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

Copyright © Cengage Learning. All rights reserved The Derivative: Algebraic Viewpoint

33 This algebraic approach is quite straightforward: Instead of subtracting numbers to estimate the average rate of change over smaller and smaller intervals, we subtract algebraic expressions. Our starting point is the definition of the derivative in terms of the difference quotient:

44 Example 1 – Calculating the Derivative at a Point Algebraically Let f (x) = x 2. Use the definition of the derivative to compute f (3) algebraically. Solution: Substituting a = 3 into the definition of the derivative, we get: Formula for the derivative Substitute for f(3) and f(3 + h). Expand (3 + h) 2.

55 Example 1 – Solution Now we let h approach 0. As h gets closer and closer to 0, the sum 6 + h clearly gets closer and closer to = 6. Thus, cont’d Cancel the 9. Factor out h. Cancel the h. As h → 0, (6 + h) → 6

66 A Function Not Differentiable at a Point

77 Recall that a function is differentiable at a point a if f (a) exists; that is, if the difference quotient [f (a + h) − f (a)]/h approaches a fixed value as h approaches 0. We mentioned that the function f (x) = | x | is not differentiable at x = 0.

88 Example 5 – A Function Not Differentiable at 0 Numerically, graphically, and algebraically investigate the differentiability of the function f (x) = | x | at the points (a) x = 1 and (b) x = 0. Solution: a. We compute

99 Example 5 – Solution Numerically, we can make tables of the values of the average rate of change (| 1 + h | − 1)/h for h positive or negative and approaching 0: cont’d

10 Example 5 – Solution From these tables it appears that f (1) is equal to 1. We can verify that algebraically: For h that is sufficiently small, 1 + h is positive (even if h is negative) and so cont’d Cancel the h. Cancel the 1s.

11 Example 5 – Solution Graphically, we are seeing the fact that the tangent line at the point (1, 1) has slope 1 because the graph is a straight line with slope 1 near that point (Figure 37). cont’d Figure 37

12 Example 5 – Solution b. If we make tables of values in this case we get the following: cont’d

13 Example 5 – Solution For the limit and hence the derivative f (0) to exist, the average rates of change should approach the same number for both positive and negative h. Because they do not, f is not differentiable at x = 0. We can verify this conclusion algebraically: If h is positive, then | h | = h, and so the ratio | h |/h is 1, regardless of how small h is. Thus, according to the values of the difference quotients with h > 0, the limit should be 1. On the other hand if h is negative, then | h | = −h (positive) and so | h |/h = −1, meaning that the limit should be −1. cont’d

14 Example 5 – Solution Because the limit cannot be both −1 and 1 (it must be a single number for the derivative to exist), we conclude that f (0) does not exist. To see what is happening graphically, take a look at Figure 38, which shows zoomed-in views of the graph of f near x = 0. cont’d Figure 38

15 Example 5 – Solution No matter what scale we use to view the graph, it has a sharp corner at x = 0 and hence has no tangent line there. Since there is no tangent line at x = 0, the function is not differentiable there. cont’d