2.3 Product and Quotient Rules

Transcription

1 .3. PRODUCT AND QUOTIENT RULES 75.3 Product and Quotient Rules.3.1 Product rule Suppose tat f and g are two di erentiable functions. Ten ( g (x)) 0 = f 0 (x) g (x) + g 0 (x) See.3.5 on page 77 for a proof. Example 14 Find x e x 0 x e x 0 = x 0 e x + x (e x ) 0 = xe x + x e x = xe x ( + x) (product rule).3. Quotient rule Suppose tat f and g are two di erentiable functions. Ten See.3.5 on page 77 for a proof. 0 = f 0 (x) g (x) g 0 (x) g (x) (g (x)) Example 15 Find y 0 for y = x + ex x. Since tis function is a quotient, we rst apply te quotient rule. x + e y 0 x =.3.3 More Examples x 0 = (x + ex ) 0 x (x + e x ) x 0 (x ) = (1 + ex ) x (x + e x ) (x) = x + x e x x xe x = x + xe x (x ) = x + ex (x ) x 3 All te applications we studied earlier wic used te derivative can now be done more quickly by using te rules of di erentiation. We look at some examples.

2 76 CHAPTER. RULES OF DIFFERENTIATION Example 16 Find (xe x ) 0 and determine were xe x is increasing. Tis function is a product of two functions, so we rst apply te product rule. (xe x ) 0 = (x) 0 e x + x (e x ) 0 = (1) e x + xe x = e x (1 + x) A function is increasing were its derivative is positive. Since e x is always positive, it follows tat (1 + x) must also be positive, tat is 1 + x > 0 or x > 1. So, xe x is increasing on te interval ( 1; 1). Example 17 Find y 0 for y = x + ex x. Since tis function is a quotient, we rst apply te quotient rule. x + e y 0 x 0 = x = (x + ex ) 0 x (x + e x ) x 0 (x ) = (1 + ex ) x (x + e x ) (x) = x + x e x x xe x = x + xe x (x ) = x + ex (x ) x 3 Example 18 Find f 00 (x) for = x 3 By de nition, f 00 (x) = (f 0 (x)) 0. So, we must rst nd f 0 (x). Terefore f 00 (x) = 3x 0 f 0 (x) = 3x (power rule) = 3 x 0 = 6x (constant multiple rule) Example 19 Te position of an object is given by (t) = 16t + 64t + 6 were is in feet and t in seconds. Find te instantaneous velocity of te object for t = 1, t =. Te instantaneous velocity of te object is 0 (t). We compute it using te rules of di erentiation. 0 (t) = 16t + 64t = 3t + 64

3 .3. PRODUCT AND QUOTIENT RULES 77 Now, te instantaneous velocity at t = 1 is 0 (1) = 3 ft/s. Te instantaneous velocity at t = is 0 () = 0 ft/s..3.4 Summary of te Rules Learned So Far Let c and n denote any constants. Suppose tat f and g are two di erentiable functions. We ave te following rules:: 1. (c) 0 = 0. (x) 0 = 1 3. (c) 0 = c () 0 4. (x n ) 0 = nx n 1 5. ( g (x)) 0 = f 0 (x) g 0 (x) (sum and di erence rules) 6. ( g (x)) 0 = f 0 (x) g (x) + g 0 (x) (product rule) 7. 0 = g (x) f 0 (x) g 0 (x) g (x) (g (x)) (quotient rule) 8. (e x ) 0 = e x Remark Wen computing derivatives, you sould always try to simplify te function rst. You sould also try to simplify your answer as muc as possible. Cances are tat after aving computed a derivative, you will ave to use it. You may need to nd were it is 0, positive, negative. If it as been simpli ed, tese tasks will be easier to perform..3.5 Proofs of te Rules Learned So Far 1. Proof of te constant rule. Let = C. We want to prove tat f 0 (x) = 0. By de nition, we ave f 0 f (x + ) (x) C C (0) = 0

4 78 CHAPTER. RULES OF DIFFERENTIATION. Proof of dx dx = 1 Let = x. Ten, by de nition f 0 f (x + ) (x) x + x = 1 3. Proof of te power rule. Note: toug te rule is true for any constant n, we only prove it in te case n is a positive integer. Let = x n, were n is a constant. Ten, by de nition f 0 f (x + ) (x) (x + ) n x n Tis computation is a little bit more di cult tan te previous ones, we do it in several steps. First, (x + ) n = x n + nx n 1 + Terefore n (n 1) x n + ::: + nx n 1 + n (x + ) n x n = x n + nx n 1 n (n 1) + x n + ::: + nx n 1 + n x n = nx n 1 n (n 1) + x n + ::: + nx n 1 + n = nx n 1 n (n 1) + x n + ::: + nx n + n 1 It follows tat f (x + ) Terefore f (x + ) = nx n 1 + = nx n 1 + n (n 1) x n + ::: + nx n + n 1 n (n 1) x n + ::: + nx n + n 1 nx n 1 + = nx n 1 Since all te oter terms contain as a factor. n (n 1) x n + ::: + nx n + n 1

