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MEMBERS OF GROUP 1 DO THI PHUONG THANH 2 BUI THI HIEN 3 NGUYEN THU TRA 4 NGUYEN THU HOAI 5 LUU DUC TUNG 6 GIAP THI THUC TRINH 7 TRUONG KHANH HUYEN 8 NGUYEN NHU QUYNH (2508) 1 CONTENTS INTRODUCTION 0.tính dao ham và các định lý liên quan
MEMBERS OF GROUP DO THI PHUONG THANH BUI THI HIEN NGUYEN THU TRA NGUYEN THU HOAI LUU DUC TUNG GIAP THI THUC TRINH TRUONG KHANH HUYEN NGUYEN NHU QUYNH (25/08) CONTENTS INTRODUCTION CHAPTER I BASIC CONCEPTS OF DERIVATIVES 1.1 Derivative at a point 1.2 One-sided derivative .3 1.3 Derivative over an interval .4 1.4 The relationship between the existence of the derivative and the continuity of the function 1.5 Math in the derivative 1.6- Derivative of elementary functions .9 1.7 Senior Derivatives 10 1.8 Meaning of derivative 11 CHAPTER : THE MEAN VALUE THEOREM 12 Fermat’s theorem: 12 Rolle’s Theorem 12 Mean Value Theorem or Lagrange Theorem 14 Cauchy’s Theorem 15 Applications 16 CHAPTER 3: TAYLOR EXPANSION AND APPLICATION 18 Taylor expansion for polynomials: 18 Taylor expansion for any function 18 Mac-Laurin expansion of some commonly used elementary fuctions 19 Taylor expansion application to calculate the limit of the function: 20 Apply Taylor expansion to calculate the higher order derivative at the point x = 20 Application of Taylor expansion with Lagrange residuals approximates the value of expression: 20 Application of taylor expansion to other forms of math 21 CHAPTER 4: L’ HOSPITAL’S RULE 23 CHAPTER 5: 24 CONCAVE DOWNWARD & UPWARD AND INFLECTION POINT 24 Concepts on concave downward , concave upward of the cave and inflection point 24 Signs of concave downward , concave upward and inflection point 26 CHAPTER 6: SUGGESTED EXERCISES 34 CONCLUSION 35 REFERENCES 35 INTRODUCTION Derivatives are a very important concept in mathematical analysis and have many applications in other sciences such as economics, mechanics, physics, and engineering Even in mathematics, the derivative is an important factor, applied to solving problems in algebra, calculus or problems in geometry that we often encounter in national math competitions and national math Olympiads economic In the high school math program, the content of derivatives and the application of derivatives play a key role, occupying a large amount of knowledge and learning time of the program Making students master the subject knowledge of derivatives is the basis for them to effectively study many content knowledges of math and some other subjects Therefore, in teaching, teachers need to determine to improve the teaching quality of the content of the topic of derivatives, both as a goal and as a necessary condition for the performance of the task of teaching the subject The application of derivatives both has the effect of reviewing and systematizing knowledge, as well as affirming the practicality of knowledge content Teaching applications of derivatives is an opportunity to develop