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

Derivatives are a very important concept in mathematical analysis and havemany applications in other sciences such as economics, mechanics, physics, andengineering Even in mathematics, the derivative is an important factor, applied tosolving problems in algebra, calculus or problems in geometry that we oftenencounter in national math competitions and national math Olympiads economic

In the high school math program, the content of derivatives and theapplication of derivatives play a key role, occupying a large amount of knowledgeand 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 todetermine 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 ofteaching the subject

The application of derivatives both has the effect of reviewing andsystematizing knowledge, as well as affirming the practicality of knowledgecontent Teaching applications of derivatives is an opportunity to developfunctional thinking for students If this skill is practiced, students will not onlymaster the mathematical knowledge system, but also contribute to training mathproblem solving skills, skills to apply math knowledge to practice, and developingmath 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 andcapacity, it is inevitable that we will have shortcomings We hope to receive yourconstructive advice We sincerely thank you!

CHAPTER I BASIC CONCEPTS OF DERIVATIVES

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 x0 ) Wesay that the function has a right derivative (or left derivative) finite at x0 if it thereexists a finite limit

(or )

We will call that the value of limit is right derivative (or left derivative) of the

Theorem (Relationship between derivative and one-sided derivative):

So that the function f(x0) has a derivative at x0 the necessary and sufficientcondition is that the function has the right derivative equal to the left derivative atthat point In that case, we have:

1.3 Derivative over an interval.

Example: Compute the following derivative:

y (x) = Solution:

With : y’(x) = -1

With : y’(x) =

= (2 – 3x + )’

= 2x - 3With : y’(x) = 1

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) = 1

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 iscontinuous at

• Note:

+) Continuous function at a point is not sure derivative at that point

+) Similarly, we state the theorem about the relationship between one-sidedcontinuity 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

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

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

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 functiony=f(u) has derivative with respect to u, then the composite function y=f[g(x)] has

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 ifthen 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 itfollows 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 ingrade 12) in the following table:

Similarly, the derivative of f’’(x) (if any) is called the third derivative of f(x) and isdenoted by f’’’(x)…

In general, the derivative of is called the n-order derivative of f(x) and is denoted

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 functionwith 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) whereQ=Q(t) is a function with derivative, then the instantaneous magnitude of thecurrent 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) attime t Consider the motion defined by the equation s=f(t), where s=s(t) is afunction with derivatives to the second order The instantaneous velocity at t of themotion is v(t)=f’(t) Taking the increment ∆t at t, then v(t) has the correspondingincrement ∆v If v’(t) = = , we call v’(t) = the instantaneous acceleration of themotion at time t Because v’(t) = should =f’’(t)

CHAPTER 2 : THE MEAN VALUE THEOREM

1 Fermat’s theorem:

Suppose f is defined in a full neighborhood of a point x0 and differentiable at x0 If

x0 is an extremum point, then it is critical for f, i.e.,

f’ (x 0 ) = 0.

Proof

To fix ideas, assume x0 is a relative maximum point and that I r (x 0 ) is a

neighborhood where f(x) ≤ f(x 0 ) for all x∈I r (x 0 ) On such neighborhood then ∆f = f(x) – f(x 0 ) ≤ 0.

If x > x 0, hence ∆x = x – x 0 > 0, the difference quotient is non-positive This

implies

Vice versa, if x < x 0, i.e, ∆x = x – x 0 < 0, then is non-negative, so

Hence,

f'(x 0 ) = = ,

so f'(x 0 ) is simultaneously ≤ 0 and ≥ 0, hence zero.

A similar argument holds for relative minima

2 Rolle’s Theorem

Let f be a function defined on a closed bounded interval [a, b], continuous on [a, b] and differentiable on (a, b) (at least) If f(a) = f(b), there exists an x 0 ∈ (a, b) suchthat

f'(x 0 ) = 0.

In other words, f admits at least one critical point in (a, b).

Rolle’s Theorem Proof.

By the Theorem of Weierstrass the range f ([a, b]) is the closed interval [m, M]

bounded by the minimum and maximum values m, M of the map:

for suitable x m , x M ∈ [a, b]

In case m = M, f is constant on [a, b], so in particular f'(x) = 0 for any x ∈ (a, b)and the theorem follows

Suppose then m < M Since m ≤ f(a) = f(b) ≤ M, one of the strict inequalities f(a) =f(b) < M, m < f(a) = f(b) will hold

If f(a) = f(b) < M, the absolute maximum point x M cannot be a nor b; thus,

x M ∈ (a, b) is an interior extremum point at which f is differentiable.

By Fermat's Theorem, we have that x M = x 0 is a critical point If m < f(a) = f(b), one proves analogously that x m is the critical point x 0 of the claim.

• The meaning geometric

In the given graph, the curve y = f(x) is continuous between x = a and x = b and atevery point, within the interval, it is possible to draw a tangent and ordinatescorresponding to the abscissa and are equal then there exists at least one tangent tothe curve which is parallel to the x-axis Algebraically, this theorem tells us that if f(x) is representing a polynomial function in x and the two roots of the equation f(x) = 0 are x = a and x = b, then there exists at least one root of the equation f’(x) = 0 lying between these values The converse of Rolle’s theorem is not trueand it is also possible that there exists more than one value of x, for which the

theorem holds good but there is a definite chance of the existence of one suchvalue.

3 Mean Value Theorem or Lagrange Theorem

Let f be defined on the closed and bounded interval [a, b], continuous on [a, b] and

differentiable (at least) on (a, b) Then there is a point x0 ∈(a, b) such that

f’(x 0 ).

