We have seen that the limit of a difference quotient can be interpreted as a rate of change, as a velocity, or as the slope of a tangent line. In addition, this limit provides solutions to two of the three basic problems stated at the beginning of the chapter.
We are now ready to introduce some terms that refer to that limit.To follow custom-ary practice, we use x in place of a and think of the difference quotient
as a function of h, with x held fixed as h tends to 0.
f(x + h) - f(x) h
DEFINITION The Derivative
For we define the derivative of f at x, denoted by , to be
If exists for each x in the open interval (a, b), then f is said to be differentiable over (a, b).
f¿(x)
f¿(x) = limh:0f(x + h) - f(x)
h if the limit exists f (x) y = f(x),
SUMMARY Interpretations of the Derivative
The derivative of a function f is a new function The domain of is a subset of the domain of f. The derivative has various applications and interpretations, including the following:
1. Slope of the tangent line. For each x in the domain of is the slope of the line tangent to the graph of f at the point (x, f(x)).
2. Instantaneous rate of change. For each x in the domain of is the instantaneous rate of change of with respect to x.
3. Velocity. If f(x) is the position of a moving object at time x, then is the velocity of the object at that time.
v = f¿(x) y = f(x)
f¿, f¿(x) f¿, f¿(x)
f¿ f¿.
EXAMPLE 4 Finding a Derivative Find the derivative of f at x, for SOLUTION To find we use a four-step process.
Step 1 Find
Step 2 Find
Step 3 Find
Step 4 Find
So if The function is a new function
derived from the function f.
f¿ f(x) = 4x - x2, then f¿(x) = 4 - 2x.
f¿(x) = limh:0f(x + h) - f(x)
h = limh:0(4 - 2x - h) = 4 - 2x f¿(x) = limh:0f(x + h) - f(x)
h .
= 4 - 2x - h, h Z 0 f(x + h) - f(x)
h = 4h - 2xh - h2
h = h(4 - 2x - h) h f(x + h) - f(x)
h .
= 4h - 2xh - h2
f(x + h) - f(x) = 4x + 4h - x2 - 2xh - h2 - (4x - x2) f(x + h) - f(x).
= 4x + 4h - x2 - 2xh - h2 f(x + h) = 4(x + h) - (x + h)2 f(x + h).
f¿(x),
f(x) = 4x - x2. f¿(x),
(Differentiability from the left or from the right is defined by using or respectively, in place of in the preceding definition.)
The process of finding the derivative of a function is called differentiation. The derivative of a function is obtained by differentiating the function.
h: 0 h: 0+,
h: 0
-Example 4 illustrates the four-step process that we use to find derivatives in this section. In subsequent sections, we develop rules for finding derivatives that do not involve limits. However, it is important that you master the limit process in order to fully comprehend and appreciate the various applications we will consider.
y
f(x) 4x x2 x 5
3
5 Slope 0
Slope 4
Slope 2 (2, 4)
(3, 3)
(0, 0)
Matched Problem 4 Find f¿(x),the derivative of f at x, for f(x) = 8x - 2x2.
EXAMPLE 5 Finding Tangent Line Slopes In Example 4, we started with the function and found the derivative of f at x to be So the slope of a line tangent to the graph of f at any point (x, f(x)) on the graph is
(A) Find the slope of the graph of f at and
(B) Graph and use the slopes found in part (A) to make a rough sketch of the lines tangent to the graph at and
SOLUTION (A) Using we have
Slope at Slope at Slope at x = 3 f¿(3) = 4 - 2(3) = -2
x = 2 f¿(2) = 4 - 2(2) = 0
x = 0 f¿(0) = 4 - 2(0) = 4
f¿(x) = 4 - 2x,
x = 3.
x = 0, x = 2, y = f(x) = 4x - x2
x = 3.
x = 0, x = 2, m = f¿(x) = 4 - 2x
f¿(x) = 4 - 2x.
f(x) = 4x - x2
Matched Problem 5 In Matched Problem 4, we started with the function Using the derivative found there,
(A) Find the slope of the graph of f at and
(B) Graph and use the slopes from part (A) to make a rough sketch of the lines tangent to the graph at x = 1, x = 2,and x = 4.
y = f(x) = 8x - 2x2,
x = 4.
x = 1, x = 2,
f(x) = 8x - 2x2.
