PARTIAL DIFFERENTIATION

The technique of differentiating multivariable function; such that we differentiate one of the variable while keeping the other variable fixed is known as Partial Differentiation and the resulting derivative is said to be Partial Derivative of the Function

 

FIRST ORDER PARTIAL DERIVATIVE

Given that , the partial derivative of f with respect x is denoted by

and is the function obtained by differentiating f with respect to x treating y as a constant .

The partial derivative of f with respect of y is denoted by

and is the function obtained by differentiating f with respect to y treating x as a constant  

 

NOTE: Given that

 

Example1

Find the partial derivative  if

SOLUTION

To compute f_x, think of f as a function of x and differentiate the sum, term by term, treating y as a constant

To compute f_ey, think of f as a function of y and differentiating term by term, treating x as a constant to get

EXAMPLE 2

Find the partial derivative of

SOLUTION

To compute f_x , think of f as a function of x and differentiate the sum, term by term, treating y as a constant

To compute f_y, think of f as a function of y and differentiating term by term, treating x as a constant to get

 

 

 

RULES FOR PARTIAL DIFFERENTIATION                         

Rule1 : The Product rule -

Rule2 : The Quotient rule

 

Rule3: The Generalized power rule

 

EXAMPLE 3

Compute the first order partial derivative for the following functions

SOLUTION

 

EXAMPLE 4: Mixing the rules together

Find the partial derivative of

SOLUTION

 EXAMPLE 5

Find the partial derivative

Should you have problem solving the questions above, you should master the techniques of differentiation. You can revisit the topic again to remind yourself the techniques. Now, let us solve the examples

SOLUTION

 

           

 

 

 

 

APPLICATION OF PARTIAL DIFFERENTIATION TO ECONOMICS

Quantity demanded and quantities supplied are affected by several factors such as consumer income, price, tax, e.t.c meaning multivariable factors affect the demand and supply. We shall focus on marginal analysis involving multivariable functions.

In Economics, the term Marginal analysis means a the method of using derivative to find or evaluate the change in value of a function resulting from a 1 – unit increase in one its variables

 

EXAMPLE 6

Conoil Nigeria Limited was able to estimate that the weekly output is given by the function  where x is the graduate workers and y is the number of non – graduate employed at their drilling plant. Currently the workforce consists of 80 graduate worker s and 130 non – graduate workers. Use marginal analysis to estimate the change in the weekly output that will result from the additional  of 1 more graduate worker if the number of non – graduate is not changed.

 

SOLUTION

COBB – DOUGLAS PRODUCTION FUNCTION

 Where . Output Q at a firm is often regarded as a function of the amount of K of capital investment and size L of the labour force

In this context, the partial derivative  is called marginal productivity of capital and measures the rate at which output Q changes with respect to capital expenditure when the size of labour is held constant. Similarly   is called marginal productivity of labour and measures the rate of change of output with respect to the labour level when capital expenditure is held constant.

 

EXAMPLE 7

A manufacturer estimates that the annual output at a certain factory is given by  where K is the capital expenditure in unit of N1000 and L is the size of the labour force in worker  – hours

(a)        Find the marginal productivity of capital  and marginal productivity of labour  when the capital expenditure is N 630, 000 and the labour level is 830 workers – hours

(b)        Should the manufacturer consider adding a unit of capital or a unit of labour to increase output more rapidly

 

 

 

 

SOLUTION

(b)    From our calculation, a unit increase in capital investment result in an increase of 10.91 units which is less than 19.33 unit increase in output that comes as a result of a increase of  unit increase of labour level. Base on this, the firm should increase the labour level by 1 worker – hour (i.e. from 830 to 831) worker – hours to increase output as quickly as possible from the current level

MARGINAL DEMAND

Substitute and Complementary goods

Two goods are said to be substitute goods if an increase in the demand for either result in a decrease for the other.

While two goods are said to be complementary goods if a decrease of either result in a decrease in the demand. Example of Complementary goods are cars and fuels, or mobile phones and phone lines. If there is more demand for mobile phone, it will lead to more demand for phone line too.

We can use partial differentiation to obtain condition for determining whether two goods are substitute or complementary

Given that  units of the first goods and  of the second demand when the unit price of goods of are

 If  given ,

 hence two goods are said to be substitute or competitive

            If  given

Hence, the two goods are said to be complementary

 have opposite sign s for any given , hence the two goods are neither complementary nor substitute .

 

EXAMPLE 8

Given that the demand function or two related products are

Use partial differentiation to determine whether the products are substitute, complementary or neither.

 

 

 

 

 

SOLUTION

EXAMPLE 9

The demand function of a pair of goods is given. Use partial differentiation to determine whether the commodities are substitute, complementary or neither

           

 SOLUTION

EXAMPLE 10

The demand function of two related commodities are given by

What can you say about the two commodities

 

 

 

 

 

 

SOLUTION

SECOND ORDER PARTIAL DERIVATIVE

Differentiating the first – order partial derivative with respect to respective variable, the resulting functions are called second – order partial derivative

Below show the summary of the four possible second – order partial derivative of a function of the two variables.

