eBook Calculus Differentiation for Science And Engineering Students And

Calculus: Differentiation

Example 1 Find the derivative for the followings: = ln 5 − 3
1
= ln 6 differentiate = ln 2 5 Simplify using law of log = 5 − 3 ∙ 5
constant = 5 ln 2 =
= 0 5
1 = 5 − 3
= 2 ln 5 = 5 ∙ 2 ∙ 2
46
1 5
= 2 ∙ 5 ∙ 5 =

2
=

Calculus: Differentiation

= ln 3 − 4 5 Simplify using law of log = ln 3 5 − 2 change Simplify using law of log
= 5 ln 3 − 4 = the form =
1
1
= 5 ∙ 3 − 4 ∙ 3 = ln 5 − 2 3

15 1
= 3 − 4 = 3 ln 5 − 2

1 1
= 3 ∙ 5 − 2 ∙ 5

5
= 3 5 − 2

= 3 ln 7 − 3 − 4 2

1
= 3 ∙ 7 − 3 − 4 2 . −3 − 8

3 −3 − 8
= 7 − 3 − 4 2

47

Calculus: Differentiation

= ln 5 − 3 2 3 2 + 4 Simplify using law of log
= ln 5 − 3 2 + ln 3 2 + 4
= 2 ln 5 − 3 + ln 3 2 + 4 2 = 2 + 2

Simplify using law of log
=

′ =2∙ 1 ∙5+ 1 ∙ 6
5 − 3 3 2 + 4
7
10 6 = ln 1 − 3 2 Simplify using law of log
= 5 − 3 + 3 2 + 4
= ln 7 − ln 1 − 3 2
2 = 2 − 2

1 1
= 7 ∙ 7 − 1 − 3 2 ∙ −6

1 6
= + 1 − 3 2

48

Calculus: Differentiation

Example 2 Find the derivative for the followings:

Quotient Rule

Product Rule ln 5 − 2 Remember !
= 2 4
5 − 2 5 − 2
2 4 ≠ 2 4

= 2 4 ln 5 − 2 = ln 5 − 2 = 2 4
−2
= 2 4 = ln 5 − 2 = 5 − 2 = 8 3


= 8 3 1 −
= 5 − 2 ∙ −2
=
−2 2
= 5 − 2
2 4 ∙ −2 − ln 5 − 2 ∙ 8 3
= 5 − 2
= + ′
2 4 2
−2
′ = ln 5 − 2 ∙ 8 3 + 2 4 ∙ 5 − 2 −4 3 + 2 ln 5 − 2 Factorize
= 5 − 2
4 4
= 8 3 ln 5 − 2 − 5 − 2 Factorize 4 8

= 4 3 2 ln 5 − 2 − 1
5 − 2 = − 5 5 − 2 + 2 ln 5 − 2

49

Calculus: Differentiation

Chain Rule

Example 3 Find the derivative for the followings:

= cos 2 3

Chain Rule = cos 2 3 =
1
= − 2 3 ∙ 6 2 =

= ln 3
= −6 2 2 3

= ln 3 =
= ×
1 1
= 3 ∙ 3 = = 2 1
= ∙ −6 2 2 3

−6 2 2 3
= × =

= 2 ∙ 1 −6 2 2 3 Simplify using reciprocal identities
= cos 2 3

2 ln 3 =
=
= −6 2 2 3

50

Calculus: Differentiation

IMPORTANT! = 2 − 3 2 − 3 Simplify using law of log
= 2 − 3 2 − 3
2 = 2 + 2
7
= 1 − 3 2 Use Product Rule

7
= 2 − 3 2 = +
= 5 − 3 2
= 5 − 3 2 Simplify using law of log


2 = 2 − 2

Use Quotient Rule

= −
2

Simplify using law of log

=

Use Chain Rule 51


= ×

Calculus: Differentiation

Differentiation Of Exponential Function

➢ Exponential functions have
the form = , then
the derivative is given by
= ( )′ =



➢ Its can be solved either using
the exponent differentiation
formula or a chain rule.

➢ A combination of either a
product rule or a quotient
rule should be used if it
involves two functions.