functional thinking for students If this skill is practiced, students will not only master the mathematical knowledge system, but also contribute to training math problem solving skills, skills to apply math knowledge to practice, and developing math thinking for students With that in mind, my group's essay presents the topic: "Differentials and applications" Although our group has made a lot of efforts, but due to limited time and capacity, it is inevitable that we will have shortcomings We hope to receive your constructive advice We sincerely thank you! CHAPTER I BASIC CONCEPTS OF DERIVATIVES 1.1 Derivative at a point Definition: Let the function y = f(x) be determined on the interval (a, b) and x0∈(a,b) If there is a limit (finite) lim ∆𝑦 ∆𝑥→0 ∆𝑥 𝑓(𝑥0 + ∆𝑥)−𝑓(𝑥0 ) = lim ∆𝑥 ∆𝑥→0 Then the limit is called the derivative of the function y= f(x) at x and denoted by f’(x0) (or y’(x0)) f’(x0) = lim 𝑓(𝑥0 + ∆𝑥)−𝑓(𝑥0 ) ∆𝑥 ∆𝑥→0 𝜋 Example: Compute the following derivative y = cosx at x0= Solution: 𝜋 𝜋 3 ∆y = cos ( - ∆𝑥) - cos 𝜋 ∆𝑥 = -2sin ( + ∆𝑦 ∆𝑥 = 𝜋 −2 sin( + ∆𝑥 ) sin ∆𝑥 ∆𝑥 ).sin 2 ∆𝑥 𝜋 ∆𝑦 ∆𝑥→0 ∆𝑥 y’( ) = lim 𝜋 ∆𝑥 = - lim sin( + ∆𝑥→0 𝜋 = -sin = ) ∆𝑥 ∆𝑥 𝑠𝑖𝑛 −√3 1.2 One-sided derivative Definition: Let the function y=f(x) be determined on the interval [𝑎, 𝑏) (or (𝑎, 𝑏]) and x0 ∈ [𝑎, 𝑏) We say that the function has a right derivative (or left derivative) finite at x0 if it there exists a finite limit lim+ ∆𝑥→0 𝑓(𝑥0 + ∆𝑥)−𝑓(𝑥0 ) ∆𝑥 (or lim− ∆𝑥→0 𝑓(𝑥0 + ∆𝑥)−𝑓(𝑥0 ) ∆𝑥 ) We will call that the value of limit is right derivative (or left derivative) of the function at x0 and denoted by f’(x0) Theorem (Relationship between derivative and one-sided derivative): So that the function f(x0) has a derivative at x0 the necessary and sufficient condition is that the function has the right derivative equal to the left derivative at that point In that case, we have: f’(x0) = f’+(x0) = f’-(x0) Proof: The function has derivative at x0: f’(x0) ↔ ∃ lim 𝑓(𝑥0 + ∆𝑥)−𝑓(𝑥0 ) ∆𝑥 ∆𝑥→0 ↔ ∃ f’+(x0) = lim+ ∆𝑥→0 = f’(x0) 𝑓(𝑥0 + ∆𝑥)−𝑓(𝑥0 ) ∆𝑥 = lim− 𝑓(𝑥0 + ∆𝑥)−𝑓(𝑥0 ) ∆𝑥→0 ∆𝑥 = f’ - (x0) Therefore, f’(x0) = f’ + (x0) = f’ - (x0) 1.3 Derivative over an interval Definition: The function y = f(x) is called the derivative over an interval (a, b) if it has derivative at all point on over an interval The function y = f(x) is said to have derivative on the over an interval [𝑎, 𝑏] if it has derivative on the interval (a, b) and it has the right derivative at a and the left derivative at b Example: Compute the following derivative: 1−𝑥 y (x) = {(1 − 𝑥) (2 − 𝑥) −(2 − 𝑥) Solution: With 𝑥 𝜖 (−∞, 1) : y’(x) = -1 𝑖𝑓 𝑥 𝜖 (−∞, 1) 𝑖𝑓𝑥 𝜖 [1, 2] 𝑖𝑓𝑥 𝜖 (2, +∞) With 𝑥 𝜖 [1, 2]: y’(x) = [(1 − 𝑥) (2 − 𝑥)]′ = (2 – 3x + 𝑥 )’ = 2x - With 𝑥 𝜖 (2, +∞) : y’(x) = Compute the following derivative at x=1 and x=2, we have: y’-(1) = lim (1−𝑥)−0 y’+(1) = lim (1−𝑥).