Every such point x 0 we shall call Lagrange point for f in (a, b).

Proof.

Introduce an auxiliary map

defined on [a, b] It is continuous on [a, b] and differentiable on (a, b), as difference

of f and an affine map, which is differentiable on all of R Note

It is easily seen that

g(a) = f(a), g(b)=f(a),

so Rolle's Theorem applies to g, with the consequence that there is a point

x 0 ∈ (a, b) satisfying

Lagrange point for f in (a, b)

• The meaning geometric of the Mean Value Theorem is clarified on the figure

above At each Lagrange point, the tangent to the graph of f is parallel to the

secant line passing through the points (a, f(a)) and (b, f(b))

4 Cauchy’s Theorem

Let f(x), g(x) be functions defined, continuous an a closed bounded interval [a, b]and g(a) ≠ g(b); assume f(x) and g(x) are differentiable on open interval (a, b) andg’(x) ≠ 0 for all x ∈ (a, b) There exists x0 ∈ (a, b) such that

Proof.

We first assume that the function of satisfies the assumptions of Lagrange’stheorem Then there exists a point c ∈ (a, b) so that

Because g’(c) ≠ 0, it follows form that g(b) – g(a) ≠ 0

Consider the function

ϕ , x ∈ [a, b]

ϕ satisfies the continuous of Rolle’ theorem:

• ϕ is continuous on closed interval [a, b]

• ϕ has a derivative on the interval (a, b):

• ϕ(a) = ϕ(b) = 0

Hence there exists at least one point x0∈ (a, b) so that ϕ’(c) = 0

From that, deduce the equality to be proved

5 Applications

5.1 Prove the existence of a solution of the equation

Problem: Let m > 0, a, b, c sastify

Prove that the equation ax2 + bx + c = 0 has a solution in the interval (0; 1).Solution:

The function

F(x) = continuous on [0; 1] with a derivative on (0; 1) and F(0) = F(1) = 0 thereexists x0∈ (0; 1) such that

F’(x0) = xm – 1(ax02 + bx0 + c) = 0

Hence, ax02 + bx0 + c = 0

5.2 Applications in solving equations

Problem: Solve the equation

5x +12x = +6x 11x

(*)Solution:

Rewrite the given equation in the form

c c

Try again, we see the values α =0,α =1

that satisfy the equation (*).This method gives 2 solutions x=1 and x=0

5.3 Application to prove inequality

Problem:

Prove:

∀x ∈ (0, +∞)Solution:

We have

Let f(x) = x

We get:

Apply Lagrange’s theorem to the function

y = ln t on [x, x+1] then there exists c ∈ (x, x+1) such that

CHAPTER 3: TAYLOR EXPANSION AND APPLICATION

1 Taylor expansion for polynomials:

Consider polynomials P(x) = anxn + an-1xn-1 + …+ a1x + a0 , then with any point xo

we have:

2 Taylor expansion for any function

2.1 Taylor expansion with Peano form remainders:

Assuming the fuction f(x) has derivative to degree n in the neighborhood

(x0 - δ; x0 + δ) of x0 Then:

The above fomular is called the Taylor expansion of the fuction f(x) to degree n at

the point xo with Peano form remainders:

2.2 Mac-Laurin expansion:

In fomular Taylor expansion with a peano residual when ,we have:

This formula is called the Mac–Laurin expansion formula

2.3 Taylor expansion with Lagrange remainders

Assume the function has derivative to dgree n+1 in neighborhood

(x0 - δ; x0 + δ) of x0 Then:

Where c is a point between x and x0

The above formula is called the Taylor expansion of the function f(x) to degree n at

point x0 with Lagrange remainders

3 Mac-Laurin expansion of some commonly used elementary fuctions

4 Taylor expansion application to calculate the limit of the function:

Example: Find the real number a such that

Solution Applying L’hopital, we have:

So,

5 Apply Taylor expansion to calculate the higher order derivative at the point

x = 0

Example: Mac-Laurin expansion of the function then calculate the derivative

Solution Applying the fomular

Therefore

6 Application of Taylor expansion with Lagrange residuals approximates the value of expression:

Example: Taylor expansion of the function up to level to at x= 31, then calculate

and error assessment

Solution We expand with Lagrange remainders:

So,

The error assessment is

7 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 Prove that

Solution Taylor expansion with Lagrange residuals has:

With a is a real number between 0 and x, b is the real number between 1 and x

Combined hypothesis, we get:

Thus,

Example 2: Let be a function that is 2 times differentiable for all x belongs to

[0;1] then Prove that

Adding the two sides of the above equation, we get

Example 3: Let be level 2 differentiable on [0;1] and satisfied f(0)=f(1)= a and

Therefore, this problem is complete

CHAPTER 4: L’ HOSPITAL’S RULE Defintion 1:

For two function f(x) and g(x) , if :

(a) = g(a) = 0

Then

Proof:

From the hypothetical g(x) ,

apply cauchy's theorem (a) = g(a) = 0 ) we have:

= = (1)

c is a point between a and x

When x then c and = A

2. Indeterminate reduction form 1 , 1, 0 Following logarithmic operation, wereturn to the form 0

3.L’hopital Rule uses a lot of derivatives, so it is necessary to remember all therules for calculating derivatives of functions

CHAPTER 5:

CONCAVE DOWNWARD & UPWARD AND INFLECTION POINT

1 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 x0

if in some neighborhood of x0 to ( c) Curve (c ) is said to be concave upward (foeconcave) 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 isconcave downward or concave upward at every point of the interval