10
Introduction to the
Derivative
Copyright © Cengage Learning. All rights reserved.
10.5
Derivatives: Numerical and Graphical Viewpoints
Copyright © Cengage Learning. All rights reserved.
Derivatives: Numerical and Graphical Viewpoints
Instantaneous Rate of Change of f(x) at x = a: Derivative
The instantaneous rate of change of f(x) at x = a is
defined as
The quantity f(a) is also called the derivative of f(x) at
x = a. Finding the derivative of f is called differentiating f.
Units
The units of f(a) are the same as the units of the average
rate of change: units of f per unit of x.
3
Derivatives: Numerical and Graphical Viewpoints
It may happen that the average rates of change
[f(a + h) – f(a)]/h do not approach any fixed number at all
as h approaches zero, or that they approach one number
on the intervals using positive h, and another on those
using negative h.
If this happens, limh→0[f(a + h) – f(a)]/h does not exist, and
we say that f is not differentiable at x = a, or f(a) does
not exist. When the limit does exist, we say that f is
differentiable at the point x = a, or f(a) exists.
4
Derivatives: Numerical and Graphical Viewpoints
It is comforting to know that all polynomials and exponential
functions are differentiable at every point.
On the other hand, certain functions are not differentiable.
Examples are f(x) = |x| and f(x) = x1/3, neither of which is
differentiable at x = 0.
5
Example 1 – Instantaneous Rate of Change: Numerically and Graphically
The air temperature one spring morning, t hours after
7:00 AM, was given by the function f(t) = 50 + 0.1t 4 degrees
Fahrenheit (0 ≤ t ≤ 4).
a. How fast was the temperature rising at 9:00 AM?
b. How is the instantaneous rate of change of temperature
at 9:00 AM reflected in the graph of temperature vs.
time?
6
Example 1(a) – Solution
We are being asked to find the instantaneous rate of
change of the temperature at t = 2, so we need to find f(2).
To do this we examine the average rates of change
Average rate of change = difference quotient
for values of h approaching 0.
Calculating the average rate of change over [2, 2 + h] for
h = 1, 0.1, 0.01, 0.001, and 0.0001 we get the following
values (rounded to four decimal places):
7
Example 1(a) – Solution
cont’d
Here are the values we get using negative values of h:
The average rates of change are clearly approaching the
number 3.2, so we can say that f(2) = 3.2.
Thus, at 9:00 in the morning, the temperature was rising at
the rate of 3.2 degrees per hour.
8
Example 1(b) – Solution
cont’d
The average rate of change of f over an interval is the
slope of the secant line through the corresponding points
on the graph of f.
Figure 24 illustrates this for the intervals [2, 2 + h] with
h = 1, 0.5, and 0.1.
Figure 24
9
Example 1(b) – Solution
cont’d
All three secant lines pass though the point
(2, f(2)) = (2, 51.6) on the graph of f.
Each of them passes through a second point on the curve
(the second point is different for each secant line) and this
second point gets closer and closer to (2, 51.6) as h gets
closer to 0.
10
Example 1(b) – Solution
cont’d
What seems to be happening is that the secant lines are
getting closer and closer to a line that just touches the
curve at (2, 51.6): the tangent line at (2, 51.6), shown in
Figure 25.
Figure 25
11
Example 1(b) – Solution
cont’d
Be sure you understand the difference between f(2) and
f(2): Briefly, f(2) is the value of f when t = 2, while f(2) is
the rate at which f is changing when t = 2. Here,
f(2) = 50 + 0.1(2)4 = 51.6 degrees.
Thus, at 9:00 AM (t = 2), the temperature was 51.6 degrees.
On the other hand,
f(2) = 3.2 degrees per hour.
Units of slope are units of f per unit of t.
This means that, at 9:00 AM (t = 2), the temperature was
increasing at a rate of 3.2 degrees per hour.
12
Derivatives: Numerical and Graphical Viewpoints
Because we have been talking about tangent lines, we
should say more about what they are.
A tangent line to a circle is a line that touches the circle in
just one point. A tangent line gives the circle “a glancing
blow,” as shown in Figure 26.
