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1 Copyright © Cengage Learning. All rights reserved.
10 Introduction to the Derivative Copyright © Cengage Learning. All rights reserved.

2 Copyright © Cengage Learning. All rights reserved.
10.5 Derivatives: Numerical and Graphical Viewpoints Copyright © Cengage Learning. All rights reserved.

3 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.

4 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.

5 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.

6 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) = t 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?

7 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 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 we get the following values (rounded to four decimal places): Average rate of change = difference quotient

8 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.

9 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

10 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.

11 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

12 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) = (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. This means that, at 9:00 AM (t = 2), the temperature was increasing at a rate of 3.2 degrees per hour. Units of slope are units of f per unit of t.

13 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

14 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

15 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

16 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

17 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)

18 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.

19 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).

20 Quick Approximation of the Derivative

21 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 with a small value of h. The value h = often works. (but see the next example for a graphical way of determining a good value to use). Rate of change over [a, a + h]

22 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]

23 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 = , the ordinary difference quotient is: Usual difference quotient

24 Example 2 – Solution cont’d This answer is accurate to ; 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 ≤ and measure the slope of the secant line joining both ends of the curve segment.

25 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. 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. Figure 29

26 Example 2 – Solution cont’d Balanced Difference Quotient: For the balanced difference quotient, we get This balanced difference quotient gives the exact answer in this case! Balanced difference quotient

27 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

28 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 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. Slope of the tangent line = derivative.

29 Leibniz d Notation

30 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

31 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.

32 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.

33 Example 3 – Solution The graph of the ball’s height as a function of time is shown in Figure 31. Figure 31

34 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) 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. = 156 – 136 = 20 ft

35 Example 3 – Solution This is just the difference quotient, so
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

36 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.

37 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.

38 The Derivative Function

39 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.

40 The 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

41 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

42 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 Graph both P and its derivative, and determine when the percentage was growing most rapidly.

43 Example 5 – Solution We obtain the graphs shown in Figure 33.
Graph of P Graph of P  Figure 33

44 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.

45 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

46 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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