CHAPTER 2 DIFFERENTIATION

2.7.2 Execute partial differential

DEFINITION Definition:

Perform partial differential to find;

a) First order partial differentiation
b) Second order partial differentiation

A) FIRST ORDER PARTIAL DIFFERENTIATION

Example: 1

If = ( , ) = + + + , then the partial derivatives are

= 8 3 3 + 10 + 7 (Note: fixed, independent variable, dependent variable)



= 6 4 2 + 5 2 + 4 3 (Note: fixed, independent variable, dependent variable)



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Example: 2

If = ( , ) = ( + ) + , then the partial derivatives are

= 12 2( 3 + 2)3 + 1 (Note: we used chain rule/ extended power rule on the first term)



= 8 ( 3 + 2)3 (Note: chain rule/ extended power rule again and second term has no )



Example: 3

If = ( , ) = , then the partial derivatives are

= + (Note: we used product rule then chain rule in second term)
(Note: no product rule but used chain rule)


= 2



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EXERCISE 16 b) = 3 + 2

Find the partial derivatives.
a) = 2 + 4 3

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


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c) = ( + )( − ) d) = 2

[Ans: = 2 , = −2 ] [Ans: = 2 , = 3 ]



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B) SECOND ORDER PARTIAL DIFFERENTIATION

Consider the case of a function of two variables, = ( ; ) ,it means that there will be a total of four
possible second order derivative as denoted for examples

2 = ( )
2


FORMULA 2 = ( )



2 = ( )



2 = ( )
2


Example: 1

Find all the second order derivatives for = ( , ) = ( ) − +

We’ll first need the first order derivatives

= −3 (3 ) − 2 3 = −3 2 3 + 10



Now, let’s get the second order derivatives

 2 = ( ) = −9 cos(3 ) − 2 3
2


 2 = ( ) = −6 3



 2 = ( ) = −9 2 3 + 10
2


 2 = ( ) = −6 3



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EXERCISE 17 b) = 4 2 − 5 4 + 3 5 − 6

Find all the second order derivatives.
a) = 3 + 2

2 2 Ans: 2 = 8, 2 = −20 3,
Ans: 2 = 6 , = 2, 2

2 2 2 = −60 2 + 60 3, 2 = −20 3 ]
[ 2 = 0, = 2 ] [ 2

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c) = 2 2 + 2 − 3 2 d) = 3 3 2 + 2

2 2 Ans: 2 = 18 2, 2 = 18 2 + 2 2 ,
Ans: 2 = 4, = 2 , 2
2 2
[ 2 = 2 − 6, = 2 ] 2 = 6 3 − 4 2 , 2 = 18 2 + 2 2 ]]
[[ 2

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e) = 2 3 + 2 f) = ln(3 + 5 )

Ans: 2 = 2 3 − 2 , 2 = 6 2 − 2 , 2 −9 2 −15
2 Ans: 2 = (3 + 5 )2 , = (3 + 5 )2 ,
2 −15
[ 2 = 6 2 + 2 , 2 = 6 2 − 2 ] 2 −25 = (3 + 5 )2 ]
2 [ 2 = (3 + 5 )2 ,

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2.8 Apply The Technique of Total
Differentiation

2.8.1 Execute total differentiation for two variables function

Definition:

Define the differential for functions of two variables

DEFFINITION ( , ), +

Let = hence the total differential for is =

which is read as



“the change in z (dz) is due partially to a change in x (dx) plus
partially due to a change in y (dy)”

For a function of two or more independent variables, the total differential of the function is
the sum over all of the independent variables of the partial derivative of the function.
The precise formula for any case depends on how many and what the variables are.
Thus we get the following examples of formulas:

Given a function ( , ), its total differential in notation is = +



Given a function ( , ), its total differential in notation is = +



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Example: 1

If = + + , determine its total differentiation

= +



= +



∴ Total differential for is


= +
= ( + ) + ( + )

Example: 2

If = − ( ) + + − , determine its total differentiation

= − + +



= − + −



∴ Total differential for is

= +



= ( − + + ) + ( − + − )


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Example: 3

If = ( , ) = − , get the change of if ( , )change from (2,3) to (2.05, 2.98)

= . − = . and = . − = − .
= −



= −



∴ Total differential for is
= +



= ( − ) + (− )
= [ ( ) − ( )]( . ) + [(− ) (− . )]
= .
Therefore increase about 0.49 unit

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EXERCISE 18

Determine the total differentiation for each function below

a) = 6 + 3 − 7 3 b) = 2 2 − 4 + 4

[Ans: = 3 − 21 2 ] [Ans: = (4 − 4 ) + (−4 + 4 3)]

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c) Given = 5 + 2 2. Calculate the d) Given = 18 − 5 3 2 − 2 3. Find the
total differential of z, when ( , ) total differential of z, if ( , ) changes
changes from (1, 2) to (1.04, 1.93) from (0.2, 0.5) to (0.25, 0.6)

[Ans: = 0.09] [Ans: = −0.1656]

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EXERCISE 19

Solve the following by Total Differentiation

a) A height of a cone is 10 and its increase b) The radius and height of a cylinder are 2
at the rate of 0.4 −1 . Its base radius is and 7 with the error of its measurement
is ± 0.03 . Determine the minimum error
8.5 and decrease at the rate for its volume.
0.2 −1. Determine the rate of change

for the cone volume.

[Ans: = −1.7 3 −1] [Ans: = 3.02 3]


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c) The base radius, of a cone is increasing at the d) The radius of a circle is decreasing at a rate
rate 0.3 −1 while the perpendicular of 7 −1. Find the rate of change of the

height, ℎ is decreasing at the rate of area for circle at the instant when the radius
0.1 −1. Determine the volume changing is 4 .

rate, , when = 3 and ℎ = 4 .

[Ans: = −6.597 3 −1] [Ans: = −175 2 −1 ]



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e) The power , consumed in a resistor is given f) If = ( , ) and = 2 2 sin 2 , find the

by = 2 watts. Determine the approximate rate of change of , when = 3 and

is and when is increasing at 3 units/s and
4
change in power, when increases by 5%
is decreasing at 0.5 units/s.
and decrease at 0. 5% if the original values

of and 50 volts and 12.5 ohms

respectively.

[Ans: = 36 ]


[Ans: = 21 ]

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