(2−𝑥) 𝑥→1−0 𝑥−0 = -1 𝑥−1 𝑥→1+0 = lim (2 − 𝑥) = -1 𝑥→1+0 Hence, y’(1) = y’-(1) = y’+(1) = -1 The same prove: y’(2) = y’-(2) = y’+(2) = −1 Therefore y’(x) = { 2𝑥 − 𝑖𝑓 𝑥 𝜖 (−∞, 1) 𝑖𝑓𝑥 𝜖 [1, 2] 𝑖𝑓𝑥 𝜖 (2, +∞) 1.4 The relationship between the existence of the derivative and the continuity of the function If the function y=f(x) has a derivative at the point 𝑥0 , then it is continuous at that point f(x) has a derivative at 𝑥0 -> f(x) is continuous at 𝑥0 Prove Assuming that f(x) has a finite derivative f’(𝑥0 ) at 𝑥0 , there is a finite limit lim 𝑥→0 𝑓(𝑥0 − 𝑥)−𝑓(𝑥0 ) 𝑥 = f’(𝑥0 ) According to the theorem, there exists a neighborhood of 𝑥0 such that: | 𝑓(𝑥0 −𝑥)−𝑓(𝑥0 ) 𝑥 | ≤ C in that neighborhood, where C is a constant From there, | 𝑓(𝑥0 )|= |𝑓(𝑥0 + 𝑥) − 𝑓(𝑥0 )| ≤ C|𝑥| Moving to the limit when x→ 𝑥0 we have lim 𝑓(𝑥0 ) =0 So, the function is continuous at 𝑥0 𝑥→0 • Note: +) Continuous function at a point is not sure ∃ derivative at that point +) Similarly, we state the theorem about the relationship between one-sided continuity and one-sided derivative 1.5 Math in the derivative 1.5.1- Derivative of sum, difference, product, quotient Suppose u(x), v(x) are functions with derivatives at x Then their sum, difference, product, quotient (non-zero denominator) is also the derivative and (u + v)’ = u’ + v’ (u - v)’= u’ - v’ (u.v)’ = u’v + uv’ 𝑢 𝑢′ 𝑣−𝑢𝑣′ 𝑣 𝑣2 ( )’= Prove 1) Let x be the increment x The corresponding increment of u is u, of v is v and of y= u + v is y We have: = u + v Make a score y x = u x Find the limit: lim + ∆𝑦 x→0 ∆𝑥 v x = lim ∆𝑢 ∆𝑥→0 ∆𝑥 + lim ∆𝑣 ∆𝑥→0 ∆𝑥 Thus, (u + v)’ = u’ + v’ 2) Similar proof we have: (u-v)’= u’-v’ 3) Let x increment x The corresponding increment of u is u, of v is v and of y is y We have: y= (u + u).(v + v) - u.v = u.v + v.u + u.v + u.v - u.v = (u).v + u.( v) + (u v) Make a score y x = u x 𝑣 + v x 𝑢 + ∆u ∆𝑥 ∆𝑣 Find the limit lim ∆𝑦 x→0 ∆𝑥 = ( lim ∆𝑥→0 ∆𝑢 ∆𝑣 ) 𝑣 + 𝑢 ( lim ∆𝑥 ∆𝑥→0 ∆𝑥 ) + ( lim ∆𝑢 ∆𝑥→0 ∆𝑥 ) ( lim v) ∆𝑥→0 Under the assumption that v has a derivative so v continuous and therefore lim v = ∆𝑥→0 On the other hand lim ∆𝑢 ∆𝑥→0 ∆𝑥 = 𝑢′ , lim ∆𝑣 ∆𝑥→0 ∆𝑥 =v’ Thus y’= u’.v + u’.0+u.v’= u’v + uv’ ->(u.v)’ = u’v + uv’ 4) Let x be the increment x The corresponding increment of u is u, of v is v Since v(x) ≠ 0, x is small enough, then v+v ≠ The corresponding increment 𝑢 of y= is y We have: 𝑣 y= 𝑢+u 𝑢 - 𝑣+v 𝑣 = 𝑢𝑣+𝑣u−𝑢𝑣−𝑢v y Make a score: x 𝑣(𝑣+v) = = 𝑣.u−𝑢.v 𝑣.(𝑣+v) u v −𝑢 x x 𝑣 𝑣.(𝑣+∆𝑣) Find the limit lim ∆𝑦 x→0 ∆𝑥 = ∆𝑢 ∆𝑣 )−𝑢.( lim ) x→0∆𝑥 x→0∆𝑥 𝑣.( lim lim 𝑣.(𝑣+∆𝑣) = x→0 𝑢 ′ Therefore, ( ) = 𝑣 𝑢′ 𝑣−𝑢𝑣′ 𝑣2 𝑢′ 𝑣−𝑢𝑣′ 𝑣2 Example 1: Calculate the derivative of the function ln (x+√𝑥 + 1) Solution We have: y’= = = 𝑥+√𝑥 +1 𝑥+√𝑥 +1 𝑥+√𝑥 +1 (1+ 2𝑥 2√𝑥 +1 𝑥+√𝑥 +1 √𝑥 +1 (x+√𝑥 + 1) ) = √𝑥 +1 Therefore, (ln (x+√𝑥 + 1))’ = √𝑥 +1 Example 2: Calculate the derivative of the function Y= (2𝑥−1)3 √3𝑥+2 (5𝑥+4)3 √1−𝑥 To make it easier to calculate, we assume that the factors in the expression of the function can be logarithmic, then: 1 ln y = 3ln(2x-1) + ln(3𝑥 + 2) – 2ln(5x+4) - ln(1-x) 𝑦′ 𝑦 = 2𝑥−1 1 + 3𝑥+2 – 5𝑥+4 − 1−𝑥 Hence: y’ = (2𝑥−1)3 √3𝑥+2 [ (5𝑥+4)3 √1−𝑥 2𝑥−1 + 2.