1. M 112 Short Course in Calculus Chapter 2 – Rate of Change: The Derivative Sections 2.2 – The Derivative Function V. J. Motto

2. The Definition of Derivative Let f(x) be a function. The derivative of f is the function whose value at x is the limit 10/18/20152

3. Consider the function f(x) = x 2 and the point (x, f(x)) as show in the diagram. Let’s define the point Q to be(x + h, f(x + h) ). Lets draw the PQ --- a secant. The slope of the secant is given by the formula below. An illustration with f(x) = x 2 10/18/20153 And the derivative can be defined as follows: That is the slope of the tangent to curve at P

4. Another graphical Illustration 10/18/20154

5. Example 1 (page 95) Estimate the derivative of the function f(x) graphed in Figure below at x = −2, −1, 0, 1, 2, 3, 4, 5. 10/18/20155

6. Solution From the graph, we estimate the derivative at any point by placing a straightedge so that it forms the tangent line at that point, and then using the grid to estimate the slope of the tangent line. For example, the tangent at x = −1 is drawn in Figure below, and has a slope of about 2, so f′(−1) ≈ 2. Notice that the slope at x = −2 is positive and fairly large; the slope at x = −1 is positive but smaller. 10/18/20156

7. Solution (continued) At x = 0, the slope is negative, by x = 1 it has become more negative, and so on. 10/18/20157

8. Solution (continued) Using the same technique we can generate the following table of estimated values for the first derivative. Table 2.5 Estimated values of derivative of function in Figure 2.18 x–2–2–1–1012345 Derivative of x62–1–1–2–2–2–2–1–114 10/18/20158

9. An Alternative Solution As an alternative approach we could use the graph to generate a table of functional values and model the function. So we have f(x) = 0.2x 3 – x 2 – 0.8x + 4 10/18/20159

10. Alternate Solution Comparison 10/18/201510 Table 2.5 Estimated values of derivative of function in Figure 2.18 x–2–2–1–1012345 Derivative of x62–1–1–2–2–2–2–1–114 Estimations using modeling Estimation using secants x-2012345 Derivative of f5.61.8-0.8-2.2-2.4-1.40.84

11. Example 2 (page 96) Plot the values of the derivative function calculated in Example 1. Compare the graphs of f′ and f. 10/18/201511

12. Basic Definitions 10/18/201512

13. Example 3 (page 97) Table 2.6 gives values of c(t), the concentration (mg/cc) of a drug in the blood stream at time t (min). Construct a table of estimated values for c′(t), the rate of change of c(t) with respect to t. 10/18/201513

14. Solution Model the function So c(t) = -1.12x 3 + 0.18x 2 + 0.51x + 0.84 10/18/201514

15. Solution (continued) Using our calculator we can generate the following table xc(t) 0.00.5096 0.10.5127 0.20.4485 0.30.3269 0.40.1180 0.5-0.1483 0.6-0.4820 0.7-0.8831 0.8-1.1351 0.9-1.8872 1.0-2.4900 The values in the table at the right are different than the ones in the textbook. The estimates in the textbook are the rate of change over an interval --- “average rate of change.” Whereas the values in the table are “instantaneous rate of change”---the value of the first derivative. However, they are “estimates” because we have modeled the function from the data. 10/18/201515