52

Calculus: Differentiation

Exponential Function

Formula Differentiation of Law of Indices / Exponent
Exponential Function

a) ( ) = × = +

÷ = −

b) ( + ) = + ∙ ( + ) =



c) =

= ∙ − 1

=

where is any value in bracket

=

d) ( ) = ∙ ′( ) 0 = 1



53

Calculus: Differentiation

How To Derive Exponential Function

= 2 = 2 7 −3

STEPS = 1 ∙ 2 = 1 ∙ 2

1) Write the exponent function 2 2 2 7 −3 7
2) Differentiate the POWER
= 2 2 = 14 7 −3 Don’t forget to
OR follow the steps
1) Write u = 2x u = 7x - 3
2) Differentiate u
STEPS
(u is power of exponent )
STEPS
1) Write
1) Write 2) Differentiate u
2) Differentiate u
OR
( ) = ′( )

54

Calculus: Differentiation

Example 1 Find the derivative for the followings:

= 6 4

= 2 = 3

= 0 differentiate = 6 ∙ 4 ∙ 4
constant

= 3 ∙ 3 = 24 4

= 3 3

= 2 −3 5
= 3 3

= 2 4 2 = 5 −3
= 4 8 3

= 2 −3 ∙ 2 = 4 8 ∙ 8 = 5 −3 ∙ −3
3

= 2 2 −3 = 32 8 15
= − 3 3
55

Calculus: Differentiation

Example 2 Find the derivative for the followings:

= 2−8 6 3 Simplify using law of index
= 6 2−8 +3 × = +
= 6 2−5
3 5−2 Simplify using law of index
= 8 4 ÷ = −

3 = 4 81 8 Simplify using law of index
8
= 5−2 −4 =


= 6 2−5 ∙ −5 1 =

3 = (81 8 )4
8
= −30 2−5 = 5−6 = 1 ∙ 8 (41)

814

3 = 3 2
8
= 5−6 ∙ −6 = 3 2 ∙ 2

9 = 6 2
= − 4 5−6

56

Calculus: Differentiation

Example 3 Find the derivative for the followings: = 2 6 − 5 change
3 the form

= 2 6 − 5 −3

8 + 2 = 2 6 ∙ 6 − 5 −3 ∙ −3
= 3
15
= 3 2 3 − 6 8 2 separate = 12 6 + 3
= 2 6 − 6 3 = 3 + 3
expand 3 (4 9
= 5 + 2 −2 3
= + 5) Factorize

= 2 6 ∙ 6 − 6 3 ∙ 3 = 5 ∙ 5 + 2 −2 ∙ −2


= 12 6 − 18 3 5 5 4
2
= 6 3 (2 3 − 3) = −

= 1 (5 7 − 4) Factorize
2

57

Calculus: Differentiation

Example 4 Find the derivative for the followings:

Chain Rule Chain Rule

Chain Rule = 4 +1

= 4 = 2

= 2 = = 4 +1 =
1
= 4 = 2 2 = = 4 4 +1 = −

=

= 4 3 =
= × = ×

1 2 2 = − ∙ 4 4 +1
= ×
= ∙
= −4 4 +1 4 +1
= ∙ 4 3 2 2
= 2
= 4 3 4

=2

58

Calculus: Differentiation

Example 5 Find the derivative for the followings: Quotient Rule

Product Rule 6 +1
= 2 4

= 6 +1 = 2 4

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


= 2 3 = 3 + 2 4 = −


= 6 3 = 4 3 + 2 3 ∙ 3 2 2

2 4 ∙ 6 6 +1 − 6 +1 ∙ 8 3
= 12 2 3 + 2 3 =
2 4 2

= + 12 4 6 +1 − 8 3 6 +1
= 4 8
′ = 3 + 2 4 ∙ 6 3 + 2 3 ∙ 12 2 3 + 2 3
4 3 6 +1(3 − 2)
= 6 3 3 + 2 4 + 24 2 3 3 + 2 3 = 4 8 Factorize

= 6 3 3 + 2 3 3 + 2 + 4 2 Factorize 6 +1(3 − 2)
= 5

59

Calculus: Differentiation

Application Of Differentiation

60

Calculus: Differentiation

a) Maximum, Minimum And Infection Point

➢ We use differentiation to determine whether a function is
increasing or decreasing.

