DIFFERNTIABLE ITS APPLICATION

THAI NGUYEN UNIVERSITY OF EDUCATION MATHEMATICS PROBLEM SEMINAR PROBLEM DIFFERNTIABLE ITS APPLICATION Supervisors Ph D NGUYEN VAN THIN Author NGUYEN NHU QUYNH Class MATHEMATICAL ANALYSIS I June, 2022 CONTENTS INTRODUCTION 3 CHAPTER I BASIC CONCEPTS OF DERIVATIVES 4 1 1 Derivative at a point 4 1 2 One sided derivative 5 1 3 Derivative over an interval 5 1 4 The relationship between the existence of the derivative and the continuity of the function 6 1 5 Math in the derivative 7 1 6 Derivative o.

THAI NGUYEN UNIVERSITY OF EDUCATION MATHEMATICS - - PROBLEM SEMINAR PROBLEM: DIFFERNTIABLE & ITS APPLICATION Supervisors: Ph.D NGUYEN VAN THIN Author: NGUYEN NHU QUYNH Class: MATHEMATICAL ANALYSIS I June, 2022 CONTENTS 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 x 0(a,b) If there is a limit (finite) = 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) = Example: Compute the following derivative y = cosx at x0= Solution: y = cos ( - ) - cos = -2sin ( + ) sin = y’() = = - = -sin = 1.2 One-sided derivative Definition: Let the function y=f(x) be determined on the interval ) (or ) and x ) We say that the function has a right derivative (or left derivative) finite at x if it there exists a finite limit (or ) 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) ∃ = f’(x0) ∃ f’+(x0) = = = 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 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: y (x) = Solution: With : y’(x) = -1 With : y’(x) = = (2 – 3x + )’ = 2x - With : y’(x) = Compute the following derivative at x=1 and x=2, we have: y’-(1) = = -1 y’+(1) = = = -1 Hence, y’(1) = y’-(1) = y’+(1) = -1 The same prove: y’(2) = y’-(2) = y’+(2) = Therefore y’(x) = 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 , then it is continuous at that point f(x) has a derivative at -> f(x) is continuous at Prove Assuming that f(x) has a finite derivative f’(at , there is a finite limit = f’( According to the theorem, there exists a neighborhood of such that: C in that neighborhood, where C is a constant From there, = C Moving to the limit when ∆x we have =0 So, the function is continuous at • 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’ ()’= 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 = + Find the limit: = + 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 = + Find the limit = + )+ Under the assumption that v has a derivative so v continuous and therefore On the other hand =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) ∆x is small enough, then v+∆v The corresponding increment of y= is We have: ∆y= Make a score: = Find the limit = = Therefore, Example 1: Calculate the derivative of the function ln (x+) Solution We have: y’= (x+) = (1+ = Therefore, (ln (x+))’ = Example 2: Calculate the derivative of the function Y= To make it easier to calculate, we assume that the factors in the expression of the function can be logarithmic, then: ln y = 3ln(2x-1) + – 2ln(5x+4) - ln(1-x) = +.3 – Hence: y’ = 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 = (hypothesis ∆u Find the limit: = Hypothetically  when ∆x0, thus ∆u= 0=0 On the other hand, under the assumption that y has a derivative with respect to u, so  when ∆u0 Therefore, = = 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 at y and we have: = Prove: Give y the increments The corresponding increment of x=g(y) is Note that if then and otherwise, then (x+-f(x)= f(x)-f(x)=0 So, we can write 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, = = 1.6- Derivative of elementary functions We summarize the basic elementary derivative calculation formula (learned in grade 12) in the following table: 1.7 Senior Derivatives Definition: Assume the function f(x) has a derivative in the interval (a,b) and assume that its derivative at x is f’(x) This derivative can have a derivative again If so, the derivative f’(x) is called