CHAPTER 2 DIFFERENTIATION

f) = 3 2 + 9


Ans: 3 + 8 8

g) = 9 ln +

9 ln


Ans: +

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FORMULA Use various differentiation techniques
to solve problems

Solve the following functions using suitable method:

Example: 1 Example: 2

= ( )( )
=

= 3 3 = cos

= 9 2 = − sin = cos = sin 2

= −sin = 2 cos 2

= −
= +
= ( 3)(− sin ) + (cos )(9 2)
2
= − 3 sin + 9 2 cos
(sin 2 )(−sin ) − (cos )(2 cos 2 )
= (sin 2 )2

− sin sin 2 − 2 cos cos 2
= (sin 2 )2

Example: 3 Example: 4

= ( − ) =

= tan = 2 − 2 = 5 = sin

= sec2 = 2 = 5 4 = cos



= × = ×

= sec2 × 2 = 5 4 × cos

= 2 sec2 = 5 sin 4 cos
= 2 sec2(2 − 2)

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EXERCISE 7

Solve the following functions using suitable method:

a) = sin (12 4 − 2

3)

Ans : 4 3 (1 4 − 3) cos (1 4 − 2
22
3)

b) = (4 2)(tan )

Ans: 4 2 sec + 8 tan
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c) = cos 2(4 2 + 2)

Ans : −16 cos(4 2 + 2) sin(4 2 + 2)

d) = 3 cos

Ans : − 3 sin + 3 2 cos
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e) = 3 2+1

cos

Ans : 6 cos +(3 2+1) sin
cos2

f) = cos 2

tan 2

Ans : −2 tan 2 sin 2 −2 cos 2 sec 2
tan2 2

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Differentiate each if the following functions:

Example: 1 Example: 2

= ( + ) = √ +

= 3 + 4 = tan = 7 + 1

= 2

= 3 3 = 2 = 7 7 + 1 = 1 −21
2

= 1
= ×
2√

3 3 × 2 7 7 + 1 1
= = × 2√

= 3 3 2 ( 3 + 4) 7 7 + 1
= 2√ 7 +

Example: 3 Example: 4

= = +



Solution: Apply the quotient rule

= 6 3 = 2 = e7x+2 = cos 2

= 18 2 2 2 7e7x+2 = −2 sin 2
= =

= −
= +
= (6 3 . 2 2 ) + ( 2 . 18 2) ]
2
= 6 2 2 (2 + 3)
= ( cos 2 . 7e7x+2)−(e7x+2 .−2 sin 2 )

(cos 2 )2

e7x+2(7 2 + 2 sin 2 )
= (cos 2 )2

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EXERCISE 8

Find for the function below.


a) = √ 1 −

Ans: −
2√ 1−

b) = 3 2

Ans: 3 2(2 2 + 3)
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c) = ln 2

Ans: 2 ( 2 ln + 1 )

d) = 7 2 ( 4 + 9)3

Ans: 14 2 ( 4 + 9)2[ 4 (6 + ) + 9 ]
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e) = e 3x
1+ e 2x

Ans: 3 (3+ 2 )
(1+ 2 )2

f) = 3
( +5)2

Ans: 3 (3 +13)
( +5)3

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Differentiate each if the following functions

Example: 1 Example: 2

= ( ) = ln √

= tan 8 = ln = sin 3 1
1
= = ln 2

= 8 2 8 1
= 3 3 = 2 ln

= 1
= ×
2

= 8 2 8 × 1 1
= 3 3 × 2
3 3
8 2 8 = 2 sin 3
= tan 8

Example: 3 Example: 4

= ( − ) =



Solution: Apply the quotient rule

= ln 2 = cos 7

= 4 3 = ln 9 − 1
− 1 = = −7 sin 7
= 12 2 = 9 −

−
=
2

= + = ( cos 7 . 1 )−(ln 2 .−7 sin 7 )


= (4 3 . −1 ) + (ln 9 − . 12 2) ] (cos 7 )2
9−
cos 7
= 4 2 ( − + 3 ln 9 − ) ( ) + 7 sin 7 ln 2
9− =

(cos 7 )2

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EXERCISE 9

Find for the function below.


a) = ln (cos + sin 3 )

Ans: 3 cos 3 −sin

cos +sin 3

b) = ln √cos

Ans: − sin = − tan
2 cos 2

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c) = (ln tan )6

Ans:6 2 (ln tan )5

tan

d) = 5 3 ln ( 1 − 3 2)

Ans: 15 2 [1−−23 22 + ln(1 − 3 2)]
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e) = 1+ln



Ans: − ln
2

f) = ln 3
2

Ans: 1−2 ln 3
3

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Apply Second Order Differentiation

FORMULA 2.3.1 Execute second order differentiation by using
various differentiation techniques for algebraic
expressions

Definition: The derivative of a derivative is called the second derivative

In the previous topic we looked at first order differential equations.
In this topic we will move on to second order differential equations.
The second order differential equations is by differentiating the first derivative.