➢ A point where a function changes from an increasing to a
decreasing function or visa-versa is known as a turning
point.

➢ Turning point is also known as stationary point or
critical point.

➢ A function is increasing if the derivative is positive, and if
the derivative is negative then the function is decreasing.

61

Calculus: Differentiation

Maximum, Minimum And Infection Point

➢ Stationary points are points on a curve where the (maximum)
gradient is zero. (infection)
(minimum)
➢ This is where the curve reaches a minimum or maximum.

➢ A maximum is a high point and a minimum is a low point
on the curve.

➢ There are three types of stationary points: maximum,
minimum and inflection/ inflexion points.

62

Calculus: Differentiation

How to find stationary point and
determine the nature of the point?

STEPS

1. At all stationary point, the gradient is zero 1. Differentiate 1st time

where =

2. Let = 0, then find the value of

2. Second derivative will tell the nature of the
curve whether it is a maximum point, a
minimum point or a point of inflection. 3. Find value of (substitute into the function
given from question )

o 2 > 0 minimum point 4. State coordinate of the turning point.
2
2
o 2 maximum point 5. Differentiate 2nd time 2 and find the value
2
< 0 of 2 (when necessary )
2
o 2 inflection point
2 = 0 6. State the nature of the point (max or min or

inflection point )

63

Calculus: Differentiation

Example 1 Find the stationary points on the graph of
= 2 + 4 + 3 and state their nature.

= 2 + 4 + 3

Step 1
Step 2 = 2 + 4

Step 3 ∴ 2 + 4 = 0
Step 4 = 0 2 = −4
Step 5 = −2
Step 6
ℎ = −2 = −2 2 + 4 −2 + 3
= −1

−2, −1

2
2 = 2 > 0

−2, −1

64

Calculus: Differentiation

Example 2 Find the stationary points on the graph of 2 stationary
= 3 − 2 2 − 4 + 5 and state their nature. points

= 3 − 2 2 − 4 + 5 Step 4 2 175
Step 5 2, −3 & − 3 , 27
2 values of
Step 1 = 3 2 − 4 − 4 2 substitute
Step 2 2 = 6 − 4 values of
ℎ = 2
Step 3 ∴ 3 2 − 4 − 4 = 0 2
= 0 2 2 2 = 6 2 − 4 = 8 > 0
ℎ = − 3 2 2
= 2 and = − 3 2 = 6 − 3 − 4 = −8 < 0

ℎ = 2 = 2 3 − 2 2 2 − 4 2 + 5

∴ = −3 Step 6 2, −3 &
2 175
2 23 22 2
ℎ = − 3 = − 3 − 2 − 3 − 4 − 3 + 5 − 3 , 27

175
∴ = 27

65

Calculus: Differentiation

Example 3 Find the stationary points on the graph of = 2 2 − 8 + 8
and state their nature. Then sketch the graph.
How to plot the graph
= 2 2 − 8 + 8
66
Step 1 ∴ 4 − 8 = 0
Step 2 = 4 − 8 4 = 8
= 2
Step 3 = 0
Step 4
Step 5 ℎ = 2 = 2 2 2 − 8 2 + 8
Step 6 = 0

2, 0

2
2 = 4 > 0
2, 0

Calculus: Differentiation

b) Rates of Change

➢ The rate of change is the rate of increasing If given = ,
or decreasing of a quantity with the respect
of time, . Then the related chain rule will be
= ×
➢ The rate of change of a function ( ) with
respect to can be found by finding the
derived function ′( ).
where is the rate of change of
➢ For example in mechanics, the rate of change
of displacement (with respect to time) is the
velocity. The rate of change of velocity (with
respect to time) is the acceleration. with the respect of time, and
is the rate of change of with the



respect of time, .

Rate increase Rate decrease
= +ve values = -ve values

67

Calculus: Differentiation

How to solve rate of change ?

Related Chain Rule STEPS

a) Rates which contain = ( ) 1. Extract ALL the given information from question.
= × Including WHAT to find.

2. Solve related formula to find value of related
variable (if the value is given or when necessary).
b) Rates which contain radius :
volume, = × 3. Write the related equation or formula (usually
WHAT rate to find ).