Tangent line to the circle at P
Figure 26
13
Derivatives: Numerical and Graphical Viewpoints
For a smooth curve other than a circle, a tangent line may
touch the curve at more than one point, or pass through it
(Figure 27).
Tangent line at P intersects graph at Q
Tangent line at P passes through curve at P
Figure 27
14
Derivatives: Numerical and Graphical Viewpoints
However, all tangent lines have the following interesting
property in common: If we focus on a small portion of the
curve very close to the point P—in other words, if we
“zoom in” to the graph near the point P—the curve will
appear almost straight, and almost indistinguishable from
the tangent line (Figure 28).
Original curve
Zoomed-in once
Zoomed-in twice
Figure 28
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Derivatives: Numerical and Graphical Viewpoints
Secant and Tangent Lines
The slope of the secant line through the points on the
graph of f where x = a and x = a + h is given by the average
rate of change, or difference quotient,
msec = slope of secant
= average rate of change
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Derivatives: Numerical and Graphical Viewpoints
The slope of the tangent line through the point on the graph
of f where x = a is given by the instantaneous rate of
change, or derivative
mtan = slope of tangent = derivative = f(a)
17
Derivatives: Numerical and Graphical Viewpoints
Quick Example
In the following graph, the tangent line at the point where
x = 2 has slope 3.
Therefore, the derivative at x = 2 is 3. That is, f(2) = 3.
18
Derivatives: Numerical and Graphical Viewpoints
We can now give a more precise definition of what we
mean by the tangent line to a point P on the graph of f at a
given point: The tangent line to the graph of f at the
point P(a, f (a)) is the straight line passing through P with
slope f(a).
19
Quick Approximation of the
Derivative
20
Quick Approximation of the Derivative
Calculating a Quick Approximation of the Derivative
We can calculate an approximate value of f(a) by using the
formula
Rate of change over [a, a + h]
with a small value of h.
The value h = 0.0001 often works. (but see the next
example for a graphical way of determining a good value to
use).
21
Quick Approximation of the Derivative
Alternative Formula: The Balanced Difference Quotient
The following alternative formula, which measures the rate
of change of f over the interval [a – h, a + h], often gives a
more accurate result, and is the one used in many
calculators:
Rate of change over [a – h, a + h]
22
Example 2 – Quick Approximation of the Derivative
a. Calculate an approximate value of f(1.5) if f(x) = x2 – 4x.
b. Find the equation of the tangent line at the point on the
graph where x = 1.5.
Solution:
a. We shall compute both the ordinary difference quotient
and the balanced difference quotient.
Ordinary Difference Quotient: Using h = 0.0001, the
ordinary difference quotient is:
Usual difference quotient
23
Example 2 – Solution
cont’d
This answer is accurate to 0.0001; in fact, f(1.5) = –1.
Graphically, we can picture this approximation as follows:
Zoom in on the curve using the window 1.5 ≤ x ≤ 1.5001
and measure the slope of the secant line joining both ends
of the curve segment.
24
Example 2 – Solution
cont’d
Figure 29 shows close-up views of the curve and tangent
line near the point P in which we are interested, the third
view being the zoomed-in view used for this approximation.
Figure 29
Notice that in the third window the tangent line and curve
are indistinguishable. Also, the point P in which we are
interested is on the left edge of the window.
25
Example 2 – Solution
cont’d
Balanced Difference Quotient: For the balanced
difference quotient, we get
Balanced difference quotient
This balanced difference quotient gives the exact answer in
this case!
26
Example 2 – Solution
cont’d
Graphically, it is as though we have zoomed in using a
window that puts the point P in the center of the screen
(Figure 30) rather than at the left edge.
Figure 30
27
Example 2 – Solution
cont’d
b. We find the equation of the tangent line from a point on
the line and its slope:
• Point (1.5, f(1.5)) = (1.5, –3.75).
Slope of the tangent line = derivative.
• Slope m = f(1.5) = –1.
The equation is
y = mx + b
where m = –1 and b = y1 – mx1 = –3.75 – (–1)(1.5) = –2.25.
Thus, the equation of the tangent line is
y = –x – 2.25.