(3𝑥+2) − 10 5𝑥+4 + 3.(1−𝑥) ] That is also the expression of the derivative y’ to find 1.5.2 Derivative of the composite function Theorem: If the function u=g(x) has derivatives with respect to x, and the function y=f(u) has derivative with respect to u, then the composite function y=f[g(x)] has derivatives with respect x and we have: y’x = y’u.u’x Prove Let x increment x The corresponding increment of u is u With that increment of u, y=f(x) has a corresponding increment of y Make a score y x = Find the limit: lim u x ∆𝑦 x→0 ∆𝑥 Hypothetically u x (hypothesis u≠ 0) = ( lim ∆𝑥→0 ∆𝑦 ∆𝑢 ) ∆𝑢 ∆𝑥 →𝑢′𝑥 when x→0, thus u= ∆𝑢 ∆𝑥 x → 𝑢′𝑥 0=0 On the other hand, under the assumption that y has a derivative with respect to u, so y u →𝑦′𝑢 when u→0 Therefore, lim ∆𝑦 x→0 ∆𝑥 =𝑦 ′ 𝑥 = ( lim ∆𝑥→0 ∆𝑦 ∆𝑢 ) ( Lim ∆𝑥 ) = 𝑦 ′ 𝑢 𝑢′ 𝑥 ∆𝑢 ∆𝑥→0 1.5.3- Derivative of the inverse function Theorem: If the function y=f(x) has a derivative y’≠ at x and has an inverse x=g(y), the inverse function has a derivative 𝑥 ′ 𝑦 at y and we have: 𝑥′𝑦= 𝑦′𝑥 Prove: Give y the increments ∆𝑦 The corresponding increment of x=g(y) is ∆𝑥 Note that if ∆𝑦 ≠ then ∆𝑥 ≠ and otherwise, ∆𝑥 = then ∆𝑦 = 𝑓(x+∆𝑥)-f(x)= f(x)-f(x)=0 So, we can write ∆𝑥 ∆𝑦 = ∆𝑦 ∆𝑥 Since the function y=f(x) has a derivative at x, it is continuous there Then it follows that the inverse function x=g(y) is continuous at y and x→0 when y→0 Therefore, 𝑥 ′ 𝑦 = lim x→0 ∆𝑦 ∆𝑥 = 𝑦′𝑥 1.6- Derivative of elementary functions We summarize the basic elementary derivative calculation formula (learned in grade 12) in the following table: −4 f(x) = (𝑥 + 1)5 ⇒ f ′(x) = (𝑥 + 1) ⇒ 𝑓”(𝑥) = − −9 (𝑥 + 1) 25 −14 36 (𝑥 + 1) 125 ⇒ f(31) = 2; 𝑓 ′ (31) = ; 𝑓 ′′′ (31) = − 16 25 512 (𝑥 − 31) − (𝑥 − 31)2 ⇒ f(x) = + 16 25 512 36 + 14 (𝑥 − 31) 125 3! (𝑐 + 1) ⇒ f ′′′ (𝑥) = 5 So, √33 = √32 + = 𝑓(32) = 2,01234375 The error assessment is 36 (32 − 31)3 = 14 14 < 125.3! (𝑐 + 1) 125 (𝑐 + 1) −6 ≈ 2,9 10 , 𝑐 ∊ (31; 32) 14 125 (31 + 1) Application of taylor expansion to other forms of math Example 1: Let function f(x) be define and has a continuous second derivative on the [0;1] such that satisfied f(0) = f(1) and |𝑓′′(𝑥)| ≤ 𝐴, ∀ 𝑥 ∊ (0; 1) Prove that |𝑓′(𝑥)| ≤ 𝐴 , ∀ 𝑥 ∊ (0; 1) Solution Taylor expansion with Lagrange residuals has: 𝑓(0) = 𝑓(𝑥) + 𝑓 ′ (𝑥)(0 𝑓 ′′ (𝑎) (0 − 𝑥)2 − 𝑥) + 𝑓(1) = 𝑓(𝑥) + 𝑓 ′ (𝑥)(1 𝑓 ′′ (𝑏) (1 − 𝑥)2 − 𝑥) + With a is a real number between and x, b is the real number between and x Combined hypothesis, we get: 𝑓 ′′ (𝑎) 𝑓 ′′ (𝑏) (1 − 𝑥)2 𝑓 (𝑥) = 𝑥 − 2 ′ 21 Thus, |𝑓′(𝑥)| = | 𝑓′ (𝑎) 𝑥2 − 𝑓′′ (𝑏) (1 − 𝑥)2 | ≤ | 𝑓′ (𝑎) 𝑥 2| + | 𝑓′′(𝑏) (1 − 𝑥)2 | 𝐴𝑥 𝐴(1 − 𝑥)2 𝐴 𝐴 𝐴 ≤ + = (2𝑥 − 2𝑥 + 1) = (2𝑥(𝑥 − 1) + 1) ≤ , ∀ 𝑥 ∊ (0; 1) 2 2 Example 2: Let 𝑓: [0; 1] ↦ ℝ be a function that is times differentiable for all x 1 belongs to [0;1] then 𝑓 ′′ (𝑥) ≤ Prove that 𝑓(0) = 2𝑓 ( ) + 𝑓(1) ≤ Solution Applying Taylor expansion at 𝑥0 = , we have: 1 1 𝑓(0) = 𝑓 ( ) − 𝑓 ′ ( ) + 𝑓 ′′ (𝑥1 ), 𝑤𝑖𝑡ℎ 𝑥1 ∊ (0; ) 2 1 1 𝑓(1) = 𝑓 ( ) + 𝑓 ′ ( ) + 𝑓 ′′ (𝑥2 ), 𝑤𝑖𝑡ℎ 