derivative ii tahun 2012.pdf

8/19/2019 DERIVATIVE II TAHUN 2012.pdf

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DERIVATIVES I

Gunarto,S.T.,M.Eng

Department of Mechanical Engineering

Muhammadiyah University of Pontianak


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• Derivative Formulas

• Trigonometric Functions

• Differentiability

• Chain Rule

• Implicit Differentiation

Topics Reviewed


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Cont….

• Differentiation is one of the two fundamental operationsof calculus.

• Differential calculus describes and analyzes change.

• The position of a moving object, the population of a city ora bacterial colony, the height of the sun in the sky, and the

price of cheese all change with time.

• Altitude can change with position along a road; the

pressure inside a balloon changes with temperature.

• To measure the rate of change in all these situations, we

introduce in this chapter the operation of differentiation.


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lntroduction to the

Derivative

Velocities and slopes are bothderivatives.

• This section introduces the basic idea of

the derivative by studying two

problems.

• The first is the problem of finding thevelocity of a moving object, and the

second is the problem of finding the

slope of the line tangent to a graph.


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Cont…

• To analyze velocity, imagine a buswhich moves due east on a straight

highway.

• Let x designate the time in secondsthat has passed since we first

observed the bus.

• (Using "x" for time rather than the

more common "t" will make it easier to

compare velocities with slopes.)


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• Suppose that after x seconds the bus has gone adistance y meters to the east (see Figure).

• Since the distance y depends on the time x, we have

a distance function y = f(x).


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• For example, if f(x) happens to be f(x) = 2x2for 0 ≤ x ≤ 5, then the bus has gone 2 .(3)2

= 18 meters after 3 seconds and 2 . (5)2 = 50

meters after 5 seconds.• The velocity of the bus at any given

moment, measured in meters per second, is

a definite physical quantity; it can bemeasured by a speedometer on the bus or

by a stationary radar device.


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• Since this velocity refers to a single instant, itis called the instantaneous velocity.

• Given a distance function such as y = f(x) =

2x2, how can we calculate theinstantaneous velocity at a specific time xo,

such as xo = 3 seconds?

• To answer this question, we will relate theinstantaneous velocity to the average velocity

during short time intervals.


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• Suppose that the distance covered ismeasured at time xo, and again at a later

time x; these distances are yo = f(xo)

and y = f(x ).• Let Δx = x - xo, designate the time

elapsed between our two measurements.

• Then the extra distance covered is y - yo,which we designate by Δy = y - yo.


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• The average velocity during the time interval Δxis defined simply as the distance travelled divided

by the elapsed time; that is, average velocity =

Δy/Δx =[ f(x) - f(xo)]/Δx.

• Since x = xo + Δx, we can also write


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• The following more general procedure is suggestedby Example 2.


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• The second problem we study is a

geometric one-to find the slope ofthe line tangent to the graph of a

given function.

• We shall see that this problem isclosely related to the problem of

finding instantaneous velocities.

• To solve the slope problem for the

function y = f(x), we begin bydrawing the straight line which

passes through the points (xo, f(xo))

and (xo + Δx, f(xo + Δx)), where Δx

is a positive number; see Figure


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Cont..

• This straight Iine is called asecant line, and Δy/Δx =

[ f(xo + Δx) - f(xo)]/Δx is its

slope.

• As Δx becomes small, xo,

being fixed, it appears that the

secant line comes close to

the tangent line, so that theslope Δy/Δx of the secant

line comes close to the slope

of the tangent line. See Fig. in

left.


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The problems involving rate of changes exist inmany area of engineering research.

Since this type of limit happens so widely, it is given

a special name – derivative The notation of derivative is f '(a).

The derivative of a function f at a given point a is

defined as: (1)


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The definition does assume that the limit exists. Inorder to extend this definition, let x = a + h,

substitute x into f '(a) and will get

(2)

The notation of a derivative can be written as: f '(a),

y ', df/dx, dy/dx, Df(x), Dxf(x).


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Explanation of the Derivative

A good way to understandderivatives is to think

about a tangent line.

According to the definitionof the tangent line to a

curve y = f(x) at point A(a,

f(a)), the tangent line can

be written as:

(3)Tangent


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Cont...

Notice that this definition is the same as the

definition of derivative f '(a).

In other words, the tangent line to y = f(x) at point

A(a, f(a)) is the line that passes through (a, f(a)) and

whose slop is equal to the derivative of f at a.

In a previous section, the concept of rate of change

was introduced. Is it related to derivative? The

answer is yes.