4. Then differentiate the related formula.
area, = × 5. State related chain rule.
6. Substitute ALL the value.
7. Solve/simplify

c) Rates which contain length :

volume, = ×


area, = ×

68

Calculus: Differentiation

Example 1 If the expression of y given by = 3 + 2 − 5 Find the
rate of change of y if increase at 0.5 Τ when = 5.

Step 1 = 5
= ? = 0.5

Step 3 = 3 + 2 − 5

Step 4 = 3 2 + 2


Step 5 Rates which contain = ( )
= ×

= ×

Step 6 3 2 + 2 × 0.5
=

Step 7 = 3 5 2+2 5 × 0.5

= 42.5 ൗ

∴ Rate of y increase at 42.5 Τ

69

Calculus: Differentiation

Example 2 The volume of a cube is decreasing at the rate of 4 −1.
Find the rate of decrease of the length when the volume is 216 cm3.



Step 1 = −4 (decreased) = 216
Step 3 = ?

Step 4
Step 5
= 3 (formula of volume) Step 6 −4 = 3 2 ×
216 = 3

= 6 Step 7 −4
= 3 2

= 3 2 1.2
=3 6 2
= 0.037 2ൗ
Rates which contain length
= × ∴ Rate of the volume decrease at 0.037 3Τ
= ×

70

Calculus: Differentiation

Example 3 The area of a circle is increasing at a constant rate of 1.2 2Τ .
Calculate the rate at which the radius, of the circle is increasing when the
perimeter of the circle is 5 .

Step 1 = 5 Step 5 Rates which contain radius
= ? = 1.2 = ×
= ×

, = 2 (formula of parameter) Step 6
∴ 2 = 5 1.2 = 2 ×
Step 2

= 2 Step 7 1.2
= 2
Step 3

Step 4 1.2
= 2 =5
= 0.24 2ൗ

∴ Rate of the area increase at 0.24 2Τ

71

Calculus: Differentiation

Parametric Equation Differentiation

➢ A parametric equation is where
the and coordinates are both
written in terms of another letter.

➢ The third variable is called a
parameter and is usually given the
letter or .

➢ The differentiation of functions
given in parametric form is carried
out using the Chain Rule.

where 1
= × =


72

Calculus: Differentiation

How to solve parametric equation
differentiation ??

STEPS
1. Differentiate each of the parametric
equations for the parameter.
2. State the related Chain Rule.
3. Substitute the resulting expression for
the parameter into the Chain Rule.
4. Simplify the equation.

73

Calculus: Differentiation

Example 1 Example 2

Find when and 3. Given = 3 2 and = 6 2 then find .

= 2 2 + 3 = 5 −

= 6 2 = 3 2

= 5 − 3 = 2 2 + 3
= −12 2 = 6 2
Step 1 1 change the
Step 1 = 5 = 4 + 3 change the = 6 2 position
1 position Step 2
Step 2 = 4 + 3 Step 3
Step 3 Step 4
Step 4 = ×
= ×
1
1 = −12 2 × 6 2
= 5 × 4 + 3
−12 2 Simplify using reciprocal identities
5 = 6 2
= 4 + 3 = −2 2
=
74

Calculus: Differentiation Example 4
Example 3
Find when = 4 and = 2 .
Given and 3 , find
2 1−7

= 4 3 = 5 3 − 2 4
= 1 − 7 = 2
= 5 3 − 3 = 4 3 = 2(1 − 7 )−1

Step 1 = 15 2 − 3 4 Step 1 = −2(1 − 7 )−2. −7 4
= 3 ∙ 3 = 4 ∙ 2
Step 2
Step 3 4 14 = 2 4
Step 4 = = 1 − 7 2 1 change the
change = 2 4 position
= 4 the position

Step 2
= × = ×

15 2 − 3 × Step 3 14 1
= 4 Step 4 = 1 − 7 2 × 2 4

15 2 − 3 7
=4 = 4 1 − 7 2

75

Calculus: Differentiation

Implicit Differentiation

76

Calculus: Differentiation Basic implicit differentiation

How to solve implicit differentiation ?? Function Differentiation

STEPS 1
1. Differentiate and with respect to
(differentiate followed by ). 1
Differentiate by using product rule

Differentiate by using quotient rule
1. Put all expression on the left side of the equal sign, while



the other expression on the right side of the equal sign.
2. Factorize expression on the left side.