SECOND ORDER DIFFERENTIATION

Second derivative = ( )



There are two main ways of denoted the second derivative, which are



Or

′′( )

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Example: 1 (Original Function)

= + −

= + − + (First Derivative) Step
1


= + − (Second Derivative) Step
2

Example: 2 (Original Function)

= +

= + (First Derivative)



= − + (Second Derivative)


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EXERCISE 10

Find second order differentiation ( ) for the following:

a) = 10 4 − 5 3 + 3 − 7 b) = 3+2
2

Ans: 2 = 120 2 − 30 2 12
2 Ans: 2 = 4

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c) = (3 − 2 )5 d) = ( + 3)(4 2 − 1)

Ans: 2 = 80(3 − 2 )3[ ] 2
2 Ans: 2 = 24 + 24

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e) = f) ( ) = 5 sin 6

+8

Ans: 2 = −16( + 8)−3 2
2 Ans: 2 = −180 sin 6

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g) = 4 + 5 h) = 2

[Ans: 2 = 16 4 ] [Ans: 2 = 2 2 (2 2 + 1) ]
2 2

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FORMULA 2.4 Demonstrate the Application of
Differentiation

2.4.1 Use differentiation to find the gradient
of the curve

2.4.2 Execute turning points/ stationary points
2.4.3 Solve maximum and minimum problems
2.4.4 Show the graph of a curve

Definition:
Process of finding the maximum or minimum value of some function

One of the most important uses of calculus is determining minimum and maximum values.
This has its applications in manufacturing, finance, and engineering and a host of other
industries. For example, Calculus is use to maximize the profit in a company, or minimize the
costs, or find the least amount of material to make a particular object.

Theory: Maximum, minimum Points of Inflection
For a function: = ( ), The value of ′( ) is the gradient at any point

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If the gradient of the functions changes sign at the Stationary Point where the
gradient of the function is zero ( ′( ) = 0), then it called a Turning Point

A Turning Point can be a local maximum or local minimum point
If the gradient of the function does not change sign at the Stationary Point, then it

is a point of inflection

The Second Derivative of ( ): "( ) is often used to test for concavity


which verify that a point is maximum or minimum.

 "( ) < , ∶ The function is maximum turning point

(concave downwards)

 "( ) = , ∶ The function may be a point of inflection

(The function changing from concave downwards to upwards or vice versa)

 "( ) > , ∶ The function is minimum turning point

(concave upwards)

It is customary to use terminology in this context. If you think of the graph as a face

reaction, then

 Concave up means a smiley face

 Concave down means a sad face

 Neither Concave up nor Concave down means a puzzled face

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Steps To Find Stationary Point
Find the stationary point on the graph of ( ) = + − and state their nature

(i.e. whether they are maxima, minimum or points of inflexion).

Step Explanation Example

Step 1 Find the derivative ′( ) = 2 + 2
o f

S t e p 2 S o l v e ′( ) = f o r t h e ′( ) = 0
critical values 2 + 2 = 0

= −1

Step 3 Plug the critical value When = −1:
i n t o t o o b t a i n t h e
function value = 2 + 2 − 3
= (−1)2 + 2(−1) − 3
= −4

Step 4 State the Stationary Therefore, Stationary point occur at
point (−1, −4)

Step 5 Find the second "( ) = 2
d e r i v a t i v e o f

Step 6 Determine the nature Since "( ) = 2 > 0,
of the point which shows (-1,-4) is a minimum point

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Example: 1

Find the turning points of the function = − + and determine whether it is a
minimum point, a maximum point or point of inflexion. Sketch the graph.

1. Find the derivative of .
= 3 2 − 3



2. Solve = for the critical values


= 0



3 2 − 3 = 0

( − 1)( + 1) = 0

= 1, = −1

3. Plug the critical values into to obtain the function values
When = 1: = 3 − 3 + 2 = (1)3 − 3(1) + 2 = 0
When = −1: = 3 − 3 + 2 = (−1)3 − 3(−1) + 2 = 4

4. State the Turning points
Therefore, Turning point occur at (1,0) and (-1,4)

5. Find the second derivative of

2 = 6
2

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6. Substitute each point in = and determine the nature of the point


When = 1: 2 = 6 = 6(1) = 6 > 0 which shows (1,0) is a minimum point
2

When = −1: 2 = 6 = 6(−1) = −6 > 0 which shows (-1,4) is a maximum point
2

7. Create data by take any points just before and after for each turning points (TP)

TP TP
-2 -1 0 1 2
0 4 2 0 4

8. Sketch the graph of cubic function

TP
TP

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EXERCISE 11
a) Find the coordinates of the stationary points of the curve = 2 2 + 8 − 3. Determine their

nature and sketch the graph of the curve.