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Cont... In a small interval [x1, x2], the

changes in x isΔx = x2 - x1

The corresponding change in y is

Δy = y2 - y1 The instantaneous rate of change

is

According to equation (3), r is the

derivative of f(x) at x1.

Rate of Change


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1.Derivative Formulas

Calculating derivative according to its definition is tedious.

Some rules have been developed for finding derivatives

without having to use the definition directly.

F(x) F '(x)

c 0

xn nxn-1

cf(x) cf '(x)

f(x) + g(x) f '(x) + g '(x)

f(x) - g(x) f '(x) - g '(x)

f(x)g(x) f '(x)g(x) + f (x)g '(x)

f(x)/g(x) (f '(x)g(x) - f (x)g '(x))/g2(x)

x-n -nx-n-1


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Cont...

• If F is a constant function, F(x) = c,then F '(x) = 0.

• If F(x) = xn, where n is a positive integer,

then F'

(x) = nxn-1

.• Assume that c is constant and f '(x) and g '(x) exist.

– If F(x) = cf(x), then F ' (x) = cf ' (x)

– If F(x) = f(x) + g(x), then F ' (x) = f ' (x) + g ' (x)

– If F(x) = f(x) - g(x), then F ' (x) = f ' (x) - g ' (x


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– If F(x) = f(x)g(x), then

F ' (x) = f ' (x)g(x) + f (x)g ' (x)

– If F(x) = f(x)/g(x), thenF ' (x) = (f ' (x)g(x) - f (x)g ' (x))/g 2 (x)

– If F(x) = x -n, where n is a positive integer, then

F '

(x) = -nx -n-1


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CASE STUDY

• As Tom plays football with agroup of friends he happens to

kick the ball into some water.

• A girl nearby finds that the ball

creates circular ripple thattravele outward.

• She wonders how fast the circle

area of the circular circles

increase in size.

• What is known:

The circular ripple travels

outward at 2 ft/sec.

Rate of increased area

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Cont...

• QuestionsHow fast does the circle of area increase after 3

seconds?

• ApproachThe rate of circular circles area change is the

derivative of the area function.


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CASE STUDY SOLUTION


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CASE STUDY SOLUTION

• A football is kicked into a pool. Itcreates a circular ripple that

traveles outward with a speed of

2 ft/sec. How fast does the

circular area enlarge after 3

seconds?

• If the radius of the circular ripple

is r and the area of the circularripple is s. The area of the circular

ripple is

s = πr 2

Circular Ripple

C t


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Cont...

• The enlarging circular area with respect to time t is s'tor ds/dt

ds/dt = d(πr 2)/dt

= d(π(r 2))/dt

= 2πr(dr/dt)

• When t = 3 seconds,

r = (2)(3) = 6

So

ds/dt = 2πr(dr/dt)

= 2π(6)(2)

= 24π ft2/sec

Di i


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Discussion

• In order to evaluate the rate of the

enlarging circular area, the concept of

derivative has been employed.

• Derivative is widely used in science

and engineering research.

• For example, how to calculate the

linear density of a non-homogeneous

rod.

• The mass from the left end to the

point x is

mass = f(x) Non-homogeneous Linear Density

Calculation

C t



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• The mass between x2 andx1 is

Δmass = f(x2) - f(x1)

• So the average density ofthe rod between x2 and x1is

average density =Δmass/Δx = ( f(x2) -

f(x1))/(x2-x1)

Non-homogeneous Linear Density

Calculation

C t



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Letting x2 approaches x1, the linear density ρ of therod at point x1 is the limit of the average density

when Δx approaches 0.

In other words, the linear density of the rod is the

derivative of the mass with respect to length. Since the concept of derivative can apply to many

problems in science and engineering, it is important

for students to understand its physical meaning.


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2 Trigonometric Functions


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2.Trigonometric Functions

In order to compute the derivatives of trigonometricfunction, some of the trigonometric limits need to be

mentioned first.

Derivatives of Trigonometric Functions


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Derivatives of Trigonometric Functions

The derivatives oftrigonometric functions

can be proven by use

of the definition of

derivative and thetrigonometric limits.

(1)

sin'x = cos x

Proof



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Proof

If f(x) = sinx, then

Cont


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Cont...

• Now that d(sinx)/dx equals cosx has been proven,what does d(sinx)/dx look like in graph? Is it the

same as cosx?

• The value of the derivative at any value of x can be

estimated by drawing the tangent at the point (x, f(x))

and estimate its slope.

• For example, at the point of x equals 0, draw the

tangent at this point and estimate its slope to be

around 1, so d(sinx)/dx = 1.

Cont


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Now plot the point (0, 1) on the graph of df/dx or f '.