𝑥2 ∊ ( ; 1) 2 Adding the two sides of the above equation, we get 1 𝑓(0) − 2𝑓 ( ) + 𝑓(1) = (𝑓 ′′ (𝑥1 ) + 𝑓 ′′ (𝑥2 )) ≤ Example 3: Let 𝑓: [0; 1] ↦ ℝ be level differentiable on [0;1] and satisfied f(0)=f(1)= a and 𝑓(𝑥) = 𝑏 Prove that max 𝑓′′(𝑥) ≥ 8(𝑎 − 𝑏) 𝑥∊[0;1] 𝑥∊[0;1] Solution Because f is continuous on [0;1] then there exists 𝑥0 ∊ [0; 1] such that 𝑓(0) = 𝑏 By Fermat’s Theorem, thus 𝑓 ′ (𝑥0 ) = Using Taylor expansion, 𝑥∊[0;1] we have: 𝑎 = 𝑓(0) = 𝑓(𝑥0 ) + 𝑓 ′ (𝑥0 ) 𝑓 ′′ (𝑥1 ) (0 − 𝑥0 ) + 𝑥0 → 𝑎 − 𝑏 𝑓 ′′ (𝑥1 ) = 𝑥0 with 𝑥1 ∊ (0; 𝑥0 ) Example 4: Let 𝑓: ℝ ↦ ℝ be a differentiable function with a positive second derivative Prove that 𝑓(𝑥 + 𝑓 ′ (𝑥)) ≥ 𝑓(𝑥) with for all real number x Solution Using Taylor expansion at 𝑥0 = 𝑥 , we get: 𝑓(𝑥 + 𝑓 ′ (𝑥) ) = 𝑓(𝑥) + 𝑓 ′ (𝑥)(𝑥 +𝑓 ′ (𝑥) 22 𝑓 ′′ (𝛿𝑥) (𝑥 + 𝑓 ′ (𝑥) − 𝑥)2 − 𝑥) + 2 This implies 𝑓(𝑥 + 𝑓 ′ (𝑥)) − 𝑓(𝑥) = (𝑓 ′ (𝑥)) + 𝑓′′ (𝛿𝑥) 2 (𝑓 ′ (𝑥)) ≥ Therefore, this problem is complete CHAPTER 4: L’ HOSPITAL’S RULE Defintion 1: For two function f(x) and g(x) ≠ , if : f(a) = g(a) = f ′(x) =𝐴 𝑥→𝑎 g′(x) lim Then lim f(x) 𝑥→𝑎 g(x) = 𝐴 Proof: From the hypothetical g(x) ≠ 0, ∀x ∈ ∆ , x ≠ a apply cauchy's theorem (f(a) = g(a) = ) we have: f(x) g(x) = f(x) − f(a) g(x)−g(a) = f(c) (1) g′(c) c is a point between a and x When x→ a then c→ a and From (1) → lim f(x) 𝑥→𝑎 g(x) f ′(x) g′(x) =A =A Definition 2: Suppose lim f(x) = lim g(x) = and ∀x big enough two function f(x) , g(x) all 𝑥→+∞ 𝑥→+∞ of them are possible and g’(x) ≠ Then lim f ′(x) 𝑥→+∞ g′(x) Hence, lim f (x) 𝑥→+∞ g(x) = 𝐴 (Finite or infinite) = 𝐴 Definition 3: Suppose there exists f ′(x) , g’(x) (g’(x)≠ 0), ( ∀x ∈ ∆, x ≠ a) lim |f(x)| = +∞, lim |g(x)| = +∞ 𝑥→𝑎 𝑥→𝑎 23 lim f ′ (x) 𝑥→𝑎 g ′(x) Then lim f(x) 𝑥→𝑎 g(x) = A =A Attention: |f (x)| If lim |g 𝑥→𝑎 (x)| = +∞ then f ′(x) has at least neighborhood of point a Then we get : lim f ′ (x) 𝑥→𝑎 g ′(x) =0 (1) Following definition ,from (1) we have : lim g(x) 𝑥→𝑎 f(x) |f (x)| = or lim |g 𝑥→𝑎 (x)| ±∞ Note: Indeterminate forms 0.∞ and ∞ - ∞ To eliminate these indeterminate forms by ∞ an algebraic transformation we return to the form or and then apply l'hopital rule ∞ Indeterminate reduction form , 1, Following logarithmic operation, we return to the form 0.∞ 3.L’hopital Rule uses a lot of derivatives, so it is necessary to remember all the rules for calculating derivatives of functions CHAPTER 5: CONCAVE DOWNWARD & UPWARD AND INFLECTION POINT Concepts on concave downward , concave upward of the cave and inflection point 24 a Concepts on concave downward, concave upward Assume taht y = f(x) is a differentiable function whose gragh is a curve (c) Defintion: A curve (c) is said to be concave downward (or convex) at the point x0 if in some neighborhood of x0 to ( c) Curve (c ) is said to be concave upward (foe concave) at x0 every point of line ( c) lies on the tangent at x0 A curve is said to be concave downward or concave upward in the interval if it is concave downward or concave upward at every point of