Ans: (−2, −11)minimum point
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b) Find the coordinates of the stationary points of the curve = 3 − 6 2 + 9 + 5. Determine
their nature and sketch the graph of the curve.

Ans: (3,5)minimum point and (1,9)maximum point
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c) Compute the stationary points for the curve = 2 3 − 3 2. Hence, identify their natures.

Ans: (1, −1)minimum point and (0,0)maximum point
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d) A curve has equation = 2 3 − 6 2 + 6. Solve the given equation to find the turning points
and their natures.

Ans: (2, −2)minimum point and (0,6)maximum point
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e) Find the stationary points for the equations = 4 3 + 3 2 − 6 and determine their nature.

Ans: 1 , − 7 minimum point and (−1,5)maximum point
(2 4)

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Calculate rates of change

DEFINITION Definition:
′( ) represents the rate of change of ( )

The idea is to find the rate of change of one quantity in terms of another quantity whose
rates of change are known.

The steps to solve an optimization problem

1. Draw a diagram if possible
2. Label the quantities
3. Find and equation that relates the quantity whose rates of change to be

found and the quantities whose rates of change are known
4. Use the chain rules / implicit to differentiate both sides of the equation

with respect to time, .
5. Solve the derivatives that will give the unknown rates of change

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Steps To Solve Rates of Change

A circle is decreasing in size at the rate of 5 / . At what rate is the radius
decreasing when the radius is 4 cm.

Step Explanation Example

Step 1 Write out what we

know from the a) = −5 cm/minute

question
b) = 4
a ) The rate that the area is
decreasing c) =?

b ) The radius at the time we want
c ) The rate that the radius is

changing

Step 2 Formula shape = 2

Start with the equation for the area
of a circle

Step 3 Differentiate the
Step 4 = 2
formula
Find
= ×
1
= 2 × −5
Then use Chain Rule

to determine the rate

of change

So that, the arrangement . Find

. Use parametric concept.


Step 5 Simplify and 5
substitute = − 2
5
= − 2 (4)
5
= − 8 @ − 0.2 cm/minute

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Example 1

The width of a rectangle increases at a rate of − . The length is four times its width.
Find the rate at which the area is increasing when it widths is

Given:

Changing rate of its width, = 2 −1


Find: 4

=? When = 5



Area of rectangle:

= 4 × = 4 2

Therefore

= 8



Then use chain rule determine the rate of change


= ×

= 8 × 2

Substitute the value of = 5 into the equation


= 8(5) × 2

= 80 −1


∴The rate at which the area is increasing when it widths is 5 is 80 −1

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EXERCISE 12

Solve the following by Chain Rule

a) The length of a square, is increasing at a rate b) A girl is blowing a spherical balloon at a rate
of 3 −1. Calculate the rate of change of of 30 3 −1. At what rate is the radius of

the area of the square at the instant when the the balloon change when it radius is 10 .
length of the square is 20

Ans: = 120 2 −1 Ans: = 0.02 −1


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c) The radius of a circle is decreasing at a rate d) The volume of huge ice is decreasing at a rate
of 7 −1. Find the rate of change of the of 300 3ℎ−1. Find the rate of decrease of
the length of its side at the instant when the
area for circle at the instant when the radius length of the cube is 5 .
is 4 .

Ans: = −5600 2 −1 Ans: = −4 ℎ−1


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e) Find the rate of change of the square area f) A spherical balloon is inflated at rate of 3

whose side is 8 cm long if the side length is cm3/s. find the increment rate of the radius

increasing at 2 cm/min. is 2 cm and 4 cm.

Ans: = 32 2 −1 Ans: = 0.0597 −1 0.0149 −1


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Understand Parametric Equation

Explanation:

Parametric derivative is a derivative in calculus that is taken when both the and variables
(traditionally independent and dependent, respectively) depend on an independent third
variable , usually thought of as “time”.

FORMULA PARAMETRIC EQUATION

If
= ( ) = ( )

Then


= ×


= ( )

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Use the chain rule to find derivative of parametric
equation for algebraic expressions

Steps To Differentiate Parametric Equation

Step Explanation Example

Step 1 D i f f e r e n t i a t e b o t h a n d The function:
with respect to the = 3 and = 4 2
parameter, t.
= 3 and = 8
Find and


Step 2 From the chain rule we 8
know that = 8 = 1


= ×

= 1

8

Hence, rearrange to

Step 3 Substitute


= ×

Step 4 Simplify = 3 × 1

8


= ×

= 3 × 1

8

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F i n d ( / ) for parametric equation below in term of t.

