Page 31 - 2024-calc4e-SE proofs-4e.indd

Sullivan 04 apcalc4e 45342 ch02 166 233 5pp August 7, 2023 12:54

Section 2.2 • The Derivative as a Function; Differentiability 181

EXAMPLE 3 Interpreting the Derivative as a Rate of Change

The surface area S (in square meters) of a balloon is expanding as a function of time t
2
(in seconds) according to S = S(t) = 5t . Find the rate of change of the surface area of
the balloon with respect to time. What are the units of S (t)?
′
Solution
© 2024 BFW Publishers PAGES NOT FINAL - For Review Purposes Only - Do Not Copy.
An interpretation of the derivative function S (t) is the rate of change of S = S(t).
′
2
S(t + h) − S(t) 5(t + h) − 5t 2
′
S (t) = lim = lim Form (2)
h→0 h h→0 h
2
2
5(t + 2th + h ) − 5t 2
= lim
h→0 h
2
2
5t + 10th + 5h − 5t 2 10th + 5h 2
= lim = lim
h→0 h h→0 h
(10t + 5h)h
= lim = lim(10t + 5h) = 10t
h→0 h h→0
Since S (t) is the limit of the quotient of a change in area divided by a change in time,
′
2
the units of the rate of change are square meters per second (m /s). The rate of change
2
of the surface area S of the balloon with respect to time is 10t m /s.
R
NOW WORK Problem 67 and AP Practice Problems 11 and 12.
2 Graph the Derivative Function
There is a relationship between the graph of a function and the graph of its derivative.

EXAMPLE 4 Graphing a Function and Its Derivative
3
Find f if f (x) = x − 1. Then graph y = f (x) and y = f (x) on the same set of
′
′
coordinate axes.
Solution
3
f (x) = x − 1 so
3
3
2
3
2
f (x + h) = (x + h) − 1 = x + 3hx + 3h x + h − 1
y Using Form (2), we find
15 f (x + h) − f (x)
f (x) = lim
′
h→0 h
10 3 2 2 3 3
3
f'(x) 3x 2 f (x) x  1 (x + 3hx + 3h x + h − 1) − (x − 1)
= lim
5 h→0 h
2
2
3hx + 3h x + h 3
3 2 1 2 3 x = lim
h→0 h
5 2 2
h(3x + 3hx + h )
= lim
10 h→0 h
2
2
15 = lim(3x + 3hx + h ) = 3x 2
h→0
Figure 9 The graphs of f and f are shown in Figure 9.
′
NOW WORK Problem 19.

© 2024 BFW Publishers PAGES NOT FINAL
For Review Purposes Only, all other uses prohibited
Do Not Copy or Post in Any Form.