3. Find (move all the expression on the left to the right side ).



77

Calculus: Differentiation

Example 2 Differentiate using
Product Rule
Example 1

Find for 3 3 + 2 = 8 − 2 4 + 5 . Find for 2 4 + 8 = 2 2 3 − 5 3.



2 4 + 8 = 2 2 3 − 5

3 3 + 2 = 8 − 2 4 + 5

Step 1 8 3 + 0 = 4 3 + 6 2 2 − 15 2

Step 1 9 2 + 2 = 0 − 8 3 + 5


Step 2 15 2 − 6 2 2 = 4 3 − 8 3
Step 3
Step 2 2 + 8 3 = 5 − 9 2 Step 4

15 2 − 6 2 2 = 4 3 − 8 3


Step 3 2 + 8 3 = 5 − 9 2 4 3 − 8 3
Step 4 = 15 2 − 6 2 2

5 − 9 2 4 3 − 2 2
= 2 + 8 3 = 3 2(5 − 2 2)

Factorize

78

Calculus: Differentiation

Example 4

Example 3

Given 2 4 − 2 = 2 − 3 2 +2 , find Find for 5 − 2 3 = 2 + 3 .



2 4 − 2 = 2 − 4 3 +2 5 − 2 3 = 2 + 3

8 3 4 3 +2 expand Step 1 5 4 − 6 3 = 2 1 3 expand
3 + 2 + 2 + 3
Step 1 − 0 = 2 −

Step 2 8 3 = 2 − 12 3 +2 − 8 3 +2 5 4 − 6 3 = 2 2 3 + 2 3 3
Step 3 + +
Step 4
2 − 8 3 +2 = 8 3 + 12 3 +2 Step 2 2 3 3 + 6 3 = 5 4 − 2 2 3
Step 3 + +
Step 4
2 − 8 3 +2 = 8 3 + 12 3 +2 3 = 5 4 − 2
2 + 3 + 6 3 +
2 3

8 3 + 12 3 +2 Factorize = 5 4 − 2
= 2 − 8 3 +2 2 2 3 − 3 3 +2 2 + 3
= + 4 3 +2 3
4 2 3 − 3 3 +2 2 + 3 + 6 3
= 2( + 4 3 +2 )
79

Calculus: Differentiation

Example 5 Differentiate using
Quotient Rule

Find for 2 4 − 15 = 2 + 3.

3

2 4 − 15 = 2 + 3
3

Step 1 8 3 − 0 = 3 (2) − 2 (3 ) + 3 2 Step 3 27 4 − 6 = 72 3 2 − 6
Step 2 3 2 Step 4

8 3 = 6 − 6 + 3 2 72 3 2 − 6
9 2 = 27 4 − 6

6 − 6 + 27 4 6 12 3 − 1
= 3 9 4 − 2
= Factorize
9 2 2 12 3 − 1
= 9 4 − 2
72 3 2 = 6 − 6 + 27 4


27 4 − 6 = 72 3 2 − 6


80

Calculus: Differentiation

How do You Solve these Problem?

2 + 2 2 2

= 2 + 2 ∙ = 2 2 ∙ 2 2 + 2 2
2 + 2

= 2 2 + 2 + 2 2 + 2 = 2 2 2 + 2 + 2 2 2 2


2 + 2 2 2
1
1 2 2 + 2 2
= 2 + 2 ∙ 2 + 2 = 2 2 ∙

2 2 2 2 2 2 2 2
= 2 + 2 + 2 + 2 = 2 2 + 2 2 = +

2+ 2 2 2
= 2+ 2 ∙ 2 + 2
= 2 2 ∙ 2 2 + 2 2


= 2 2+ 2 +2 2+ 2 = 2 2 2 2 + 2 2 2 2 81



Calculus: Differentiation

Partial Differentiation

82

Calculus: Differentiation

a) First Order Of Partial Differentiation

➢ If ( , ) is a function of two variables, there are two first order partial differentiation
of that is partial derivative of with respect to and the partial derivative of with
respect to

➢ The first partial derivatives of the equation = ( , ) be a function with two variables

o with respect to : = OR


o with respect to : = OR


83

Calculus: Differentiation

How to solve 1st order partial
differentiation ??

