Copyright © Cengage Learning. All rights reserved. 4 Techniques of Differentiation with Applications

Copyright © Cengage Learning. All rights reserved. 4.1 Derivatives of Powers, Sums, and Constant Multiples

3 Shortcut Formula: The Power Rule

4 To find the derivative of the function we can write, f (x) = x 2 f (x) = 2x f (x) = x 3 f (x) = 3x 2 This pattern generalizes to any power of x: Theorem 4.1 The Power Rule If n is any constant and f (x) = x n, then f (x) = nx n – 1. Quick Example If f (x) = x 2, then f (x) = 2x 1 = 2x.

5 Example 1 – Using the Power Rule for Negative and Fractional Exponents Calculate the derivatives of the following: Solution: a. Rewrite as f (x) = x –1. Then f (x) = (–1)x –2 b. Rewrite as f (x) = x –2. Then f (x) = (–2)x –3

6 c. Rewrite as f (x) = x 0.5. Then f (x) = 0.5x –0.5 Alternatively, rewrite f (x) as x 1/2, so that cont’d Example 1 – Solution

7 By rewriting the given functions in Example 1 before taking derivatives, we converted them from rational or radical form (as in, say, and ) to exponent form (as in x –2 and x 0.5.) Shortcut Formula: The Power Rule

8 Another Notation: Differential Notation

9 Differential notation (“d-notation”) is based on an abbreviation for the phrase “the derivative with respect to x.” For example, we learned that if f (x) = x 3, then f (x) = 3x 2. When we say “f (x) = 3x 2,” we mean the following: The derivative of x 3 with respect to x equals 3x 2.

10 Differential Notation; Differentiation means “the derivative with respect to x.” Thus, [f (x)] is the same thing as f (x), the derivative of f (x) with respect to x. If y is a function of x, then the derivative of y with respect to x is (y) or, more compactly, To differentiate a function f (x) with respect to x means to take its derivative with respect to x. Another Notation: Differential Notation

11 Quick Example In Words Formula The derivative with respect (x 3 ) = 3x 2 to x of x 3 is 3x 2. Another Notation: Differential Notation

12 The Rules for Sums and Constant Multiples

13 The Rules for Sums and Constant Multiples We can now find the derivatives of more complicated functions, such as polynomials, using the following rules: Theorem 4.2 Derivatives of Sums, Differences, and Constant Multiples If f and g are any two differentiable functions, and if c is any constant, then the sum, f + g, the difference, f – g, and the constant multiple, cf, are differentiable, and [f ± g] (x) = f (x) ± g (x) [cf ] (x) = cf (x). Sum Rule Constant Multiple Rule

14 In Words: The derivative of a sum is the sum of the derivatives, and the derivative of a difference is the difference of the derivatives. The derivative of c times a function is c times the derivative of the function. Differential Notation: The Rules for Sums and Constant Multiples

15 Quick Example The Rules for Sums and Constant Multiples

16 Example 2 – Combining the Sum and Constant Multiple Rules, and Dealing with x in the Denominator Find the derivatives of the following: a. f (x) = 3x 2 + 2x – 4 b. c. Solution: a. Rule for sums Rule for differences

17 b. Notice that f has x and powers of x in the denominator. By rewriting them in exponent form (that is, in the form constant  power of x): We are now ready to take the derivative: cont’d Rational form Exponent form Example 2 – Solution

18 Example 2 – Solution Rational form cont’d Exponent form c. Rewrite f (x) using exponent form as follows: Rational form Exponent form

19 Example 2 – Solution cont’d Now recall that the derivative of | x | is Thus, Rational form Simplify

20 The Derivative of a Constant Times x and the Derivative of a Constant If c is any constant, then: Rule Quick Example Proof of the Sum Rule

21 Application

22 Example 4 – Gold Price You are a commodities trader and you monitor the price of gold on the spot market very closely during an active morning. Suppose you find that the price of an ounce of gold can be approximated by the function G(t) = 5t 2 – 85t + 1,762 (7.5 ≤ t ≤ 10.5), where t is time in hours.