the interval b Inflection points Definition: The point which concave upward arc is called the inflection point of the gragh Note: At inflection point , the tangent goes through the gragh So, C is an inflection point 25 Signs of concave downward , concave upward and inflection point Theorem : let y = f(x) be the function which has the second derivative on the interval (a,b) (1) If f’’(x) < for any x∈ (a;b), then the gragh of function is concave downward on the interval x f’’(x) y = f(x) a b Concave downward (2) If f’’(x) > for any x∈ (a;b), then the gragh of function is concave upward on the interval x f’’(x) y = f(x) a b + Concave upward Proof: Assume that x0 ∈ (a;b) By expanding Taylor function y = f(x) in the neighborhood of x0 , we get : y = f(x) =f(x0) + f’(x0)(x-x0) + 𝑓′′(𝛿)(𝑥−𝑥0 )2 2! where 𝛿 ∈ (x; 𝑥0 ) Let y be the coordinate of the point with coordinate x on the tangent to the line y = f(x) at x0, we have : y = f(x0) + f’(x0)(x-x0) Hence : y – y = 𝑓′′(𝛿) 2! (𝑥 − 𝑥0 )2 If f’’(x) < , we have y < y’ ,that is in the neighborhood of x0 the curve is below the tangent , the curve is concave upward +) Geometric meaning: If f’’(x) < , f’(x) is decreasing that is the slope of the tangent to the curve decreased, the curve is concave downward, similarly if the slope of the tangent to the curve increase, the curve is concave upward 26 Theorem 2: Let y = f(x) be the function which has the second derivative on the interval (a,b) and x0 ∈ (a;b) If f’’(x) changes its sign when x goes through x , then the point M0 (x0; f(x0)) is an inflection on point of the gragh of the given function Illustrative examples Example 1: Using Taylor expansion, calculate the following limit:a) L1 = lim x − x2 ln(1 + ) x → x x x Solution: We have ln(1 + ) = − 1 + 0( ), ( x → +) 2x x 1 1 x − x2 ln(1 + ) x + x2 ln(1 + ) = x − x2 − + 0( ) x x x x 2x 1 = + 0( ) x Thus lim x − x2 ln(1 + ) = lim + 0( ) = x → x x→ x b) L2 = lim x→ 1 arcsin x − sin x ex + ln(1 − x) − 3 Solution: Expand to x because the numerator only needs to be x to be non-zero We have 1 arcsin x − sin x = x + x3 + O( x3 ) − x − x3 + O( x3 ) x 6 27 1 1 ex + ln(1 − x) − = + x + x2 + x3 + O( x3 ) + ( − x) − ( − x)2 + ( − x)3 + O( x3 ) − − x3 Thus L2 = lim x→ 1/ 3x3 = −2 −1/ 6x3 Example 2: Suppose f ( x) C ( R), f ( k) (0) = k=0,1,2 and x 0, k = 1,2, Prove that f ( x) = with x f ( k ) ( x) with Solution: Taylor expansion of the function f ( x) with x = 0, x noti f k (0) = k N n−1 f ( x) = k= = f k (0) k f ( n) n x + x , 0 x k! n! ( n) f () xn n! Because f ( n+1) ( x) 0, x so f ( n) ( x) with x Therefore f ( n) (0) f ( n) ( x) Thus f ( x) = f ( n) () n ( n) x f ( x) xn , n n! n! (1) On the other hand, applying Taylor's formula at point x we have n−1 f (2x) = k= f ( k) ( x) k f ( n) (1 ) n n−1 f ( k) ( x) x + x , x 1 2x k! n! k! k= (2) From (1) and (2) f (2x) nf ( x) n This happens when f ( x) = we have something to prove Example 3: For what value of x is the following approximate formula: x2 cos x − error up to 0.0001 Solution: Applying Taylor's formula to Lagrange remainders cos x = − x2 sin x + x, 4! ( 1) 28 x2 sin x x4 cos x − − = x 0,0001 2 4! 