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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.
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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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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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The first derivative test
Maximum point
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()
=
(, )
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Maximum point
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The first derivative test
< = >
′() + 0 -
m
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Test table
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The first derivative test
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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The first derivative test
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< = >
′() - 0 +
Slope
Test table
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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
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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
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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
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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…
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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
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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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Example in rate of change
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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
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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−
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Example in rate of change
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
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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
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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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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).
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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
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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, =
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= +
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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