= + = + = =
Differentiate ,and Differentiate ,and Differentiate ,and Differentiate ,and
held as CONSTANT held as CONSTANT
COPY back COPY back

When and When and
stands alone comes together

If given , = If given =
Then and
Then and


84

Calculus: Differentiation

How to differentiate ??

Example 1 Example 2

Given = 3 + 5 2 − 8. Find and Given = 3 + 5 2 − 8 3 2. Find and


When and
When and comes together
stands alone

respect to = 3 2 + 0 − 0 Differentiate ,and
respect to held as CONSTANT
respect to = 3 2 + 0 − 24 2 2 Differentiate ,and
= 3 2 Differentiate ,and COPY back
held as CONSTANT
= 3 2 − 24 2 2 Differentiate ,and
= 0 + 10 − 0 COPY back
respect to
= 10 = 0 + 10 − 16 3

= 10 − 16 3

85

Calculus: Differentiation

Example 3 Find and for the followings:

= 2 + 2 1 − 3
= 2 − 3 2 + 2 − 6
expand

= 4 + 5 − 3 2
= 2 − 6 + 0 − 0

= 2 − 6

4 3 = 4 + 2 4 2 − 5

= + 0 + 0 = 0 − 3 2 1 +0−6

= 4 3 = 4 + 8 3 2 − 0 = 3 2 − 6


= 4 + 8 3 2
= 0 + 0 − 6
= 0 + 4 4 − 5 4 = 3 + 2 − 5
= −6

= 4 4 − 5 4 = 3 + 1 2 − 0

= 3 + 2

= 0 − 2 2 − 5

= −2 2 − 5

86

Calculus: Differentiation

Example 4 Find and for the followings: = 3 + 2 5 + 4 − 3



= 5 3 + 2 4 3 2 + 0 +4−0


= 15 2 3 + 2 4 + 4

= 5 + 3 − 2 +3 = 2 + 2 + 4 − 3 = 5 3 + 2 4 0 + 2 +0−3

1
= 5 + 0 − 2 +3 = 2 + 2 ∙ 2 + 4 − 0 = 10 3 + 2 4 − 3

2 2
= 2 + 2 + 4
= 5 − 2 2 +3
1
= 5 (1) − 2 +3 3 = 2 + 2 ∙ 2 + 0 − 3

2
= 5 − 3 2 +3 = 2 + 2 − 3

87

Calculus: Differentiation

Example 5 Find and for the followings: ℎ

= 4 2 = 3 + 3

= −3 2 3 + 3
= 8

Product Rule

= 2 4 − 4 2 3 + 3 + 5

= 8 3 − 3 + 3 ∙ 8 + 4 2 ∙ −3 2 3 + 3 +0 4 2 3 + 3 must

use product rule for derivative with
respect to and

= 8 3 − 8 3 + 3 + 12 4 3 + 3

3 + 3 ∙ 4 2 + 4 2 ∙ −3 3 + 3 +5 ℎ
= 0 −
= 4 2 = 3 + 3
= −4 2 3 + 3 + 12 2 3 + 3 + 5
4 2 3
= = −3 + 3

88

Calculus: Differentiation

Example 6 Find and for the followings: ℎ

= 4 2 + 3 = 5 − 3

Quotient Rule
= 8 = 5

= 2 − 4 2 + 3 + 5
5 − 3

5 − 3 ∙ 8 − 4 2 + 3 5 + 5 4 4 2+3 must use quotient rule for
= 0 − 5 − 3 2 5 − 3

=− 40 2 − 8 3 − 20 2 − 15 + 5 4 derivative with respect to and
5 − 3 2

=− 20 2 − 8 3 − 15 + 5 4
5 − 3 2

5 − 3 ∙ 3 − 4 2 + 3 −3 2 ℎ

= 2 − 5 − 3 2 +0 = 4 2 + 3 = 5 − 3

= 2 − 15 − 3 3 + 12 2 2 + 9 3 = 3 = −3 2

5 − 3 2

= 2 − 15 + 12 2 2 + 6 3

5 − 3 2 89

Calculus: Differentiation

b) Second Order Of Partial Differentiation

➢ Given a function ( , ), the function is said to be differentiable if
and exist.