24 x 0,0001 − 24 = 0,1 24 0,222 We have: -3 Example 4: Approximation with error ε = 10 value A = ln (1,05) In this section, we will use Taylor's formula with Lagrange remainders to calculate Solution: Set f ( x ) = ln(1 + x ) = x − x + x + + ( −1) n −1 n n x + Rn It is necessary to calculate A = ln(1,05) ie we choose x =0,05, the constant c in the Lagrange remainder R lies between and 0,05 n Rn = f ( n +1) (c) n +1 ( −1) n 0,05n +1 x = ,0 c 0,05 (n + 1)! (c + 1) n +1 ( n + 1)! We must find n to|R |≤10 -3 n c 0, 05 + c 1, 05 Rn Thus 0, 05 n +1 ( n + 1)! = ( n + 1)!20 n +1 10 (1 + c ) n −3 = n=2 1000 A = ln(1,05) 0,05 − (0,05) = 0,05 − 0,00125 = 0,04875 0,49 Example 5: Find the differential of the function a) y = cos 3x.sin 2x b) y=f(x)=sin √x+ cos √xy=f(x)=sin x+ cos x Solution : a) y = cos 3x.sin 2x y’ = (cos 3x)’sin 2x + cos 3x(sin 2x)’ = – 3sin 3x.sin 2x + 2cos 3x.cos 2x 29 Thus dy = (– 3sin3x.sin2x + cos3x.cos2x) dx b) y=f(x)=sin √x+ cos √xy=f(x)=sin x+ cos x f'(x)=12√xcos√x−12√xsin√x=12√x (cos √x− sin √x) f'(x)=12xcosx−12xsinx=12xcosx−sinx Thus dy=12√x(cos√x−sin√x)dx Example 6: Calculation limit a) lim x→ Since cos x 2x − (cos x) ( 2x − ) =− sin x → − x → 2 according to L 'Hospital's rule lim x→ cos x =− 2x − We can briefly write it as follows cos x (cos x) sin x = lim = lim− =− 2 x → 2x − x→ ( 2x − ) x→ lim 2 ex − e− x ex + e− x = lim =2 x → ln(1 + x) x→ 1+ x 0 0 b) lim c) −1 ln x − x + x lim − = lim = lim x →1 x − ln x x→1 ( x − 1) ln x x→1 ln x + x − x 1− x −1 = lim = lim =− x →1 x ln x + x − x →1 ln x + 2 1 lim+ x→0 x d) 1 Set u = x tan x tan x , ln u = tan x(− ln x) = − ln x , we have cos x sin x 30 ln x x lim+ ln x = lim+ − = lim+ − x→ x→ x→ cos x − sin x sin x sin x = lim+ sin x =0 x→ x cos x lim u = lim+ eln x = Therefore Thus lim+ x→0 x x→0+ x→0 tan x =1 Example 7: Solve the equation 5x + 12 x = x + 11x (*) Solution: Rewrite the given equation in the form 12 x − 11x = x − x Assuming the equation has a solution then 12 − 11 = 6 − 5 (**) Consider the function f (t ) = (t + 1) − t (t>0) f (t ) is continuous and has a derivative on (0; +) and f ( t ) = (t + 1) −1 − t −1 On the other hand, from (*) we have So by Rolle's theorem c ( 5;11) such that f (c) = −1 − c −1 (c+ 1) =0 = = −1 −1 =c = (c+ 1) Try again, we see the values = 0, = that satisfy the equation (*) This method gives solutions x=1 and x=0 Example 8: Let a < b < c, prove that 31 3a a + b + c − a + b2 + c − ab − bc − ca 3b a + b + c + a + b + c − ab − bc − ca 3c Solution Consider the function f ( x) = ( x − a)( x − b)( x − c) f (a) = f (b) = f (c) = According to Lagrange's theorem there exists a x1 b x2 c such that: f (c) − f (b) = (c − b) f '( x1 ) f '( x1 ) = f '( x2 ) = f (a) − f (b) = (a − b) f '( x1 ) , f '( x) = 3x − 2(a + b + c) x + ab + bc + ca x1 = x2 = a + b + c − a + b + c − ab − bc − ca a + b + c + a + b2 + c − ab − bc − ca Therefore, from a x1 b x2 c then 3a a + b + c − a + b2 + c − ab − bc − ca 3b a + b + c + a + b2 + c − ab − bc − ca 3c Example 9: Given the sequence of real numbers (𝑥𝑛 ) determined by: x1 = 2007 x = + xn , n N * n +1 xn2 − Find the limit of the sequence as n approaches infinity Solution: We have xn 3, n N * Consider the function f(x) = f '( x) 2 x 3+ x2 −1 , we have : We have f ( x) = x x = + x x2 − ( x − 3) = x2 x2 −1 x − 3x = −1 + 15 x= ( x − 3x) − 2( x − 3x) − = 2 x − 3x = 2 ( x − 1)3 , x ( 3; +) If (xn) has a limit, then that limit is the larger solution of the equation f ( x) = x f '( x) = − 32 3 + 15 , According to Lagrange's theorem, there is always or (a; xn ) satisfying f ( xn ) − f (a) = f '(cn ) xn − a Set a = xn +1 − a = f ( xn ) − f (a) = f '(cn ) xn − a lim( 2 ) n x1 − a = , thus limxn = a = 2 xn − a ( + 15 33 2 ) n x1 − a CHAPTER 6: SUGGESTED EXERCISES Exercise 1: Expanding the function f ( x) = e2x− x according to the positive integer power of x to x Exercise 2: Find the third derivative at x = 0, where f(x) = ex.sinx Exercise 3: Expanding f ( x) = + x2 − x (x 0) to a positive integer power of 1 to x x Exercise 4: Applying Taylor expansion, calculate the following limits: tan x − sin x x →0 sin x + x cos x − + x x →0 ln(1 + x) − x L1 = lim L4 = lim ex − x + x −1 L5 = lim x →0 sin xchx − shx x − sin x L2 = lim x →0 x sin x L3 = lim x →0 ( ) ln + x3 − 2sin x + x cos x L6 = lim tan x − sin x e x + ln(1 − x) − x →0 − x3 − Exercise 5: Estimate the absolute error of the following approximation a) ex = + x + b) 1+ x 1+ x2 xn + + , x 2! n! x x2 − , 0 x 1 Exercise 6: Suppose f ( x) C −1,1, f ( k) (0) = k = 1,2 and exists the number ( 0,1) such that sup f ( k) ( x) k k! k Prove that f ( x) = on −1,1 34 CONCLUSION In the current university’s advanced math program, the calculus module plays an important role and is distributed right from the first semester In which differential and application help us to study many properties of functions Differential equations play a very important role in physics, engineering, biology and economics Differential has many wide applications in life eg: - Temperature change in certain time - Velocity of an object that can fall freely in a time interval -The current through the circuit for a certain time - The volatility of the stock market over a certain period of time -The temperature increases with the density in the gas tank During the research process due to limited time, after completing the essay, my group continued to study more deeply about differentials and applications of differentials We are looking forward to the teacher’s comments to help learn more Our article is more complete Thank you sincerely REFERENCES [1 ].Analytical textbook of Thai Nguyen University of education [2].Nguyen The Hoan, Tran Van Nhung, Differential Equations Exercises [3].Nguyen Dinh Phu (2004), Differential Equations, Vietnam National University, Ho Chi Minh City 35