the second derivative of f(x) and is denoted by f’’(x) Similarly, the derivative of f’’(x) (if any) is called the third derivative of f(x) and is denoted by f’’’(x)… In general, the derivative of is called the n-order derivative of f(x) and is denoted by [ 1.8 Meaning of derivative 1.8.1 Geometric meaning For a function y=f(x) defined on (a, b) and derivative at Let (C) be the graph of that function The derivative of the function y=f(x) at the point is the slope of the tangent T to (C) at the point M ( 1.8.2 Physical meaning • Transportation time Instantaneous velocity is the derivative of position with respect to time Consider the linear motion defined by the equation s=s(t) where s=s(t) is a function with derivatives The instantaneous velocity of the motion at time is the derivative of the function s=s(t) at : v ( • Instant intensity If the amperage Q transmitted in the conductor is a function of time: Q=Q(t) where Q=Q(t) is a function with derivative, then the instantaneous magnitude of the current at time is the derivative of the function Q=Q(t) at I (Q’ ( 1.8.3- The mechanical meaning of the second derivative The second derivative f’’(t) is the instantaneous acceleration of the motion s=f(t) at time t Consider the motion defined by the equation s=f(t), where s=s(t) is a function with derivatives to the second order The instantaneous velocity at t of the motion is v(t)=f’(t) Taking the increment ∆t at t, then v(t) has the corresponding increment ∆v If v’(t) = = , we call v’(t) = the instantaneous acceleration of the motion at time t Because v’(t) = should =f’’(t) 10 CHAPTER 5: CONCAVE DOWNWARD & UPWARD AND INFLECTION POINT Concepts on concave downward , concave upward of the cave and inflection point 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 x 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 22 So, C is an inflection point 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) (2) a b Concave downward 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 : 23 y = f(x) =f(x0) + f’(x0)(x-x0) + where 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 = ( If f’’(x) < , we have y < y’ ,that is in the neighborhood of x 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 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) 1  L1 = lim x − x2 ln(1+ ) x→∞ x  Solution: We have 1 1 ln(1+ ) = − + 0( ), (x → +∞) x x 2x x 24 1  1 ⇒ x − x2 ln(1+ )x + x2 ln(1+ ) = x − x2  − + 0( ) x x x   x 2x 1 = + 0( ) x Thus 1 1  1 lim x − x2 ln(1+ ) = lim + 0( ) = x→∞ x  x→∞  x  arcsin x − sin x x→ ex + ln(1− x) − L2 = lim b) Solution: Expand to x3 because the numerator only needs to be x3 to be non-zero We have 1     arcsin x − sin x =  x + x3 + O(x3 ) ÷−  x − x3 + O(x3 ) ÷ : x3 6     1 1     ex + ln(1− x) − =  1+ x + x2 + x3 + O(x3) ÷+  (− x) − (− x)2 + (− x)3 + O(x3) ÷−     ≈ − x3 1/ 3x3 = −2 x→ −1/ 6x3 L2 = lim Thus Example 2: Suppose x > 0, k = 1,2, f (x) ∈ C∞ (R), f (k) (0) = ∀k=0,1,2 Prove that f (x) = with and f (k) (x) ≥ with x> Solution: Taylor expansion of the function f (0) = ∀k ∈ N k 25 f (x) with x = 0, x > noti n−1 f (x) = ∑ k= = (n) fk (0) k x + xn, < θ < x k! n! (n) f (θ)xn n! Because Thus f (n+1) (x) ≥ 0, x > so f (n) (x) with x> f (n) (θ) n (n) f (x) = x ≥ f (x)xn, n∈ Ν n! n! Therefore f(n) (0) < (n) (x) (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) (θ1) n n−1 f (k) (x) x + x ≤∑ , x < θ1 < 2x k! n! k! k= From (1) and (2) (2) ⇒ f (2x) ≤ nf (x) ∀n∈ Ν 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 ≈ 1− error up to 0.0001 Solution: Applying Taylor's formula to Lagrange remainders cos x = 1− x2 sinθx + x, 4! ( < θ < 1)  x2  sinθx x4 cos x −  1− ÷ = x ≤ ≤ 0,0001 2 4! 