By repeating this procedure at different points, the

graph of f ' can be drawn and will resemble a cosine

curve.

The correction of the graph is confirmed by formula

(1).

Using the definition of derivative and thetrigonometric limits, the formula

d(cosx)/dx = -sinx (2)

can be proved.

Cont


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• According to the formula (1) and (2), the derivative of

tangent function can be found.

d(tanx)dx = sec2 x

Proof • If f(x) = tanx

Cont


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• The derivatives of the csc, sec, and cot are listed

below:

d(secx)/dx = secx tanx

d(cscx)/dx = -cscx cotxd(cotx)/dx = -csc2x

Cont


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The follow table is a list of most commonly used

derivatives of the trigonometric functions

Number Derivatives of the trigonometric functions

1 d(sinx)/dx = cosx2 d(cosx)/dx = -sinx

3 d(tanx)/dx = sec2x

4 d(secx)/dx = secx tanx

5 d(cscx)/dx = -cscx cotx

6 d(cotx)/dx = -csc2x

CASE STUDY


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CASE STUDY

As Curtis drives from McDonalds

at 5 m/sec, his daughter notes

that as the distance between the

car and the light increases, the

shadow of the car gets longer. She knows that this is due to the

changing angle made by the

pole, the car and the light.

She is curious about how fastthis angle changes one second

after starting to leave.

Car's shadow

Cont...



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What is known:

• The height of the pole is

4 m.

• The height of the car is1.5 m.

• The speed of the car is 5

m/sec.

The Relationship among

the Pole, the Light and

the Car

Cont...


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Questions

How fast does the angle that is made by the pole,

the light and the car change 1 second after they

start driving?

Approach

Derivative can be employed in this case.

CASE STUDY SOLUTION


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CASE STUDY SOLUTION

• The speed of the car is 5 m/sec.

• The height of the pole and the

car is 4 and 1.5 meters

respectively.

• How fast does the the angle θ

change 1 second after the carstarts to move.

• Let the angle made by the pole,

the light and the car be θ.

• Let the distance from the pole tothe car be x.

• Let the length of the shadow be

y.

The Relationship among

the Pole, the Light and

the Car

Cont...


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• ΔABC is similar to ΔADE, so

1.5/4 = y/(x + y)

y = 0.6x

x + y = 1.6x

• In trigonometry, the

relationship between x + y and

θ is:

tanθ = (x + y)/4= 1.6x/4

= 0.4x

Trigonometry for Finding

sec2θ

Cont...


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• Differentiate this relationship with respect to time:

sec2θ(dθ/dt) = 0.4dx/dt

dθ/dt = 0.4(dx/dt)/sec2θ

• Notice that the rate of change of the angle, dθ/dt is afunction of both the car speed, dx/dt and the current

angle.

• In this case dx/dt = 5 is given, but sec2

θ must bedetermined

Cont...


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• 1 second after the car starts,

x = (1)(5) = 5

x + y = 1.6(5) = 8

sec2

θ(AD/4)2

= (


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3. Differentiability

• Consider for example, the function y

= 3x2 + 2x - 1, then dy/dx = 6x + 2,which is reasonable.

• When x < -1/3, dy/dx or tanθ is

negative and when x > -1/3, dy/dx or

tanα is positive.• When x= 1/3, the slope of the

tangent line is horizontal and thus

equal to 0.

• These correspond to its derivativefunction figure and it is obvious that

the function is differentable at all

points.

y = 3x2 + 2x + 1

dy/dx = 6x + 2

Cont...y = (x2)0.5y = (x2)0.5


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Cont...

• A function is differentiable at a

point if the derivative of the

function exists at that point.

• A function is differentiable on

an interval if it is differentiable

at every point in the interval,

as already concluded.

• According to the abovedefinition, y = 3x2 + 2x - 1

function is differentiable.

3x2 + 2x - 1 is differentiable

y = (x2)0.5

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• Now, consider another example, y = (x2)0.5. Is y

differentiable in this case at all points?

• The definition of derivative gives

• In order to applying the definition to the y = (x2)0.5

function, Δy/Δx need to be find first.• At x = 0,

Cont...


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The left hand limit is:

The right hand limit is:

The left hand limit is not equal to the right hand limit.Thus does not exist, in other words, the

function y = (x2)0.5 is not differentiable when x = 0.

(x2)0.5 is not Differentiable at

all Points

Differentiability and Continuity

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e e t ab ty a d Co t u ty

• If a function is differentiable at

point b, then this function is

continuous at this specific point.

Proof:

• When x approaches b, the

difference between f(x) and f(b)

is yx - yb. Since x ≠ b, yx - yb can

be written as:Differentiability and Continuity I

Cont...