5 .3. PRODUCT AND QUOTIENT RULES Proof of te constant multiple rule. Let C be a constant and suppose tat f is di erentiable, prove tat (C) 0 = Cf 0 (x). Since we are assuming tat f is di erentiable, it means tat f 0 (x) exists and f 0 f (x + ) (x). Let us de ne F (x) = C. We need to prove tat F 0 (x) = Cf 0 (x). By de nition, F 0 F (x + ) F (x) (x) Cf (x + ) C (since F (x) = C) C ((x + ) ) (factor C) f (x + ) = C (since C is a constant) = Cf 0 (x) (de nition of te derivative) 5. Proof of te sum rule Suppose tat f and g are two di erentiable functions. Let F (x) = + g (x). We want to prove tat F 0 (x) = f 0 (x) + g 0 (x). By de nition F 0 (x) F (x + ) F (x) f(x + ) + g(x + ) f(x) g(x) f(x + ) f(x) + g(x + ) g(x) f(x + ) f(x) + g(x + ) g(x) (since F (x) = + g (x) ) f(x + ) f(x) g(x + ) g(x) + (property of its) = f 0 (x) + g 0 (x) (de nition of te derivative) 6. Proof of te di erence rule Te proof is very similar to te proof of te sum rule. We leave it as an exercise. 7. Proof of te product rule Suppose tat f and g are two di erentiable functions. Let F (x) = g (x). We want to prove tat F 0 (x) = f 0 (x) g (x) + g 0 (x). By de nition F 0 F (x + ) F (x) (x) f (x + ) g (x + ) g (x) (de nition of F )

6 80 CHAPTER. RULES OF DIFFERENTIATION It elps wen we know te result to prove. We are trying to obtain f 0 (x) g (x)+ g 0 (x). If we tink in terms of te de nition of te derivative, tat means tat in te numerator we must ave g (x) (f (x + ) )+ (g (x + ) g (x)) tat is, if we expand it, we realize tat we must ave in te numerator g (x) f (x + ) g (x) + g (x + ) g (x) g (x). Tis is more terms tan wat we currently ave in te numerator. Tis means tat we need to insert new terms. Of course, we cannot simply insert wat is missing, we would cange te problem. We use te standard trick of adding someting and subtracting it immediately. Tis way we aven t canged anyting. We ten rearrange te terms. We decide to insert te following term: g (x + ) + g (x + ) g (x + ). You can see tat tis term is simply 0, so we ave not canged anyting. Here is wat we obtain after inserting tis term: F 0 f (x + ) g (x + ) g (x + ) + g (x + ) g (x) (x) g (x + ) (f(x + ) f(x)) + (g(x + ) g(x)) g (x + ) (f(x + ) f(x)) + (g(x + ) g(x)) g (x + ) (f(x + ) f(x)) (g(x + ) g(x)) + (property of its) Let us evaluate eac it separately. g (x + ) (f(x + ) f(x)) = g (x + ) f (x + ) Since g is di erentiable, by a teorem studied in class, we also know tat it is continuous. Terefore, g (x + ) = g (x). Te second it is, by de nition, f 0 (x). So,we see tat g (x + ) (f(x + ) f(x)) = f 0 (x) g (x) Similarly, we ave (g(x + ) g(x)) = g 0 (x) and terefore F 0 (x) = f 0 (x) g (x) + g 0 (x) 8. Proof of te quotient rule Suppose tat f and g are two di erentiable functions. Let F (x) = g (x).

7 .3. PRODUCT AND QUOTIENT RULES 81 We want to prove tat F 0 (x) = f 0 (x) g (x) g 0 (x) (g (x)). By de nition F 0 F (x + ) F (x) (x) f (x + ) g (x + ) g (x) f (x + ) g (x) g (x + ) g (x + ) g (x) and combined fractions) (reduced to te same denominator As in te proof of te product rule, we realize tat we are missing terms to obtain wat we need. So, we insert some extra terms. Tis time, we insert g (x) + g (x). We obtain F 0 (x) f (x + ) g (x) g (x) + g (x) g (x + ) g (x + ) g (x) g (x) (f(x + ) f(x)) (g(x + ) g(x)) g (x + ) g (x) g (x) (f(x + ) f(x)) g (x + ) g (x) (g(x + ) g(x)) g (x + ) g (x) We evaluate eac it separately. As in te proof of te product rule, g (x + ) = g (x). So, we ave g (x) (f(x + ) f(x)) f(x + ) f(x) = g (x) g (x + ) g (x) g (x + ) g (x) f(x + ) f(x) = g (x) Similarly, we obtain tat (g(x + ) g(x)) g (x + ) g (x) = g (x) f 0 (x) (g (x)) = f 0 (x) g (x) (g (x)) = g(x + ) g(x) g (x + ) g (x) g(x + ) g(x) = = g 0 1 (x) (g (x)) = g0 (x) (g (x)) 1 g (x + ) g (x) 1 g (x + ) g (x)

8 8 CHAPTER. RULES OF DIFFERENTIATION And terefore, combining te two its (don t forget te minus sign in between), we obtain te desired result..3.6 Problems 1 1. Find a rule for x n. Find a rule for ( np x) 0 F 0 (x) = f 0 (x) g (x) g 0 (x) (g (x)) 0 3. Prove te di erence rule. 4. Be able to do problems suc as # 1, 3, 5, 7, 11, 13, 17, 1, 7, 37, 4, 44, 48 on pages