topic 3 introduction -differentiation

7/28/2019 Topic 3 Introduction -Differentiation

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MODULE OUTCOMES:

MO1 Identify basic mathematical concepts, skills and mathematical techniques for

algebra, calculus and data handling.

MO2 Apply the mathematical calculations, formulas, statistical methods and calculus

techniques for problem solving in industry.

MO3 Analyze calculus and statistical problems in industry.



LEARNING OUTCOMES

At the end of this topic, student should be able to :

Identify the rules of differentiation Apply the rules of differentiation to find the derivative of a given

function

Solve simple differentiation problems

Differentiate higher order differentiation

Identify the nature of critical points: maximum, minimum andinflection points

Determine the coordinates of the maximum point and minimumpoint

Solve basic application of differentiation in economic and

engineering concepts Define the integration as anti-derivative

Use the basic rules of integration

Solve simple integration problems

Solve basic application of integration

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Differentiation

Rules of DifferentiationRule 1 : The Derivative of a Constant

Example 1Differentiate

Solution

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0cdx

d

3 y

0dx

dy



Rule 2 : The General Power Rule

Example 2

Differentiate

Solution

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1 nn

nx xdx

d

4 x y

3

4 xdx

dy



Rule 3 : The Derivative of a Constant times a

Function

Example 3

Differentiate

Solution

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)()( xkf xkf dx

d

23 x y

xdx

dy

23 x6



Rule 4 : The Derivative of a Sum or a Difference

Example 4

Differentiate

Solution

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dxdv

dxduvu

dxd )(

x x x f 73)(

2

xdx

d

xdx

d

x f 73)(2'

76 x



Rule 5 : The Product Rule

Example 5

Differentiate

Solution

Let

Therefore, using the formula for Product Rule

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dx

duv

dx

dvuuv

dx

d )(

)1)(26( 32 x x x y

x xu 26 2 13 xv

212 xdxdu 23 x

dx

dv

dx

duv

dx

dvu

dx

dy

)212)(1()3)(26( 322 x x x x x

212830 34 x x x



Rule 6 : The Quotient Rule

Example 6

Differentiate

Solution

Let

Therefore, using the formula for Quotient Rule

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2)(

v

dx

dvu

dx

duv

v

u

dx

d

1372

x x y

72 xu 13 xv

xdx

du2 3

dx

dv

2

2

)13()3)(7()2)(13(

x x x x

dxdy

2

2

)13(2123

x x x



Rule 7 : The Chain Rule

Example 7

Differentiate

Solution

Let

Therefore, using the formula for Chain Rule

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dx

du

du

dy

dx

d

42 )5( x y

5

2

xu

4u y

xdx

du2

34u

du

dy

x x xudxdy 2)5(424 323

32)5(8 x x



Rule 8 : The Power Rule

Example 8

Differentiate

Solution

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dx

du xun xu

dx

d nn 1))(())((

42 )5( x y

)5()5(4 2142

xdx

d xdx

dy

x x 2)5(432

32)5(8 x x



Exercise1. Find for the following functions :

(a)

(b)

(c)

(d)

(e)

(f)(g)

(h)

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dx

dy

54 x y

)7(14 32 x x x y

13

72 2

x

x y

25)7( x x y

3

81

x y

52 )21)(34( x x x y

x x y 23

3

2

x

x y



Differentiate Higher Order Derivatives

0Derivative of a function is a function

0Differentiate the derivative of a function andthus obtain the higher derivative of thefunction.

or ′ or is called the first derivative

or ′′ or is called the secondderivative

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dx

dy ' f

2

2

dx

yd '' f



Example 9

Given , find and

Solution

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5423)( 234 x x x x x f )(' x f )('' x f

5423)(

234 x x x x x f

21236)(

42612)(

2''

23'

x x x f

x x x x f



Maximum and Minimum ValuesDefinition

0 Critical Point

0 A point on curve where the gradient is zero (0) that is:

= 0 or

is not defined.

0 3 types of critical point:

0 Maximum point

0 Minimum point

0 Inflection point

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Maximum Point Minimum Point

= 0

is not defined

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Maximum and minimum

points

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http://upload.wikimedia.org/wikipedia/commons/1/1e/Extrema_example.svg



Continue..

0 If

= 0 the critical point is also known as the

stationary point.

0 The maximum and minimum points are also calledturning points.

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Continue..

0 If > 0 then the function is increasing in the

interval , .

0 If < 0 then the function is decreasing in the

interval , .

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Example 10

0 Find the intervals on which the function = 4 + 3 is increasing and decreasing.

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Solution

= 4 + 3

= 2 4 = 2( 2)

If < 2, < 0 (-ve). Therefore, is decreasing.

It follows that ′ < 0 if ∞ < < 2 → is decreasing.