24  We have: ⇒ x ≤ 0,0001− 24 = 0,1 24 < 0,222 -3 Example 4: Approximation with error ε = 10 value A = ln (1,05) 26 In this section, we will use Taylor's formula with Lagrange remainders to calculate f ( x ) = ln(1 + x ) = x − Solution: Set 2 x + 3 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 n -3 ≤ c ≤ 0, 05 ⇒ ≤ + c ≤ 1, 05 ⇒ ≥ ⇒ Rn ≤ Thus 0, 05 n +1 ( n + 1)! = ( n + 1)!20 n +1 ≤ 10 (1 + c ) −3 = n 1000 ⇒n=2 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 Thus dy = (– 3sin3x.sin2x + cos3x.cos2x) dx b) y=f(x)=sin √x+ cos √xy=f(x)=sin x+ cos x 27 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 lim a) π x→ cos x 2x − π (cos x)′ Since ( 2x − π ) ′ =− sin x π → − x → 2 lim according to L 'Hospital's rule π x→ cos x =− 2x − π We can briefly write it as follows cos x (cos x)′ sin x = lim = lim− =− π π π 2 x→ 2x − π x→ ( 2x − π ) ′ x→ lim 2 ex − e− x ex + e− x = lim =2 x→ ln(1+ x) x→0 1+ x lim b)  0  0÷   −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 c) tan x  1 lim  ÷ x→ 0+  x  d) tan x  1 u=  ÷  x Set , lnu = tan x(− ln x) = − ln x cos x sin x 28 , we have ln x x limln x = lim+ − = lim+ − + x→ x→ x→0 cos x − sin x sin x sin x = lim+ sin x =0 x→ x cos x lim u = lim+ elnx = x→ 0+ Therefore x→ tanx Thus  1 lim  ÷ x→ 0+  x  =1 Example 7: Solve the equation 5x + 12x = x + 11x (*) Solution: Rewrite the given equation in the form 12 x − 11x = x − x Assuming the equation has a solution 12α − 11α = 6α − 5α on and then (**) Consider the function (0; +∞) α f (t ) = (t + 1)α − t α (t>0) f (t ) is continuous and has a derivative f ′ ( t ) = α  (t + 1)α −1 − t α −1  On the other hand, from (*) we have So by Rolle's theorem f ′(c ) = 29 ∃ c ∈ ( 5;11) such that α −1 ⇒α  − cα −1   (c + 1) =0 α = α = ⇒ ⇒ α −1 α −1 =c α =  (c+ 1) Try again, we see the values α = 0, α = that satisfy the equation (*) This method gives solutions x=1 and x=0 Example 8: Let a < b < c, prove that 3a < a + b + c − a + b + 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 f (a ) − f (b ) = (a − b ) f '( x1 ) such that: f (c ) − f (b) = (c − b) f '( x1 ) ⇒ f '( x1 ) = f '( x2 ) = , ⇒ x1 = f '( x) = 3x − 2(a + b + c ) x + ab + bc + ca x2 = Therefore, from a < x1 < b < x2 < c a < x1 < b < x2 < c a + b + c − a + b + c − ab − bc − ca a + b + c + a + b2 + c − ab − bc − ca then 3a < a + b + c − a + b + c − ab − bc − ca < 3b < a + b + c + a + b + c − ab − bc − ca < 3c Example 9: Given the sequence of real numbers () determined by:  x1 = 2007   x = + xn , ∀n ∈ N *  n +1 xn2 −  Find the limit of the sequence as n approaches infinity 30 Solution: We have xn > 3, ∀n ∈ N * x 3+ x −1 Consider the function f(x) = ⇒ f '( x ) < 2 f '( x ) = − , we have : ( x − 1)3 , ∀x ∈ ( 3; +∞) If (xn) has a limit, then that limit is the larger solution of the equation f ( x) = x x f ( x) = x ⇔ x = + x2 ⇔ ( x − 3) = x −1 x2 −1 We have  x − 3x = −1 + 15 ⇔ ⇔x= 2 2 x − x =  ⇔ ( x − 3x) − 2( x − 3x ) − =  a= Set + 15 satisfying , According to Lagrange's theorem, there is always or f ( xn ) − f (a) = f '(cn ) xn − a ⇒ xn +1 − a = f ( xn ) − f (a) = f '(cn ) xn − a < lim( 2 ) n x1 − a = , thus limxn = a = 2 xn − a < < ( + 15 31 2 ) n x1 − a (a; xn ) 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 x to x3 f (x) = 1+ x2 − x (x > 0) to a positive integer power of Exercise 4: Applying Taylor expansion, calculate the following limits: + x cos x − + x x →0 ln(1 + x) − x tan x − sin x x →0 sin x L4 = lim L1 = lim L2 = lim x →0 L3 = lim x →0 ex − x + x −1 x → sin xchx − shx x − sin x x sin x ( L5 = lim ) ln + x3 − 2sin x + x cos x L6 = lim e x + ln(1 − x) − x →0 tan x − sin x − x3 − Exercise 5: Estimate the absolute error of the following approximation ex = 1+ x + a) b) x2 xn + + , ≥ x ≥ 2! n! x x2 1+ x ≈ 1+ − , ≥ x ≥ Exercise 6: Suppose α ∈ ( 0,1) f (x) ∈ C∞ [ −1,1] , f (k) (0) = k = 1,2 such that sup f (k) (x) ≥ α kk! ∀k ∈ Ν 32 and exists the number Prove that f (x) = on [ −1,1] 33 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 34 ... 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... 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...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,