(x  10)
x  10
5 lim
x:0
5(x2  20x  300)
x  10
lim
x:10
[(5x2  100x  10,000)  11500]
x  10
lim
x:10
S(x)  S(10)
x  10
lim
x:10
S
x

S(20)  S(10)
20  10

14,000  11,500
10
 250 magazines per day
SOLUTION
Find marginal cost and
marginal revenue
5
Marginal Cost
Suppose C  C(x) is the cost of producing x units. Then the derivative C(x) is
called the marginal cost.

We interpret the marginal cost as follows. Since
it follows for small values of h that
That is to say,
C(x)  cost of increasing production from x to x  h
h
C(x)  C(x  h)  C(x)
h
C(x)  lim
h:0
C(x  h)  C(x)
h
The Definition of a Derivative 281
In most practical situations x is very large. Because of this, many economists let h 
1, which is small compared to large x. Then, marginal cost may be interpreted as
C(x)  C(x  1)  C(x)  cost of increasing production by one unit

EXAMPLE 7 Finding Marginal Cost
Suppose that the cost in dollars for a weekly production of x tons of steel is given by
the function:
(a) Find the marginal cost.
(b) Find the cost and marginal cost when x  1000 tons.
(c) Interpret C(1000).
(a) The marginal cost is the derivative C(x).We use the difference quotient of C(x) to
find C(x).
C(x) 
1
10
x2  5x  1000
(b) We evaluate C(x) and C(x) at x  1000
The cost when x  1000 tons is C(1000)  (1000)2  5  1000  1000  $106,000
The marginal cost when x  1000 tons is C(1000) 1000  5  $205/ton
(c) C(1000)  $205 per ton means that the cost of producing one additional ton of
steel after 1000 tons have been produced is approximately $205. ?
Note that the average cost of producing one more ton of steel after the 1000th ton is
 $205.10/ton
We observe that the average cost differs from the marginal cost by only 0.1 dollar/ton,
which is less than th of 1%. Note, too, that the marginal cost is easier to compute
than the actual average cost.
1
20
  1
10  10012  5  1001  1000   1
10  10002  5  1000  1000
C
x

C(1001)  C(1000)
1001  1000
1
5
1
10
SOLUTION
 lim
h:0
1
5
xh  1
10
h2  5h
h
 lim
h:0
 1
5
x 
1
10
h  5 
1
5
x  5
 lim
h:0
 1
10
(x2  2xh  h2)  5x  5h  1
10x2  5x
h
 lim
h:0
 1
10 (x  h)2  5(x  h)  1000   1
10 x2  5x  1000
C(x)  lim h
h:0
C(x  h)  C(x)
h
282 Chapter 4 The Derivative of a Function