If > 2, > 0 (+ve). Therefore, is increasing.

It follows that > 0 if 2 < < ∞ → is increasing.

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The first derivative test

0Maximum point

0Refer the following graph, point A is a

maximum point. The slope of the curve ispositive (+ve) to the left of point (, )

and negative to the right of the point

(, ).

0 If = () have a maximum point at = ,

then = 0 at = .

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Maximum point

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()

=

(, )



Maximum point

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< = >

′() + 0 -

m

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Test table




0Minimum point

0Refer the following graph, point B is a

minimum point. The slope ′() changesfrom negative (-ve)to the left of point

(, ) and the curve is positive (+ve)

to the right of the point (, ).

0 If = () have a minimum point at = ,then = 0 at = .

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Minimum point

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< = >

′() - 0 +

Slope

Test table



Example 11

0 Determine the maximum and minimum points of the

function = 4 using the First Derivative Test.

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Solution

= 4

= 4 2

Critical point when = 0, 4 2 = 0, = 2

= 2 is a maximum number, so that the maximumpoint is 2, 2 = (2, 4). The minimum point does

not exist.

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< = >

′() + 0 -

Slope



Second derivative test

0 Suppose is twice differentiable at a stationary point

.

0 If > 0, then has a relative minimum at .

0 If < 0, then has a relative maximum at .

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Step on finding max/min points:

1. Find y’(x) and y’’(x) .

2. Find x when y’(x)=0 or where y’ doesn’t exist .

3. Substitute x in y’’(x).

4. Do 2nd derivative test:

i. y’’(x) < 0 max point at x .

ii. y’’(x) > 0 min point at x .

iii. y’’(x) = 0 test failed, use 1st derivatives test.

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METHODS



Example 12

Locate and describe the relative extrema of = 2.

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Solution

= 4 4 = 4 4 = 4 1

= 4( 1)( + 1)

= 12 4

Solving ′() yields the critical points = 0, = 1 , = 1.

Then, check for

0 = 12(0) 4 = 4 < 0

1 = 12(1)4 = 8 > 0

1 = 12(1)4 = 8 > 0 Thus, there is relative maximum at = 0, and there are

relative minima at = 1 and = 1.

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Example 13

0 Use the second derivative test to find the minimum

and maximum point, if exist for the function

= 3 + 2

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Solution

= 3 6 = 3 2

= 6 6

The critical point when

= 0; = 0 and = 2. 0 = 6 0 6 < 0; shows that there is a relative

maximum at = 0.

Then (0, 0 ) is a maximum point = (0, 2).

2 = 6 2 6 > 0; shows that there is a relativeminimum at = 2.

Then 2, 2 is a minimum point = (2, 2).

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Inflection point

0 At the critical point = , when:

< 0 to the left of = ,

and

< 0 to the right of =

= 0 at = is called the point of inflection.

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Inflection

point

Concave up

Concave down

(0.0)

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Example 14

0 Determine the maximum, minimum or inflection

points for the function

= + 3 + 3

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Solution = + 3 + 3 = 3 + 6

= 6 + 6

= 6

At critical point, = 0

3 + 6 = 0

3 + 2 = 0

= 0 = 2

= 3 = 7

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Solution

When = 0,

= 6 0 + 6 = 6 (+ve)

Therefore (0,3) is a minimum point.

When =-2,

= 6 2 + 6 = 6 (-ve)

Therefore (2, 6) is a maximum point.

At inflection point,

= 0

6 + 6 = 0

= 1 = 5

Since

≠ 0 therefore (-1,5) is an inflection point.

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Exercise

Determine the maximum, minimum or inflection

points if they exist for each of the function below:

a. = 3 9

b. = 6

c. =

+

6 + 8

Next, sketch the graph of the function.

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0 Rate of change and motion:

0 Mathematics modeling

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Application of differentiation



Rate of change

0 The speed at which a variable changes over a specificperiod of time.

0 Rate of change for related quantities:

0 If = (), then

represents the change in y with

respect to the change in x . For example,

= 5 means

y increases by 5 units for each unit increase in x and

= 5 means y decreases by 5 units for each increase

in x .

0 In cases where three variables are involved say

= () and = (), then

=

×

.

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0 In economics we will see that the rate of change as follows:

0 If the total revenue function, () and the total cost function,() and the total profit function, () is given by

= ()

0 And differentiating the total profit function is given by

= ()

0 For maximum profit, = 0, thus = 0

0 The relationship of revenue:

= ×

0 When the quantity will change in time, t then the function willnote as below:

= ()

0 From product rule we found that = + as a rate of change of revenue in time.

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Continue…



Example in rate of change

1. Given that = 3 4 where x and y are two

related variables, find the rate of change of x when x

= 2 if y increases at a rate of 2 unit −.