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Thus

Therefore,

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Co t• This proves that when y is

differentiable at point b, it iscontinuous at that point.

• Function y = 3x2 + 2x - 1 is

differentiable in its variable

range, and so it iscontinuous.

• If a function is continuous, is

it differentiable? The answeris may not be. For example

y = (x2)0.5 is continuous, but

it was shown that it is not

differentiable.

Differentiability and Continuity II

Differentiability and ContinuityNot Differentiable at a -- A Corner



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y y

Function y = (x2)0.5 is

not differentiable at 0

because its left hand

limit is not equal to its

right hand limit.

Normally, if a function's

graphic has a corner or

kink (loop), then thefunction is not

differentiable.

Not Differentiable at a -- A Corner

Not Differentiable at a -- Kink or Loop

Cont...


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If a function's graphic is

discontinuous, then this functionis not differentiable. since the

function's left and right hand

limits are different.

When the curve of a continuousfunction has a vertical tangent

line at point a, the slope of the

tangent line is infinity. This

means

Not Differentiable at a -- Function

Discontinuous

Not Differentiable at a -- Vertical

Tangent Line

Case StudyTilt Ice Hockey StickTilt Ice Hockey StickTilt Ice Hockey Stick


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y

After a game, Alan put his ice hockey

stick against a vertical wall, but thebottom of the stick slid away from the

wall.

He wonders how fast the slippeddistance changes with respect to the

angle made by the wall and the stick.

What is known:

• The length of the stick is 53 in.

• The angle made by the wall and the

stick is θ = 45o.

• The slipped distance is y.

Tilt Ice Hockey Stick

Cont...


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Questions

• How fast does the slipped distance change with

respect to the angle that is made by the wall and the

stick is 45o or π/4?

Approach

• Derivative with respect to angle can be used in this

case to calculate the rate of slippage.

CASE STUDY SOLUTION


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• A 53 in ice hockey stick rests

against a vertical wall.• The bottom of the stick slides

away from the wall.

• How fast does the distance, y

change with respect to the angle, θ,which is made by the wall and the

stick, when θ is equals 45o?

Assumption:

• The angle made by the wall andthe stick is θ.

• The slipped distance is y.

Cont...y = 53 sinθy = 53 sinθ


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According to the diagram,

sinθ = y/53

Thus

y = 53 sinθ

y = 53 sinθ

Cont...dy/dx = 53 cosθ


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Derivative y with respect to θ

dy/dθ = d(53 sinθ)/dθ

=53 cosθ

When θ = 45o,

dy/dθ = 53 cosθ = 37.5 in/rad

• The derivative of y = 53 sinθ is dy/dθ =

53 cosθ when θ is in the range of (0,

π/2).• Since y is differentiable, it is

continuous, this is shown in the plot of

y = 53 sinθ.

dy/dx = 53 cosθ

Cont...


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• In previous sections, various

formulas have been introduced tocalculate derivatives.

• However, it is not possible to have a

formula for every possible situation.• Thus, basic formulas like d(xn)/dx =

nxn-1 need to be generalized so they

can be used in a variety of cases.

• Suppose y = (x2 + 1)3, how can

dy/dx be calculated using derivative

formula?

Derivative Formula

Cont...



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• Let f(u) = u3, where u = g(x) = x2 + 1.

• Then

• The derivative of f and g can be calculated according

to derivative formula.

• A rule is needed to calculate the derivative of Fwhich is a composite function.

• This rule is named as Chain Rule.

4.Chain Rule


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The Chain Rule states:

• If the derivatives of g(x) and f(g(x))

exist, and F = f g is the composite

function defined by F(x) = f(g(x)), then

the derivative of F(x) exists and isgiven by

• F'(x) = f '(g(x))g'(x)

• In other words, If both y = f(u) and u =

g(x) are differentiable functions, then

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• When y = (x2 + 1)3, y

relates to the power

function of x2 + 1, this is a

special case of chain rule.

• If n is any real number

and u = g(x) is

differentiable, then

Cont...



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• Returning to the example y = (x2 + 1)3, dy/dx can be

easily be determined.

• Let y = f(u) = u3 and u = g(x) = x2 + 1, then

• so

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CASE STUDY


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• Albert, astronomy amateur, notices

that the brightness of Delta Cepheiperiodically increases and decreases.

• He wonders how fast does its

brightness changes at time equals1day?

What is known:

• The brightness cycles every 5 days

• The brightness of the star is modeled

by the function:

B(t) = 4 + 0.35 sin(2πt/b)

• The time t is 1 day

Brightness of Delta Cephei

Cont...