Example: 1

= + and = +

= 2 3 + 4 t Step
1
= 6 2 +4 = 2 + 8


= 2

Step 1
2 = 2


= ×

= (6 2 + 4) 1 Step
× 2 3

= 6 2 +4 = 3 2 +2

2

Example: 2

= − and =

= 2 5 − t = ln
1
= 10 4 − 1 =



=


= ×

= (10 4 − 1) ×


= 10 5 − t

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Example: 3

= ( + ) and = −

= 4 e(5t+3) = 8 − 6 3

= 20 e(5t+3) = −18 2


1
= − 18 2


= ×

= 20 e(5t+3) 1
× − 18 2

= − 20 e(5t+3)
18 2

= − 10 e(5t+3)
9 2

Example: 4

= + and = −



= 4 5 + 3 = 2 − sin 6

5 = 2 − 6 6

= 4 4 + 3 2



= 2 −6 1 6



= ×

= 4 4 + 3 2 1
× 2 − 6 6

= 4 4 + 3 2 = 2 (4 2 + 3)

2 −6 cos 6 2( −3 6 )

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EXERCISE 13

F i n d ( / ) for parametric equation below in term of t.
a) = 3 2 + t6 and = 15 2 + 6t

Ans: ( 1+ 4 )

5 +1

b) = 5 4− 2 and = 2 3 − 2

2

Ans: e(4t+5)
(2 2+3 −1 )

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c) = 3 2 − 2 + 1 and = ln 2

Ans: t (3t – 1)

d) = 2 5 and = 3 2 − 5

Ans: −5 5

3

e) = e(4t+5) and = 4 + 2 3 − 2

2

Ans: (10 2−1)
2(3 2−1)

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Apply Implicit Differentiation

Use the chain rule to find derivative of
parametric equation for algebraic expressions

The Process of finding the derivative of a dependent variable in an implicit function by
differentiating each term separately.

Previously we meet equations here is expressed explicitly in term of only.

DEFINITION IMPLICIT DIFFERENTIATION

For example, if
= −  explicitly as a function of [ = ( )]

Then the derivative of is
To find

=



However, not every equations expressed explicitly in term of
only, such as:

+ =  implicitly as a function of
[ ( , ) = ( , )]

To find derivatives of such expressions, we need to use implicit differentiation.
Implicit differentiation is a function in which the relationship between and is
not clear.
To get the differentiation of these equations, the way is to differentiate each
term individually and reorganized the results obtained.

85

Example: 1

Find the expression for if + − =


To find to the equation, we differentiate each term of x as


+ − =



+ − =



We rearrange to collect all terms involving together



+ = +



+ = ( + )



∴ = +



Example: 2 (Involves Product Rule)

Find the expression for if + =



To find to the equation, we differentiate each term of x as


( ) + ( ) = ( )


[ ( ) + ( )] + ( ) =


( ) + + =



We rearrange to collect all terms involving together



( + ) = −



∴ = −
+

86

EXERCISE 14

Differentiate each of the following with respect to and find


a) 6 3 − 2 = 1 b) 2 − 2 + 3 = 5 2

9 2 2 + 3
[Ans: = ] [Ans: = 12 ]

87

c) 4 − 2 + 10 = 2 d) 2 2 + 2 2 − 3 = 4

10y + 4 −4 − 2 2
[Ans: = −10 + 2 ] [Ans: = 4 − 3 2 ]

88

e) 5 2 − 3 = 3 f) = +

3 2 sin − 10 1 −
[Ans: = − 3 cos − 3 ] [Ans: = − 1 ]

89

g) 2 + 2 = 3 + ln h) sin = 3

4 3 + 7 − 2 ]
[Ans: = [Ans: = ]
2

90

EXERCISE 15

Determine the value of at the point given.


a) 2 + 5 − 4 2 + 3 + 21 = 0 b) 2 + ( − ) 3 = 9 at point (1,3)
at point (2,3)

9 5
[Ans: = 21] [Ans: = 6]

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c) 2 + 4 = 4 + 2 at point (-1,1) d) 3 −4 − 3 2 = 4 2at point (0,-1)

4 8
[Ans: = 5] [Ans: = 3 ]

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2.7 C o n s t r u c t P a r t i a l D i f f e r e n t i a t i o n

2.7.1 Define partial differentiation

DEFINITION Definition:

A partial function is the process of finding one of the partial derivatives
of a function of several variables

For functions of two variables the notation simply becomes = ( ; )

and are two independent variables
is the dependent variable
If we want to differentiate = ( ; ), we have to decide whether we are differentiating
with respect to or differentiating with respect to
Use symbol “ ” (usually call “partial” or “rounded d” or “del d” or “curly d” ) for partial
derivatives of , which are

is read as “partial derivative of (or ) with respect to ”, and means differentiate


with respect to holding constant

means differentiate with respect to holding constant


Another common notation is the subscript notation:

 means


 means


Note that we cannot use the dash ‘ symbol for partial differentiation because it would not be
clear what we are differentiating with respect to

93