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Solution



Solution

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:

=

×

= (3 4)

= 3 4 , =

= 3,

= 2

=

×

= 2 × 3

= 6

Subs u = 3x-4

= 6 3 4

= 18 24

:

= 2,

=

×

1 8 2 4 = 2 ×

18 24 = 2

18 24 = 2

=

2

18 24

S : Given x= 2

= 2(18 24)

=2

3624

= 212

=1

6

E l i t f h




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Volume of sphere is V =4

3πr. If the volume

is increasing at a rate of 10 cm

s−

,Find the rateat which the radius, r is increasing when r = 5cm.

Solution



Solution

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:

=

×

=4

3,

=

4

33

= 4

:

= 10,

=

×

4 = 1 0 ×

4 = 10

(4

)=10

=

10

4

:

= 5

=

10

4

=

10

4(5)

=10

100

=

cm−




2. Suppose that a product currently sells for RM25,

with the price increasing at the rate of RM2 per year.

At this price, consumers will buy 150 thousand

items, but the number being bought is decreasing at the rate of 8 thousand per year. At what rate is the

total revenue changing? Is the total revenue

increasing or decreasing?

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Solution



SolutionThe basic relationship of revenue:

= ×

Suppose that the quantities are changing in time, then = ()

Where () is revenue, () is quantity sold and () is the price, all at timet . From product rule we found that:

= + ′()

From the question above:

The initial price, 0 = 25.The rate of change of the price, 0 = 2 per yearThe initial quantity, 0 = 150 (thousand items)The rate of change of quantity, 0 = 8 (thousand items per year andnote that the negative sign denotes a decrease in Q)

Thus,

0 = 8 25 + 150 2 = +100 ( ).Since the rate of change is positive the revenue is increasing at present. Thismay be a surprise since one of the two factors in the equation is decreasing,and the rate of decrease of the quantity is more than the rate of increase inthe price.

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3. A total revenue function is given as = 40 2.

Find the second derivative of the function and comment on

the result.

Solution:

= 40 4

= 4 (the critical point is a maximum point)

This shows that the function of the total revenue will bring

maximum revenue at the critical point. That is the curve of the function at the critical point will be at the highest peak.

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Motion



Motion0 Motion is a change in position of an object with respect to time

0In mechanics the second derivative of the displacement of an object with respect to time is its acceleration and this is used in themathematical modelling of problems in mechanics using the law:

× =

0 What information does the second derivative of a function give us?Graphically, we get a property called concavity. One important application of the second derivative is acceleration.

0 Acceleration is the instantaneous rate of change of velocity.Consequently, if the velocity of an object at time t is given by v (t ),then the acceleration is:

= =

.

0 Since derivatives are always rates of change, an equivalent definition is: acceleration is the derivative of velocity.

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http://en.wikipedia.org/wiki/Position_(vector)

http://en.wikipedia.org/wiki/Position_(vector)



Example: 1

0 Suppose that the height of a skydiver t seconds after jumping

from an airplane is given by = 640 20 16 feet.

Find the person’s acceleration at time t .

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Solution

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1:

= 640 20 16

= = 20 16 2

= 20 32

2:

= = 32 ft/s2

(negative sign is shown as the direction

decreases by 32 ft/s every second due to gravity).



Example 2:

0 A particle moves along a straight line and passes

through a fixed point A. Its displacement, s from A

given by s= 6 + 2, where t is the time, in

seconds, after passing through A. Find theacceleration / motion function of the particle.

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Solution

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:

= 6 + 2

=

= 6 + 2 6

:

=

= 2 12

Therefore, the acceleration function of the particle

Is = 2 12



Mathematics modeling

0 It will illustrate how to apply the methods of calculus

to problems requiring you to find the maximum or

minimum.

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Example in Mathematics

Modeling1. A rectangular sheet of metal having dimensions 20cm by

12cm has squares removed from each of the four corners

and the sides bent upwards to form an open box.

Determine the maximum possible volume of the box.

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x

x

x

x

x

x

x

x

(20 – 2x)

(12 – 2x)12 cm

20 cm

Solution

0 Volume of box, =



= +

0

= + = for a turning point.

0 Hence, + = , i.e. + =

0 Using quadratic formula,

0 =−(−)± (−)−()()

()

= . .

0 Since the breadth is then = . is not possible and neglected.

0 Hence, = . .

0

= +

0 When = . ,

is negative, giving a maximum value.

0 The dimension of the box are:

0 Length = 20 – 2(2.427) = 15.146cm

0 Breadth = 12–

2(2.427) = 7.146cm0 Height = 2.427cm

0 Maximum volume = (15.146)(7.146)(2.427) = 262.7 cm3

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topic 3 introduction -differentiation

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