23 Example 4 – Gold Price (See Figure 2. t = 8 represents 8:00 am.) G(t) = 5t 2 – 85t + 1,762 Source: Figure 2

24 a. According to the model, how fast was the price of gold changing at 8:00 am? b. According to the model, the price of gold (A) increased at a faster and faster rate (B) increased at a slower and slower rate (C) decreased at a faster and faster rate (D) decreased at a slower and slower rate between 7:30 and 8:30 am. cont’d Example 4 – Gold Price

25 Example 4(a) – Solution Differentiating the given function with respect to t gives G (t) = 10t – 85. Because 8:00 am corresponds to t = 8, we obtain G (8) = 10(8) – 85 = –5. The units of the derivative are dollars per hour, so we conclude that, at 8:00 am, the price of gold was dropping at a rate of $5 per hour.

26 Example 4(b) – Solution From the graph, we can see that, between 7:30 and 8:30 am (the interval [7.5, 8.5]),the price of gold was decreasing. Also from the graph, we see that the slope of the tangent becomes less and less negative as t increases, so the price of gold is decreasing at a slower and slower rate (choice (D)). cont’d

27 Example 4(b) – Solution We can also see this algebraically from the derivative, G (t) = 10t – 85: For values of t less than 8.5, G (t) is negative; that is, the rate of change of G is negative, so the price of gold is decreasing. Further, as t increases, G (t) becomes less and less negative, so the price of gold is decreasing at a slower and slower rate, confirming that choice (D) is the correct one. cont’d

28 An Application to Limits: L’Hospital’s Rule

29 An Application to Limits: L’Hospital’s Rule The limits are those of the form lim x→a f (x) in which substituting x = a gave us an indeterminate form, such as L’Hospital’s rule gives us an alternate way of computing limits such as these without the need to do any preliminary simplification. It also allows us to compute some limits for which algebraic simplification does not work. Substituting x = 2 yields Substituting x = yields

30 An Application to Limits: L’Hospital’s Rule Theorem 4.3 L’Hospital’s Rule If f and g are two differentiable functions such that substituting x = a in the expression gives the indeterminate form then That is, we can replace f (x) and g(x) with their derivatives and try again to take the limit.

31 An Application to Limits: L’Hospital’s Rule Quick Example Substituting x = 2 in yields Therefore, l’Hospital’s rule applies and

32 Example 5 – Applying L’Hospital’s Rule Check whether l’Hospital’s rule applies to each of the following limits. If it does, use it to evaluate the limit. Otherwise, use some other method to evaluate the limit.

33 Setting x = 1 yields Therefore, l’Hospital’s rule applies and We are left with a closed-form function. However, we cannot substitute x = 1 to find the limit because the function (2x – 2)/(12x 2 – 6x – 6) is still not defined at x = 1. Example 5(a) – Solution

34 Example 5(a) – Solution In fact, if we set x = 1, we again get 0/0. Thus, l’Hospital’s rule applies again, and Once again we have a closed-form function, but this time it is defined when x = 1, giving Thus cont’d

35 Example 5(b) – Solution Setting x = yields so Setting x = again yields so we can apply the rule again to obtain Note that we cannot apply l’Hospital’s rule a third time because setting x = yields the determinate form 4/ = 0. Thus, the limit is 0. cont’d

36 Example 5(c) – Solution Setting x = 1 yields 0/0 so, by l’Hospital’s rule, We are left with a closed-form function that is still not defined at x = 1. Further, l’Hospital’s rule no longer applies because putting x = 1 yields the determinate form 1/0. cont’d

37 Example 5(d) – Solution Setting x = 1 in the expression yields the determinate form 1/0, so l’Hospital’s rule does not apply here. Using the methods of limits and continuity, we find that the limit does not exist. cont’d