MATHEMATICS 1 CONTENTS Derivatives for functions of two variables - - PowerPoint PPT Presentation

β–Ά
mathematics
SMART_READER_LITE
LIVE PREVIEW

MATHEMATICS 1 CONTENTS Derivatives for functions of two variables - - PowerPoint PPT Presentation

Mar 19, 2024 266 likes β€’472 views

Partial derivatives BUSINESS MATHEMATICS 1 CONTENTS Derivatives for functions of two variables Higher-order partial derivatives Derivatives for functions of many variables Old exam question Further study 2 DERIVATIVES FOR FUNCTIONS OF TWO

slide-1
SLIDE 1

1

BUSINESS MATHEMATICS

Partial derivatives

slide-2
SLIDE 2

2

CONTENTS Derivatives for functions of two variables Higher-order partial derivatives Derivatives for functions of many variables Old exam question Further study

slide-3
SLIDE 3

3

DERIVATIVES FOR FUNCTIONS OF TWO VARIABLES Can we find the extreme values of a function 𝑕 𝑦, 𝑧 of two variables 𝑦 and 𝑧? β–ͺ e.g., 𝑕 𝑦, 𝑧 = 𝑦3𝑧 + 𝑦2𝑧2 + 𝑦 + 𝑧2 Try 𝑕′ 𝑦, 𝑧 = β‹― β–ͺ this fails! Derivative of a function 𝑔 𝑦 of one variable: 𝑔′ 𝑦 = 𝑒𝑔 𝑦 𝑒𝑦 = lim

β„Žβ†’0

𝑔 𝑦 + β„Ž βˆ’ 𝑔 𝑦 β„Ž Generalization into partial derivative of 𝑕 𝑦, 𝑧 of 2 variables: πœ–π‘• 𝑦, 𝑧 πœ–π‘¦ = lim

β„Žβ†’0

𝑕 𝑦 + β„Ž, 𝑧 βˆ’ 𝑕 𝑦, 𝑧 β„Ž

slide-4
SLIDE 4

4

DERIVATIVES FOR FUNCTIONS OF TWO VARIABLES So differentiate β–ͺ 𝑕 𝑦, 𝑧 = 𝑦3𝑧 + 𝑦2𝑧2 + 𝑦 + 𝑧2 with respect to 𝑦, keeping 𝑧 fixed β–ͺ

πœ–π‘• πœ–π‘¦ = 3𝑦2𝑧 + 2𝑦𝑧2 + 1

and with respect to 𝑧, keeping 𝑦 fixed β–ͺ

πœ–π‘• πœ–π‘§ = 𝑦3 + 2𝑦2𝑧 + 2𝑧

Clearly, in this case

πœ–π‘• πœ–π‘¦ β‰  πœ–π‘• πœ–π‘§

β–ͺ therefore, never write 𝑔′ for a function of two variables!

slide-5
SLIDE 5

5

DERIVATIVES FOR FUNCTIONS OF TWO VARIABLES So we define the partial derivative of 𝑔 with respect to 𝑦 πœ–π‘” 𝑦, y πœ–π‘¦ = lim

β„Žβ†’0

𝑔 𝑦 + β„Ž, 𝑧 βˆ’ 𝑔 𝑦, 𝑧 β„Ž And similar with respect to 𝑧 πœ–π‘” 𝑦, y πœ–π‘§ = lim

β„Žβ†’0

𝑔 𝑦, 𝑧 + β„Ž βˆ’ 𝑔 𝑦, 𝑧 β„Ž

slide-6
SLIDE 6

6

DERIVATIVES FOR FUNCTIONS OF TWO VARIABLES

πœ–π‘” 𝑦, 𝑧 πœ–π‘¦ πœ–π‘” 𝑦, 𝑧 πœ–π‘§

slide-7
SLIDE 7

7

DERIVATIVES FOR FUNCTIONS OF TWO VARIABLES Alternative notations β–ͺ

πœ–π‘” πœ–π‘¦

β–ͺ

πœ–π‘” 𝑦,𝑧 πœ–π‘¦

β–ͺ 𝑔′

𝑦

β–ͺ 𝑔′

1

β–ͺ 𝑔

𝑦

β–ͺ 𝑔

1

β–ͺ πœ–π‘¦π‘” β–ͺ etc.

Not important to remember, but important to recognize so, basically a lot of choice, but never write 𝑒𝑔

𝑒𝑦 or 𝑔′

slide-8
SLIDE 8

8

DERIVATIVES FOR FUNCTIONS OF TWO VARIABLES The partial derivative in a point is a number β–ͺ e.g.,

πœ–π‘” 𝑦,𝑧 πœ–π‘¦ 𝑦,𝑧 = 2,βˆ’5 = βˆ’3

The partial derivative over a range of points is a function of 𝑦 and 𝑧 β–ͺ e.g.,

πœ–π‘” 𝑦,𝑧 πœ–π‘¦

= 2𝑦 + 3𝑧 βˆ’ 6

slide-9
SLIDE 9

9

DERIVATIVES FOR FUNCTIONS OF TWO VARIABLES Example: Cobb-Douglas production function β–ͺ decribing how a firm’s output π‘Ÿ depends on capital input (𝐿) and labour input (𝑀) π‘Ÿ 𝐿, 𝑀 = 𝐡 Γ— 𝐿𝛽 Γ— 𝑀𝛾 β–ͺ where 𝐡, 𝛽, and 𝛾 are positive constants Marginal productivity of capital:

