VSLT (V&T) DOCUMENT

TOPIC 1 :
PARTIAL
DIFFERENTIATION

❑1.1 First Order Partial Derivatives 1
❑1.2 Second Order Partial Derivatives
❑1.3 Extremum of Functions of Two

Variables
❑1.4 Lagrange Multiplier

1.1 – First Order Partial Derivatives

COURSE FRAMEWORK AND STUDENT LEARNING TIME (SLT)

a) Compute the first order partial derivatives
involving two variables x and y

PHD PExp

dy =V du + U dv 1) Salin semula soalan
dx dx dx 2) Darab DIFF kuasa

PHB du dv PKuasa
dx dx
dy V −U 1) Kuasa datang ke depan
2) Salin semula dalam kurungan
= 3) Kuasa berkurang satu
dx V 2 4) Darab DIFF dalam kurungan

PLn

dy = f (x) 2
dx f (x)

1. Functions of several variables consist of few
variables in a single function.

For example : a) f (x, y) = x2 + y 2 − x − y

( )b) Z (x, y) = ln x − y 2

2. The first order partial derivative is obtained
by differentiating the function with respect
to a particular variable, by keeping the
other variables fixed.

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The First Order Partial Derivative

Notations

Description ‘ del ’ operator f operator

Partial derivative of f z fx
x x
z = f ( x, y)

with respect to x :

Partial derivative of f z fy
y y
z = f ( x, y)

with respect to y :

We know that f (x, y) = Z then 4
fx (x, y) = Z x and f y (x, y) = Z y

EXAMPLE 1: f (x, y) = 2x2 + 3x y − 6 y2

Find fx and fy for Partial derivative of f
with respect to y
Partial derivative of f
with respect to x

 fx ( x, y) = 4x + 3y  fy ( x, y) = 3x −12 y 5

EXAMPLE 2:

Find f and f for each of the following :
x y

(a) f ( x, y) = 2x2 + 3y − 6y2 (b) f ( x, y) = x2 y3 + x4 y

Ans : f = 4x , f = 3 −12 y Ans : f = 2xy3 + 4x3 y , f = 3x2 y2 + x4 6
x y x y

EXAMPLE 3:

Find partial derivatives of f ( x, y) with respect to x

and y

(a) f ( x, y) = ( x − 2y)3 [Recall : The Power Rule]

(b) f ( x, y) = 4x2 + y2 [Recall : Square Root Functions]
(c) f ( )x, y = ex2y+2 [Recall : The Exponential Functions]
(d ) f ( x, y) = ln ( x + 3y)2
[Recall : Natural Logarithmic Functions]

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Find partial derivatives of f ( x, y) with respect to x

and y

(a) f ( x, y) = ( x − 2y)3 [Recall : The Power Rule]

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Find partial derivatives of f ( x, y) with respect to x
and y

(b) f ( x, y) = 4x2 + y2 [Recall : Square Root Functions]

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Find partial derivatives of f ( x, y) with respect to x
and y
(c) f ( )x, y = ex2y+2 [Recall : The Exponential Functions]

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Find partial derivatives of f ( x, y) with respect to x

and y

( ) ( ) ( )d f x, y = ln x + 3y 2 [Recall : Natural Logarithmic Functions]

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EXAMPLE 4: [Using Product Rule : KAT+VU]

Find fx for each of the following .

(a) f ( )x, y = x e2 2x2 +4 y (b) f ( x, y) = ex−2y ln (3x)

Ans :

( )( )a fx = e2x2 +4 y (b) ex−2 y 1 ln (3x)
4x3 + 2x , fx =  x + 12

EXAMPLE 5: [Using Quotient Rule : KAT+VU]

Find f y for each of the following . (b) f ( x, y) = ln y

(a) f ( x, y) = 3x − 5y 3y − 2x

2xy −1

Ans :

(a) fy = 5 − 6x2 , (b) fy = 3y − 2x − 3y ln y 13

(2xy −1)2 y (3y − 2x)2

EXAMPLE 6:

If z = ln x2 + y2, show that x z + y z = 1
x y

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EXAMPLE 7: x − 2y  z z 
x2 − 2y2  x y 
z = . Show that 1  −  = 1
+2 x2 − 2y2 2
( )Given that x2 y2

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1.2 – Second Order Partial Derivatives

COURSE FRAMEWORK AND STUDENT LEARNING TIME (SLT)

a) Compute the second order partial derivatives
involving two variables x and y

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❑ The second order partial derivative is obtained by differentiating the
first order partial derivative with respect to a particular variable, by
keeping the other fixed.

Description Notations
‘ del ’ operator
f operator

Partial derivative of fx  ( fx ) = 2 f 2z f xx
with respect to x : x x2 x2 f yy
Partial derivative of f y 2z f xy
with respect to y : ( ) fy = 2 f y 2
Partial derivative of fx y 2 2z f yx
with respect to y : x yx
Partial derivative of f y 2z 19
with respect to x :  ( fx ) = 2 f xy
y yx

( ) fy = 2 f
xy
x

EXAMPLE 8:

Given that f ( x, y) = 3x2 − 2xy + 5y2 .