➢ If we can differentiate z= ( , ), therefore its 1st order derivatives
can be differentiated again and we can delineate the 2nd order partial
derivatives of z= ( , ) as follows:

90

Calculus: Differentiation

How to solve 2nd order partial
differentiation ??

Partial respect to respect to
Derivative
1st order If given , = If given =
2nd order First order
2 2 First order
2nd order 2 2 and and

respect to respect to Second order
and
2 2 and Second order
2 and 2

Note !!! 2

The answer for 2z should be the same as 2z 2 and 2
y x y x
2

91

Calculus: Differentiation

Example 1 How to differentiation ??

Given = 3 3 + 2 3 − 5 3 2. Find 2 , 2 and 2
2 2

Differentiate = 9 2 + 0 − 15 2 2 Differentiate = 0 + 6 2 − 10 3 First Order Partial
Differentiation
respect to respect to
= 9 2 − 15 2 2 = 6 2 − 10 3 Second Order Partial
Differentiate Differentiate Differentiation
2 = 18 − 30 2 2 = 12 − 10 3
respect to 2 respect to 2
(2nd time) (2nd time)
Differentiate 2 = 0 − 30 2 Differentiate 2 = 0 − 30 2



respect to respect to

= −30 2 = −30 2

The answer SHOULD be the same 92

Calculus: Differentiation

Example 2 Find 2 , 2 and 2 for the followings:
2 2

= 5 + 5 6 + 4 + 25 = 2 +3 + 3 − 4

= 2 +3 (2) + 3 1 4 (3) = 2 +3 3 1
− + 3 − 4 (−4)
= 5 + 5 4 6 + 0 + 0 6 5 5
= 0 + + 4 + 0 3 3 2 +3 4
2 2 +3 − −
= 5 + 5 4 6 = 6 5 5 + 4 = + 3 4 = − 3 4

2 = 0 + 20 3 6 2 30 4 5 2 = 2 2 +3 (2) − 3 2 = 3 2 +3 (3) + 4 2 (−4)
2 2 2 3 − 4 2 (3) 2 3 − 4
= + 0
16
= 20 3 6 = 30 5 4 = 4 2 +3 − 9 = 9 2 +3 − 3 − 4 2
3 − 4 2
2 2 2 4
= 0 + 30 5 4 = 30 4 5 + 0 = 3 2 +3 2 − 3 − 4 2 (3)

= 30 4 5 = 30 4 5 2 = 2 2 +3 (3) + 3 12
3 − 4 2 (−4) 3 − 4 2
= 6 2 +3 −

= 6 2 +3 − 12
3 − 4 2

93

Calculus: Differentiation
Example 3

, = 2 4 + 3 2 + 3

= 2 4 − 3 2 + 3 ∙ 6 6 3 2 + 3 must use = 2 4 ∙ 4 − 3 2 + 3 ∙ 3 2 3 2 3 2 + 3 must
= 2 4 − 6 3 2 + 3 product rule for 2nd order = 8 4 − 3 2 3 2 + 3 use product rule for 2nd order
derivative with respect to
derivative with respect to

= 0 − 3 2 + 3 ∙ 6 + 6 ∙ 3 2 + 3 ∙ 6 = 8 4 ∙ 4 − 3 2 + 3 ∙ 6 + 3 2 3 2 + 3 ∙ 3 2

= −6 3 2 + 3 − 36 2 3 2 + 3 = 32 4 − 6 3 2 + 3 − 9 4 3 2 + 3

= 2 4 ∙ 4 − 6 3 2 + 3 ∙ 3 2 = 8 4 − 3 2 3 2 + 3 ∙ 6
= 8 4 − 18 2 3 2 + 3 = 8 4 − 18 2 3 2 + 3

94

Calculus: Differentiation Total Differentiation

➢ Total derivatives are often used in 95
related rates problems; for example,
finding the rate of change of volume
when two parameters are changing
with time.

➢ A function = ( , , ) and with
, , being functions of , Then the
total derivatives of with respect to
is given by.


= ∙ + ∙ + ∙