πœ–π‘Ÿ πœ–πΏ

πœ–π‘Ÿ πœ–πΏ = 𝐡 Γ— 𝛽 Γ— πΏπ›½βˆ’1 Γ— 𝑀𝛾 β–ͺ when 0 < 𝛽 < 1,

πœ–π‘Ÿ πœ–πΏ is a decreasing function of 𝐿

β–ͺ diminishing marginal returns

slide-10
SLIDE 10

10

EXERCISE 1 Given is 𝑔 𝑦, 𝑧 = 𝑦2𝑓2𝑧. Find

πœ–π‘” πœ–π‘¦ and πœ–π‘” πœ–π‘§.

slide-11
SLIDE 11

12

EXERCISE 2 Given is 𝑔 𝑦, 𝑧 = 𝑦𝑧. Find

πœ–π‘” πœ–π‘¦ and πœ–π‘” πœ–π‘§.

slide-12
SLIDE 12

14

HIGHER-ORDER PARTIAL DERIVATIVES

Recall the second derivative β–ͺ

𝑒 𝑒𝑦 𝑒𝑔 𝑦 𝑒𝑦

=

𝑒2𝑔 𝑦 𝑒𝑦2

= 𝑔′′ 𝑦 Four possibilities for function 𝑕 𝑦, 𝑧 : β–ͺ

πœ– πœ–π‘¦ πœ–π‘• πœ–π‘¦

=

πœ–2𝑕 πœ–π‘¦2

β–ͺ

πœ– πœ–π‘§ πœ–π‘• πœ–π‘§ = πœ–2𝑕 πœ–π‘§2

β–ͺ

πœ– πœ–π‘§ πœ–π‘• πœ–π‘¦

=

πœ–2𝑕 πœ–π‘§πœ–π‘¦

β–ͺ

πœ– πœ–π‘¦ πœ–π‘• πœ–π‘§ = πœ–2𝑕 πœ–π‘¦πœ–π‘§

Alternative notations: β–ͺ

πœ–2𝑕 πœ–π‘¦πœ–π‘§, πœ–π‘• 𝑦,𝑧 πœ–π‘¦πœ–π‘§ , 𝑕′′ 𝑧𝑦, 𝑕′′ 21, 𝑕𝑧𝑦, 𝑕21, πœ–π‘¦π‘§π‘•, etc. so, never 𝑒2𝑕

𝑒𝑦2 or 𝑕′′

slide-13
SLIDE 13

15

HIGHER-ORDER PARTIAL DERIVATIVES Example: 𝑕 𝑦, 𝑧 = 𝑦3𝑧 + 𝑦2𝑧2 + 𝑦 + 𝑧2

β–ͺ

πœ–2𝑕 πœ–π‘¦2 = 6𝑦𝑧 + 2𝑧2

β–ͺ

πœ–2𝑕 πœ–π‘§2 = 2𝑦2 + 2

β–ͺ

πœ–2𝑕 πœ–π‘¦πœ–π‘§ = 3𝑦2 + 4𝑦𝑧

β–ͺ

πœ–2𝑕 πœ–π‘§πœ–π‘¦ = 3𝑦2 + 4𝑦𝑧

For almost all functions

πœ–2𝑕 πœ–π‘¦πœ–π‘§ = πœ–2𝑕 πœ–π‘§πœ–π‘¦

β–ͺ and certainly for all functions we encounter in business and economics

slide-14
SLIDE 14

16

EXERCISE 3 Given is 𝑔 𝑦, 𝑧 = 4𝑦3𝑧2 βˆ’ 3𝑧4𝑓2𝑦. Find

πœ–2𝑔 πœ–π‘¦πœ–π‘§ in

𝑦, 𝑧 = βˆ’1,0 .

slide-15
SLIDE 15

18

HIGHER-ORDER PARTIAL DERIVATIVES

Likewise, we can define third-order β–ͺ

πœ– πœ–π‘¦ πœ– πœ–π‘¦ πœ–π‘” 𝑦,𝑧 πœ–π‘¦

=

πœ–3𝑔 πœ–π‘¦3

β–ͺ

πœ– πœ–π‘§ πœ– πœ–π‘§ πœ–π‘” 𝑦,𝑧 πœ–π‘¦

=

πœ–3𝑔 πœ–π‘§2πœ–π‘¦

β–ͺ how many are there? β–ͺ how many are different? And even higher-order partial derivatives β–ͺ

πœ–π‘œπ‘” πœ–π‘¦π‘œ

β–ͺ

πœ–π‘œπ‘” πœ–π‘¦π‘œβˆ’1πœ–π‘§

β–ͺ etc.

slide-16
SLIDE 16

19

DERIVATIVES FOR FUNCTIONS OF MANY VARIABLES Let 𝑔 𝑦1, 𝑦2, 𝑦3, … , π‘¦π‘œ We can form π‘œ first-order partial derivatives β–ͺ

πœ–π‘” πœ–π‘¦1 , πœ–π‘” πœ–π‘¦2 , … , πœ–π‘” πœ–π‘¦π‘œ

and many many second-order partial derivatives

slide-17
SLIDE 17

20

EXERCISE 4 Given is 𝑕 𝐲 =

1 π‘œ σ𝑗=1 π‘œ

𝑦𝑗. Find

πœ–π‘• πœ–π‘¦4.

slide-18
SLIDE 18

22

OLD EXAM QUESTION 27 March 2015, Q1b

slide-19
SLIDE 19

23

OLD EXAM QUESTION 22 October 2014, Q1h

slide-20
SLIDE 20

24

FURTHER STUDY Sydsæter et al. 5/E 11.1-11.2 Tutorial exercises week 2 partial derivatives higher-order partial derivatives partial derivatives graphically