giải tích 1 lê xuân đại derivatives sinhvienzone com

Lê Xuân Đại HCMUT-OISP DERIVATIVES HCMC — 2016... Lê Xuân Đại HCMUT-OISP DERIVATIVES HCMC — 2016... Lê Xuân Đại HCMUT-OISP DERIVATIVES HCMC — 2016... Lê Xuân Đại HCMUT-OISP DERIVATIVES H

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E LECTRONIC VERSION OF LECTURE

Dr Lê Xuân Đại

HoChiMinh City University of Technology Faculty of Applied Science, Department of Applied Mathematics

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1 D ERIVATIVES

2 H IGHER DERIVATIVES

3 L INEAR APPROXIMATIONS AND D IFFERENTIALS

4 R ATES OF CHANGE AND R ELATED RATES

5 M AT L AB

Dr Lê Xuân Đại (HCMUT-OISP) DERIVATIVES HCMC — 2016 2 / 55

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The tangent line to the curve y = f (x) at the point

P(a, f (a)) is the line through P with slope

m = lim

x→a

f (x) − f (a)

provided that this limit exists.

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Suppose an object moves along a straight lineaccording to an equation of motions = f (t),wheresisthe directed distance of the object from the origin atthe timet.In the time interval fromt = atot = a + h

the change in position isf (a + h) − f (a).The average velocityover this time interval is

average velocity=f (a + h) − f (a)

h

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Now suppose we compute the average velocitiesover shorter and shorter time intervals[a, a + h].Welet happroach0.Theinstantaneous velocityv(a) attimet = ais defined by

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Suppose that a ball is dropped from the upper

ground

1 What is the velocity of the ball after5seconds?

2 How fast is the ball travelling when it hits the ground?

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3 The velocity after5sisv(5) = 9.8 × 5 = 49m/s

4 Since the observation deck is450m above theground, the ball will hit the ground at the timet1

whens(t1) = 450 ⇒ 4.9.t12= 450 ⇒ t1 =

q

450 4.9 ≈ 9.6s ⇒ v(t1) = 9.8t1≈ 94m/s.(the velocity of the ball as ithits the ground)

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The derivative of a function f at a number a ,

denoted by f0(a), (read: f prime of a ) is

f0(a) = lim

h→0

f (a + h) − f (a)

if this limit exists.

Other Notations: f0(a) = y0(a) = dy

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THEOREM 1.1

A function y = f (x) is differentiable at a if and only if the left-hand and the right-hand derivatives of f at a

exist and are equal

f0(a) = f−0(a) = f+0(a) (6)

DEFINITION1.4

A function y = f (x) is differentiable on an open interval (a, b) [or (a, ∞) or (−∞,a) or(−∞,∞)] if it is differentiable at every number in the interval.

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Conclusion: f is not differentiable ata = 0.

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2 p

x·d)y =pn

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DIFFERENTIATION FORMULASII

3 Derivatives of exponential functions

y = a x (a > 0,a 6= 1) ⇒ y0= a x ln a.

Special case:y = e x ⇒ y0= e x,sinceln e = 1

4 Derivatives of logarithmic functions

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DIFFERENTIATION FORMULASIII

Derivatives of trigonometric functions

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THE CHAIN RULE

THEOREM 1.2

If function u = u(x) is differentiable at x and function

y = f (u) is differentiable at u(x) then the composite function y = f ◦ u = f (u) = f (u(x)) is differentiable at x

and y0 is given by the product

y0(x) = f0(u(x)).u0(x). (7)

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If s = s(t) is the position function of an object that moves in a straight line, we know that its first derivative represents the velocity v(t) of the object as a function of time:

v(t) = s0(t)

The instantaneous rate of change of velocity with respect to time is called the acceleration a(t) of the object Thus the acceleration function is the

derivative of the velocity function and is therefore the second derivative of the position function:

a(t) = v0(t) = s00(t)

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Iff (x) andg(x)haven−th derivatives then

c1f (x) + c2g(x), c1, c2∈ Ralso hasn−th derivative and

(c1f (x) + c2g(x)) (n) = c1f (n) (x) + c2g (n) (x) (9)

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LEIBNIZ’S FORMULA.