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Questions

• How fast is the brightness changing after a day?

Approach

• Applying the chain rule to find the derivative of thebrightness.

CASE STUDY SOLUTION


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• The brightness of Delta Cephei

increases and decreases according toits brightness function

B(t) = 4 + 0.35sin(2πt/b).

• The brightness of the star b is 5 days.

How fast does its brightness changesat time equals one day?

• Substitute the average brightness of

the star 5 days into the brightness

function,• B(t) = 4 + 0.35 sin(2πt/5)

• Let f(t) = 2πt/5, then

• B(t) = 4 + 0.35 sin(f(t))B(t) = 4 + 0.35 sin(2πt/5)

Cont...


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Using the to Chain Rule,

the rate of changes for thebrightness of Delta Cephei

is

dB(t)/dt = 0.7 cos(2πt/5)/5

Cont...


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After a day the brightness is

Cont...


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In the last section, the Chain Rule is introduced. It

can calculate the derivative of a function when it is

can be expressed in terms of another expression,

such as y = (x +1)2 sin(x + 1).

Chain Rule Example

Cont...



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Suppose y is defined by a relation with x

and it is hard to express the function y in

terms of x.

For example, x3 + y3 - 4xy = 0. In order to

draw the diagram of function x3 + y3 - 4xy =

0, the diagram of function z = x3 + y3 - 4xy

is plotted.

The function z intersects the x-y plane at z

= 0.

The function expression and diagram

shows that y is hard to be rewritten in term

of x.

In these cases, the derivative of y can be

calculated by with the Implicit

Differentiation method.

Implicit Differentiation Example

Cont...



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Function z = x3 + y3 - 4xy Function x3 + y3 - 4xy = 0

5.Implicit Differentiation


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• The implicit differentiation method

states:• An equation f(x,y) = 0 defines y

implicitly as a function of x.

• In order to to find the derivativeof y, differentiate both sides of the

original equation f(x, y) = 0 and

solve the resulting equation fordy/dx.

• This differentiation method is

known as implicit differentiation

Implicit Differentiation

Cont...



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• For example, given x3 + y3 - 4xy = 0, find dy/dx.

• Differentiate both side of x3 + y3 - 4xy = 0 with

respect to x, gives

• (1)

• Since y is implicitly defined by x, d(y3)/dx is not 0.

Consider z = y3 and apply Chain Rule,

• (2)

Cont...


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Recall that if F(x) = f(x)g(x), then F '(x) = f '(x)g(x) + f

(x)g '(x).

The formula can be used to calculate d(4xy)/dx.

Consider f(x) = x and g(x) = y, thus

(3)

Cont...


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Substitute equation(2) and (3) into (1),

so

• 3x2 + 3y2dy/dx - (4y + 4xdy/dx) = 0

• (3y2 - 4x)dy/dx = 4y - 3x2

• dy/dx = (4y - 3x2)/(3y2 - 4x)

CASE STUDY


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In his arts and crafts class, Tony

is given a piece of paper andasked to make a rectangular box.

The length of the box he created

is 3 times of its width. His friends made boxes with

various lengths, widths and

heights.

He wonders how the heightchanges affects the width.Rectangular Box

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What is known:

• The surface area of the box is constant at 100 in2.

• The length is 3 times of its width.

Questions

What is the rate of change of the width with respect to

the enlarge of height?

Approach

• Find the relationship between the width and the

height.

• Apply implicit differentiation method

CASE STUDY SOLUTION


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• Recall Tony made a rectangular box with 100 in2

surface area in his arts and crafts class.

• The length of his box is 3 times of its width.

• He wondered how the height changes affects width's

enlargement?

• Let the length of the box be y, the width be x and the

height be z.

• Let the surface area of the box be s.

• The surface area of the box can be calculated as

s = 2xy + 2xz + 2yz

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• Since the length is 3 times of the box's width,

substitute y = 3x into the equation.

• s = 2x(3x) + 2xz + 2z(3x)

• = 6x2 + 2xz + 6zx

• = 6x2 + 8xz

• Because the height z is implicitly defined by its width

x, the expression, the height z changes with respect

to its width x (dz/dz), can be calculated using implicit

differentiation method.

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• Differentiate both side of surface area equation with respect

to xds/dx = d(6x2 + 8xz)/dx

• The left hand side of the equation, ds/dx is zero since the

surface area is constant at 100 in2

.• The right hand side is

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• The left hand side equals the right hand side, so

12x + 8z + 8xdz/dx = 0

• Rearranging gives

dz/dx = -(3x + 2z)/2x = -1.5 + z/x

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derivative ii tahun 2012.pdf

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