 

EXAMPLE 11

Find the second – order partial derivative

SOLUTION

 

 

 

 

 

 

EXAMPLE 12

Find the 4 second – order partial derivative function

SOLUTION

 

SOME IMPORTANT THEOREMS

1.         Young’s theorem: It states that given a function  that is cross partial derivatives are obtainable and is continuous they will also be equal that is

EXAMPLE 13

Verify whether young’s theorem hold for the following function

SOLUTION

EXAMPLE 14

2.EULER’S THEOREM: States that if  is a homogeneous function of degree n   then

What Euler’s theorem implies is this given a differentiable homogeneous function of specified degrees, Euler’s Theorem required the weighted sum of the first partial of the function to be equal to the original function times its degree of homogeneity.

Where the weight are the independent variable of the function.

For a homogeneous differentiable function of variable  we have

EXAMPLE 14

Use Euler’s theorem to determine the degree of homogeneity of the following functions

 

SOLUTION

 

3.         JACOBIAN THEOREM: One of the benefit of partial derivative is that it provide a means of test whether there exist d\functional (linear or non linear) dependence among a set of n – functions in a n – variables. This is whole idea behind Jacobian’s theorem

Jacobian Determinant is composed of all first order partial derivative of a system

arranged in order sequence. It is denoted by

 

JACOBIAN MATRIX AND DETERMINANT

If all the first order partial determinant of n – variable system arranged in an orderly sequence are placed in a square matrix, the resulting matrix is know as Jacobian Matrix and denoted by J and when the determinant is evaluated the result is know as Jacobian Determinant or Jacobi for short denoted by

Given

 

JACOBIAN THEOREM

Given the function

Jacobian Theorem can be stated that, is there an existence of functionally dependence among a set of  n – functions  if the Jacobian determinant  is zero

 

Example 15

Use the Jacobian’s theorem to verify the functional dependence of mathematical relations

                       

SOLUTION

Take the first order Partial Derivative

EXAMPLE 16

Use the Jacobian Theorem to test the existence the existence of functional dependence between the following pairs of function

 

SOLUTION

Obtain the first order partial derivatives

 

4.         LAPLACE’S EQUATION: The function  is said to satisfy Laplace’s Equation if . Any function which satisfy this equation that equation is in general called harmonic function

 

EXAMPLE 17

Determine whether the given function satisfy Laplace’s equation

SOLUTION

 

 

 

TOTAL DIFFERENTIAL OF A FUNCTION

In our study of  Partial differentiation will not be limited to measurement of Partial Variation in the dependent variable of a multivariable, but we want to look and evaluate the total effect on the dependent variable in a function.  When all the independent variable of function change simultaneously at some given rates.

            Suppose the function  was changed as a result of small change in x and y. The small increments in both independent variables are arbitrary assigned as . The function  as a result of change in x and y

The function  z = f (x ,y) as a result of change in x and y. The more general case of a function of

n – independent variable

. The total differential of this function can be written as

 

EXAMPLE 18

Find the total differential given

SOLUTION

EXAMPLE 19

Find the total differential given

SOLUTION

 

TOTAL DERIVATIVES

 EXAMPLE 20

Find the total derivative . Given that

SOLUTION

 

 

 

DIFFERENTIAL OF IMPLICIT FUNCTION

 EXAMPLE 21

Find

SOLUTION

 

OPTIMIZING FUNCTION OF TWO VARIABLE

Supposing the manufacturer noodles produces two types of noodles, the onion flavour and the chicken flavour. The cost of producing of x – units of onion flavour and y – units of chicken flavour is given by the function . The manufacturer is trying to find the production level , that result in minimal cost or the production expenditure on capital (  ) and labour  that will maximum output given by . This is what we try to learn in this section, that is optimising a multivariable function.

   

RELATIVE EXTREMA

The function  is said to have a relative maximum at  in the domain of f  if  for all point  in a circular disk centered at p. Similarly if   for all  in a circular disk at Q,  then  has a minimum at

 

                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                               

 

CRITICAL POINT

The point (a,b) in the domain of  for which

 

 

CRITICAL POINTS AND RELATIVE EXTREMA

SADDLE POINT

A saddle point is a where there is relative maximum is one direction and relative minimum is in the other direction

 

THE SECOND PATIAL DERVIVATIVE TEST

For function of two variables, a necessary condition that  has relative maximum or minimum is

EXAMPLE 21

Find all the critical point for the function  and classify each as a relative maximum, a relative minimum or saddle point.

SOLUTION

EXAMPLE 22

Examine  for extrema

SOLUTION

 

EXAMPLE 23

Examine the function  for extrema

SOLUTION

 

 

 

 

 

 

 

SUMMARY 

For minima Extrema

For maxima Extrema

For Saddle point

 

APPLICATION TO ECONOMICS

EXAMPLE 24

A Beverage shop carries two competing beverages – ovaltine and Milo. The owner of the store can obtain both types at a cost of N2 per beverage and estimate that if Ovaltine beverage are sold for x naira apiece and Milo beverage for y naira apiece, consumer will buy 40– 50x + 40y Ovaltine beverage and 20 + 60x –70y Milo beverage each day. How should the owner price the beverages in order to generate the largest profit?