The four-step process used in Example 4 is summarized as follows for easy reference:
PROCEDURE The four-step process for finding the derivative of a function f:
Step 1 Find Step 2 Find Step 3 Find Step 4 Find lim
h:0
f(x + h) - f(x)
h .
f(x + h) - f(x)
h .
f(x + h) - f(x).
f(x + h).
E
XPLORE& D
ISCUSS1
In Example 4, we found that the derivative of is In Example 5, we graphed f(x) and several tangent lines.(A) Graph f and on the same set of axes.
(B) The graph of is a straight line. Is it a tangent line for the graph of f?
Explain.
(C) Find the x intercept for the graph of What is the slope of the line tangent to the graph of f for this value of x? Write a verbal description of the rela-tionship between the slopes of the tangent lines of a function and the x inter-cepts of the derivative of the function.
f¿.
f¿ f¿
f¿(x) = 4 - 2x.
f(x) = 4x - x2
EXAMPLE 6 Finding a Derivative Find the derivative of f at x, for SOLUTION We use the four-step process to find
Step 1 Find
Step 2 Find
Combine like terms.
Step 3 Find
Step 4 Find
= 1
2x + 2x = 1
22x x 7 0
h lim:0
f(x + h) - f(x)
h = limh:0 1
2x + h + 2x f¿(x) = limh:0f(x + h) - f(x)
h .
= 1
2x + h + 2x h Z 0
= h
h(2x + h + 2x)
= x + h - x
h(2x + h + 2x)
= 2x + h - 2x h
2x + h + 2x 2x + h + 2x f(x + h) - f(x)
h = 2x + h - 2x
h f(x + h) - f(x)
h .
= 2x + h - 2x
f(x + h) - f(x) = 2x + h + 2 - (2x + 2) f(x + h) - f(x).
f(x + h) = 2x + h + 2 f(x + h).
f¿(x).
f(x) = 2x + 2.
f¿(x),
We rationalize the numerator (Appendix A, Section A-6) to change the form of this fraction.
Combine like terms.
Cancel.
So the derivative of is a new function.
The domain of f is Since is not defined, the domain of is a subset of the domain of f.
(0, q), f¿
f¿(0)
30, q).f(x) = 2x + 2 f¿(x) = 1>(22x),
Matched Problem 6 Find f¿(x) for f(x) = 1x + 4.
EXAMPLE 7 Sales Analysis A company s total sales (in millions of dollars) t months from now are given by Find and interpret S(25) and Use these re-sults to estimate the total sales after 26 months and after 27 months.
S¿(25).
S(t) = 1t + 2.
SOLUTION The total sales function S has the same form as the function f in Example 6. Only the letters used to represent the function and the independent variable have been changed. It follows that and also have the same form:
Evaluating S and at we have
So 25 months from now, the total sales will be $7 million and will be increasing at the rate of $0.1 million ($100,000) per month. If this instantaneous rate of change of sales remained constant, the sales would grow to $7.1 million after 26 months,
$7.2 million after 27 months, and so on. Even though is not a constant func-tion in this case, these values provide useful estimates of the total sales.
S¿(t) S(25) = 225 + 2 = 7 S¿(25) = 1
2225 = 0.1 t = 25,
S¿
S¿(t) = 1
22t f¿(x) = 1 22x S(t) = 2t + 2 f(x) = 2x + 2
f¿ S¿
Matched Problem 7 A company s total sales (in millions of dollars) t months from now are given by Find and interpret S(12) and . Use these results to estimate the total sales after 13 months and after 14 months. (Use the derivative found in Matched Problem 6.)
S¿(12) S(t) = 1t + 4.
In Example 7, we can compare the estimates of total sales by using the deriva-tive with the corresponding exact values of S(t):
Exact values Estimated values
For this function, the estimated values provide very good approximations to the exact values of S(t). For other functions, the approximation might not be as accurate.
Using the instantaneous rate of change of a function at a point to estimate val-ues of the function at nearby points is an important application of the derivative.