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Leibniz d Notation
29
Leibniz d Notation
We introduced the notation f(x) for the derivative of f at x,
but there is another interesting notation. We have written
the average rate of change as
Average rate of change
As we use smaller and smaller values for x, we approach
the instantaneous rate of change, or derivative, for which
we also have the notation df/dx, due to Leibniz:
Instantaneous rate of change
30
Leibniz d Notation
That is, df/dx is just another notation for f(x). Do not think
of df/dx as an actual quotient of two numbers: remember
that we only use an actual quotient f/x to approximate
the value of df/dx.
31
Example 3 – Velocity
My friend Eric, an enthusiastic baseball player, claims he
can “probably” throw a ball upward at a speed of
100 feet per second (ft/s). Our physicist friends tell us that
its height s (in feet) t seconds later would be
s = 100t – 16t2. Find its average velocity over the interval
[2, 3] and its instantaneous velocity exactly 2 seconds after
Eric throws it.
32
Example 3 – Solution
The graph of the ball’s height as a function of time is shown
in Figure 31.
Figure 31
33
Example 3 – Solution
cont’d
Asking for the velocity is really asking for the rate of change
of height with respect to time. (Why?) Consider average
velocity first. To compute the average velocity of the ball
from time 2 to time 3, we first compute the change in
height:
s = s(3) – s(2) = 156 – 136 = 20 ft
Since it rises 20 feet in t = 1 second, we use the defining
formula speed = distance/time to get the average velocity:
Average velocity
from time t = 2 to t = 3.
34
Example 3 – Solution
cont’d
This is just the difference quotient, so
The average velocity is the average rate of change of height.
To get the instantaneous velocity at t = 2, we find the
instantaneous rate of change of height. In other words, we
need to calculate the derivative ds/dt at t = 2.
Using the balanced quick approximation described earlier,
we get
35
Example 3 – Solution
cont’d
In fact, this happens to be the exact answer; the
instantaneous velocity at t = 2 is exactly 36 ft/s.
36
Leibniz d Notation
Average and Instantaneous Velocity
For an object moving in a straight line with position s(t) at
time t, the average velocity from time t to time t + h is the
average rate of change of position with respect to time:
The instantaneous velocity at time t is
In other words, instantaneous velocity is the derivative of
position with respect to time.
37
The Derivative Function
38
The Derivative Function
The derivative f(x) is a number we can calculate, or at
least approximate, for various values of x. Because f(x)
depends on the value of x, we may think of f as a function
of x. This function is the derivative function.
39
The Derivative Function
Derivative Function
If f is a function, its derivative function f is the function
whose value f(x) is the derivative of f at x.
Its domain is the set of all x at which f is differentiable.
Equivalently, f associates to each x the slope of the
tangent to the graph of the function f at x, or the rate of
change of f at x.
The formula for the derivative function is
Derivative function
40
The Derivative Function
Quick Example
Let f(x) = 3x – 1. The graph of f is a straight line that has
slope 3 everywhere. In other words, f(x) = 3 for every
choice of x; that is, f is a constant function.
f(x) = 3x – 1
f(x) = 3
41
Example 5 – An Application: Broadband
The percentage of United States Internet-connected
households that have broadband connections can be
modeled by the logistic function
where t is time in years since the start of 2000. Graph both
P and its derivative, and determine when the percentage
was growing most rapidly.
42
Example 5 – Solution
We obtain the graphs shown in Figure 33.
Graph of P
Graph of P
Figure 33
43
Example 5 – Solution
cont’d
From the graph on the right, we see that P reaches a peak
somewhere between t = 4 and t = 5 (sometime during
2004).
Recalling that P measures the slope of the graph of P, we
can conclude that the graph of P is steepest between t = 4
and t = 5, indicating that, according to the model, the
percentage of Internet-connected households with
broadband was growing most rapidly during 2004.
Notice that this is not so easy to see directly on the graph
of P.
44
Example 5 – Solution
cont’d
To determine the point of maximum growth more
accurately, we can zoom in on the graph of P using the
range 4 ≤ t ≤ 5 (Figure 34).
Graph of P
Figure 34
45
Example 5 – Solution
cont’d
We can now see that P reaches its highest point around
t = 4.5, so we conclude that the percentage of
Internet-connected households with broadband was
growing most rapidly midway through 2004.
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