 

SOLUTION

Cost price for Ovaltine beverage is N2 apiece

Selling price for beverage for Ovaltine beverage apiece is Nx

Profit apiece on each Ovaltine beverage is N (x – 2)

Profit on Ovaltine P_o = (40 –50x + 40y) (x – 2)

Cost Price for Milo beverage is N 2 apiece

Selling price for Milo beverage is N y apiece

Profit apiece on Milo beverage is N (x – 2)

Profit on Milo P_M = (20 + 60x – 70y) (x – 2)

Total Profit = (Profit from Milo sales) + (Profit from Ovaltine sales)

The Profit Function will be

EXAMPLE 25

A company produces x units of commodity A and y units of commodity B. All the units can be sold for p =100 – x Naira per unit of A and q =100 – y Naira per units of B. The cost (in Naira) of producing these unit is given by the joint – cost function C(x, y) = x²  + xy + y². What should x and y be to maximize profit.

 

SOLUTION

Revenue for selling commodity A = (100 – x )x =100x – x²

Revenue for selling commodity B = (100 – x )x =100y – y²

Total Revenue for selling A and B = 100x – x² + 100y – y²

The profit Function = Total Revenue – Total Cost

Gives x = 20 and y = 20

Obtain the second order partial

 

CONSTRIANED OPTIMIZATION

METHOD OF LAGRANGE MULTIPLIER

In allocation of resources, managers are faced with restriction or constraint. Such constraint can be payment of wages and salary to work, allocation of funds to advertisement, training of workers, acquirement of new equipments, while trying to maximize profit and minimize   cost within a fixed budget.

For example, the Production Manager of Nestle foods Nigeria Limited is constrained to stay with a fixed budget of 1,000,000 Naira, may wish to allocate this funds between new equipment and training of his staff in order to maximize the future sales of their products. If x denotes the amount of funds allocated to acquiring new equipments, y, the amount allocated to training of staff and f (x, y) the corresponding number of product that will be sold.

The Production Manager would like to maximize the sale function f (x, y) subject to the budget constraint that is x + y = 1,000,000

In this section, we shall introduce another method called method of Lagrange Multipliers, in which a third variable called the multiplier assist us to solve constrained Optimization Problems

The method of Lagrange Multiplier uses the fact that any relative extrenum of the function f (x, y) subject to the constraint g (x, y) = k

Must occur at the critical point (a, b) of the function

Where l is the Lagrange Multiplier

EXAMPLE 26

Find the minimum value of  subject to the constraint

SOLUTION

Set the constraint to zero, that is

           

Set up Lagrange’s function

 

EXAMPLE 27

Find the maximum and minimum values of 

SOLUTION

 

Note:

If a function f  has constrained maximum, it will have the largest of the critical values of  f (a, b)

If a function f  has constrained minimum, it will have the smallest of the critical values of  f (a, b)

 

EXAMPLE 28

A cattle rearer wishes to fence off a rectangular pasture along the bank of a river. The area of the pasture is to be 12800 square metres and no fencing is needed along the riverbank. Find the maximum dimension of the pasture that will require the least among

 

SOLUTION

 

                       

 

 

 

 

 

 

 

Let f denote the amount of fencing required, f(x ,y) = 2x + y

Our aim is to minimize f , since the area must be 6400 square metres, that is the constraint

g (x,y) = x y = 12800

The Langragean function

APPLICATION TO ECONOMICS

(A) MAXIMIZATION OF UTILITY

            The satisfaction or utility a consumer derives from consuming x – items and y – items can be written as U (x,y). We can apply Lagrange Multiplier to determine of many units of each items the consumer should purchase while staying within a fixed budget

 

EXAMPLE 29

Mr Tawose has N650 Naira to spend on two items. The first of which cost N4 per unit and second N10 per unit. Suppose that the utility derived by the consumer from x unit of the first item and y unit of the second is given by

(a)                How many unit of each commodity should the consumer buy to maximize utility

(b)                Compute the lagrange multiplier

SOLUTION

The total cost of buying x units of the first items at N4 per unit and y units of the second items at N10 per unit is 4x + 10y = 560. Since the consumer has only N560, the aim will be to maximize utility U (x, y) subject to  the budgetary constraint is 4x + 10y = 560

The Objective Function is , while the Constraint Function is 4x + 10y = 560

Set the constraint to zero 4x + 10y – 560 = 0

The Lagrange Function

For the consumer to maximize his satisfaction or utility, the consumer buy 35 units of x – items and 42 units of y – items 

Substitute x =35, y =  42 into equation (1)

 

(B) ALLOCATION OF FUNDS

EXAMPLE 30

If N x  thousand is spent on labour and N y thousand is spent on equipment, the output at the factory of Matrix Nigeria Limited will be  units. If N12,000 is available, how should this be allocated between labour and equipment to generate largest possible output.

 

SOLUTION

The objective function =

The constraint is

Set up the lagrangian function