Find fxx, f yy, fxy and f yx,

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EXAMPLE 9:

Find 2z , 2z ,   z  for the following functions .
x2 y 2 y  x 

(a) z = x 2 y 3 + x 4 y

z xx = 2 y 3 + 12x 2 y, z xy = 6xy 2 + 4x3 , z yy = 6x 2 y 22

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EXAMPLE 9 :

Find 2z , 2z ,   z  for the following functions .
x2 y 2 y  x 

(b) z = e3xy + 4x 2 y

z xx = 9 y 2e3xy + 8 y, z xy = 9xye3xy + 3e3xy + 8x , z yy = 9x 2e3xy 24

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1.3 – Extremum Of Functions Of Two Variable

COURSE FRAMEWORK AND STUDENT LEARNING TIME (SLT)

a) Determine the stationary point (a, b) by solving simultaneous

equations up to third degree polynomials.

b) Determine the nature of stationary point (a, b) by using the second
order partial derivatives test.
c) Solve problems related to economics and business.

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A necessary condition for the existence of a

stationary point (critical point) of a function

f (x, y): fx = 0 fy = 0

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Discriminant
(D-Test Formulae)

The type (nature) of a stationary point
can be determined by using the second
order partial derivative test ( D test).

D(x, y) = f xx (x, y)  f yy (x, y) −[ f xy (x, y)]2

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D ( x, y) = fxx ( x, y) • f yy ( x, y ) −  fxy ( x, y )2

D(x, y)  0 fxx (x, y)  0 maximum
fxx (x, y)  0 exist
Min/max
exist minimum
exist

D(x, y)  0 Saddle point exist

D(x, y) = 0 Test fails

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EXAMPLE 10: Find the stationary point (critical point )
of the function f (x, y) = 2xy + y2 + 2x2 +10

and determine its nature.

Find stationary point :

1: Find fx , fy :

fx (x, y) = 2 y + 4x f y (x, y) = 2x + 2 y

2: Let fx = 0, f y = 0. Solve simultaneously

x=0 y=0

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3: Conclusion :State stationary point.

(0,0) is a stationary point

Determine its nature:

1: Find f xx , f xy , f yy

f xx = 4 f xy = 2 f yy = 2

2: Using D test formulae

D(x, y) = f xx (x, y)  f yy (x, y) − [ f xy (x, y)]2

At point (0,0)

D(0,0) = 4  2 − [2]2 = 4  0 (min/max)

f xx = 4  0 (minimum exist) (0,0) is a minimum point
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EXAMPLE 11:

Find the maximum, minimum or / and saddle point(s)

of the functions. f (x, y) = x3 + y 3 − 3xy

(0, 0) is a saddle point, (1,1) is minimum point 32

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EXAMPLE 12:

A company produces two products, X and Y. How many
units should be produced for each product in order to
maximise the profit, if the profit functions is given

as P(x,y) = -3x2 + 4xy +100x-5y2-30 y +1500. How

much is the maximised profit?

20 units product X and 5 units product Y.The maximise profit is RM2425.00 34

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EXAMPLE 13:

The price function for producing x units of product X is

is p(x) = 200 − 1 x. The revenue function for producing y

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units of product Y is R(y) = 160 y − 1 y 2. The combine cost

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function of the two products is C(x,y) = 100x + 70 y + 1 xy.

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Find

(a) the maximum profit to produce the two products
and the quantity of each product,

(b) the price of product Y at maximum profit.

The maximise profit is RM14500[200 units product x and 100 units product y] 37
The price of product y is RM135.00

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1.4 – Lagrange Multiplier

COURSE FRAMEWORK AND STUDENT LEARNING TIME (SLT)

a) Formulate the problem related to maximize or minimize Z = f ( x, y)
subject to the constraint g ( x, y) = 0
*Restrict g ( x, y) to linear only

b) Determine the critical points ( x, y) by solving the simultaneous

equations.
*Do not use the second order partial derivative test (discriminant) to prove
maximum or minimum point.

c) Solve problems related to economics and business.

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Lagrange multiplier

The below constrained optimization problem
can be solved by using the Lagrange Multiplier
Method

Maximise/minimise Z = f (x, y)

Subject to g(x, y) = 0

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STEP 1: Determine the new function.

F(x, y, ) = f (x, y) + [g(x, y) = 0]

Lagrange
Multiplier

STEP 2: Determine the critical point (x, y, )

by solving the simultaneous equations.

Fx (x, y, ) = 0 Fy (x, y, ) = 0 F (x, y, ) = 0

STEP 3: Determine the minimum or / and

maximum value(s) by substituting the

value(s) into f (x, y) 42

EXAMPLE 14: Max value f (4,2) = 25

Maximise f (x, y) = −x2 − y 2 + 45 subject to

g(x, y) = 2x + y −10.

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EXAMPLE 15: Min value f 10 ,− 5  = 59
 3 3 3

Minimise f (x, y) = x2 + 2 y 2 + 3 subject to x − y = 5.

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