Iff (x) andg(x)haven−th derivatives thenf (x).g(x)

also hasn−th derivative and

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SOME BASIC FORMULAS

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µ 1

x − 2

¶(n)

−14

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By substitutingα = −1,a = 1,b = ±2,we have

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Find the n− th derivative of f (x) = x2cos 2x.

Using Leibniz’s formula, we have

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1 f (x) ≈ f (a) + f0(a)(x − a) is called the linear approximation or tangent line approximation

of f at a.

line, that is,

L(x) = f (a) + f0(a)(x − a)

is called the linearization of f at a.

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Find the linearization of the function f (x) =px + 3 at

a = 1 and use it to approximate the numbersp3.98

andp4.05.Are these approximations overestimates or underestimates?

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p 3.98 =p0.98 + 3 ≈7

4 +0.98

4 = 1.995

⇒p3.98 < 1.995and

p 4.05 =p1.05 + 3 ≈7

4 +1.05

4 = 2.0125

⇒p4.05 < 2.0125Our approximates are overestimates

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THE GEOMETRIC MEANING OF DIFFERENTIALS

Thereforedyrepresents the amount that the tangent

line rises or falls (the change in the linearization),

whereas∆y = f (x + ∆x) − f (x)represents the amountthat the curvey = f (x)rises or falls whenxchanges by

an amount dx =∆x.

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∆y ≈ dybecomes better as∆x becomes smaller.

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RATES OF CHANGE

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RATES OF CHANGE

Ifx changes fromx1 tox2,then the change inxis

∆x = x2− x1 and the corresponding change iny is

∆y = f (x2) − f (x1 ).The difference quotient

∆y

∆x =

f (x2) − f (x1 )

x2− x1

is theaverage rate of change ofywith respect to x

over the interval[x1, x2].

The instantaneous rate of change ofy with respect to

x or the slope of the tangent line atP(x1, f (x1))is

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The position of a particle is given by the equation

s = f (t) = t3− 6t2+ 9t,

1 Find the velocity at time t. What is the velocity after 2s? When is the particle at rest? When is the particle moving forward (that is, in the positive direction) and backward?

during the first five seconds.

is the particle speeding up, slowing down?

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1 The velocity function

It moves backward then1 < t < 3.

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2 We need to calculate the distances travelled bythe particle during the time intervals[0, 1], [1, 3]and[3, 5] separately.

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3 The acceleration is the derivative of the velocityfunction:

a(t) = v0(t) = s00(t) = 6t − 12 ⇒ a(4) = 12m/s2

4 The particle speeds up when the velocity ispositive and increasing (it meansv(t)and a(t)areboth positive) and also when the velocity is

negative and decreasing (it meansv(t)and a(t)

are both negative) In other words, the particlespeeds up when the velocity and accelerationhave the same sign

v(t).a(t) > 0 ⇔ (3t2− 12t + 9)(6t − 12) > 0

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RELATED RATES

* If we are pumping air into a balloon, boththe volume and the radius of the balloonare increasing and their rates of increaseare related to each other

* In a related rates problem the idea is tocompute the rate of change of one quantity

in terms of the rate of change of anotherquatity

* The procedure is to find an equation thatrelates the two quantities and then use theChain Rule to differentiate both sides withrespect to time

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Air is being pumped into a spherical balloon so that its volume increases at a rate of 100cm3/s. How fast is the radius of the balloon increasing when the

SOLUTIONLetV (t)be the volume of the balloonand letr(t)be its radius We start by identifying twothings

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2 theunknown: the rate of increase of the radiuswhen the diameter is50cm ⇒ dr

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If we putr = 25and dV

dt = 100in this equation, weobtain

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MATLAB: DERIVATIVES

1 Derivatives: diff(f ) or diff(f,x).Example:syms x;

diff (xˆ2 + 2) ⇒ans=2*x

2 Then−th derivative: diff(f,n) or diff(f,x,n).

⇒ ans = 12 ∗ exp(xˆ2 + 1) + 48 ∗ xˆ2 ∗ exp(xˆ2 + 1) +

16 ∗ xˆ4 ∗ exp(xˆ2 + 1).

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THANK YOU FOR YOUR ATTENTION

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