Calculus of multivariable functions

Lê Xuân Trường

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Outline

Partial derivatives and total differential

Definition of partial derivatives

Intepretations of partial derivatives
Higher-order partial derivatives
Total differential

Applications

Relative Minimums And Maximums

Absolute Minimums and Maximums
Lagrange Multipliers

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Partial derivatives

Definition
Let f (x, y ) be a function of two variables.

The partial derivative of f with respect to x at th point (x0 , y0 )

∂f f (x0 + h, y0 ) − f (x0 , y0 )
(x0 , y0 ) ≡ fx (x0 , y0 ) := lim
∂x h→0 h
The partial derivative of f with respect to y at th point (x0 , y0 )

∂f f (x0 , y0 + h ) − f (x0 , y0 )
(x0 , y0 ) ≡ fy (x0 , y0 ) := lim
∂y h→0 h

Remark: To calculate ∂ f /∂ x we treat y as a constant.

To calculate ∂ f /∂ y we treat x as a constant
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Partial derivatives

Examples: Find the partial derivatives in following cases

a/ z = f (x, y ) = 3x 2 y 3 + xexy
p
b/ z = g (x, y ) = x 2 + 3y − sin(2x + y 2 )

Giải
fx (x, y ) = 6xy 3 + (1 + xy )exy

fy (x, y ) = 9x 2 y 2 + x 2 exy

x
gx (x, y ) = p − 2 cos(2x + y 2 )
x 2 + 3y
3
gx (x, y ) = p − 2y cos(2x + y 2 )
2
2 x + 3y
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Intepretations of partial derivatives

The rates of change of the function z = f (x, y )

The rate of change of z with respect to x when y held fixed

f (x0 + ∆x, y0 ) − f (x0 , y0 )

fx0 (x0 , y0 ) ≈
∆x
The rate of change of z with respect to y when x held fixed

f (x0 , y0 + ∆y ) − f (x0 , y0 )
fy0 (x0 , y0 ) ≈
∆y

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Intepretations of partial derivatives

(Partial) marginal functions

Marginal function with respect to x

Mzx (x, y ) = fx (x, y ) ≈ f (x + 1, y ) − f (x, y )

Marginal function with respect to y

Mzy (x, y ) = fy0 (x, y ) ≈ f (x, y + 1) − f (x, y )

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Intepretations of partial derivatives

Example: Maginal Productivity

A manufacturer of a popular toy has determined that the production
function is √
Q = LK ,
where
L is the number of labour-hours per week
K is the capital required for a weekly production of Q gross of the toy

(one gross is 144 units)

Determine the marginal productivity functions, and evaluate them when

L = 400, K = 16.

Interpret the results.

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Intepretations of partial derivatives

Partial (Elasticity) of the function z = f (x, y )

Elasticity with respect to x

x x ∆z
Ezx (x, y ) = fx0 (x, y ) ≈ .
f (x, y ) f (x, y ) ∆x

Elasticity with respect to y

y y ∆z
Ezy (x, y ) = f 0 (x, y ) ≈ .
f (x, y ) y f (x, y ) ∆y

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Intepretations of partial derivatives

Example: Income Elasticity of Demand

Assume that the demand function of a product is given by

Q = 100 − 2P + PA + 0, 1Y

where Q is the quantity demanded and

P is the price
PS is the price of subtitue goods
Y is income
Find the income elasticity of demand in the case

P = 10, PS = 12, Y = 1000.

If the income increases 5% then let estimate the percentage change in quan-
tity demanded.
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Higher-Order Partial Derivatives

Second-Order Partial Derivatives

Given a function z = f (x, y ). The second-order partial derivatives of f
are defined as follows
∂ 2f ∂
fxx (x, y ) ≡ (x, y ) := (fx (x, y ))
∂ x2 ∂x
∂ 2f ∂
fyx (x, y ) ≡ (x, y ) := (fx (x, y ))
∂y∂x ∂y
∂ 2f ∂
fxy (x, y ) ≡ (x, y ) := (fy (x, y ))
∂ x∂ y ∂x
∂ 2f ∂
fyy (x, y ) ≡ (x, y ) := (fy (x, y ))
∂y2 ∂y

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Higher-Order Partial Derivatives
Example: Find the four second-order partial derivatives of

z = f (x, y ) = 2x 2 y + 6x 1/3 y 2/3

Solution Since
fx (x, y ) = 4xy + 2x −2/3 y 2/3
we have
4 4
fxx (x, y ) = 4y − x −5/3 y 2/3 fyx (x, y ) = 4x + x −2/3 y −1/3 .
3 3
Also, since
fy (x, y ) = 2x 2 + 4x 1/3 y −1/3
we have
4 4
fyy (x, y ) = − x 1/3 y −4/3 fxy (x, y ) = 4x + x −2/3 y −1/3 .
3 3
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Higher-Order Partial Derivatives

Remark: Under suitable conditions we have

fxy (x, y ) = fy x (x, y ).

In this course we always assume that this is the case for all the functions
that we consider.

Example: Given the function w = w (x, y , z defined by

w = 2x exp (xy + z 2 )

Find the value of

∂ 3w
(0, 1, 0)
∂ y ∂ z∂ x

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Total differential

Definition

Given the function z = f (x, y ). Let dx and dy represent changes in x

and y , respectively.

The total differential of z at (x, y ) is defined by

df (x, y ) = fx (x, y )dx + fy (x, y )dy

The second-order differential of z given by

d 2 f (x, y ) = fxx (x, y )dx 2 + 2fxy (x, y )dxdy + fyy (x, y )dy 2

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Total differential

Applications
Approximation for the change in z

df (x, y ) ≈ ∆z := f (x + dx, y + dy ) − f (x, y ).

Linear approximation of two variable functions

f (x, y ) ≈ f (x0 , y0 ) + fx (x0 , y0 )(x − x0 ) + fy (x0 , y0 )(y − y0 ).

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Total differential

Example
Given the function

z = f (x, y ) = 6x 2/3 y 1/2

a/ Find the total differential df (1000; 100).

b/ Using above result to find the approximate value of

f (998; 101, 5)

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Maxima and Minima for functions of Two Variables

Global and Local Extrema

The function z = f (x, y ) has a local maximum at (x0 , y0 ) if

f (x0 , y0 ) ≥ f (x, y )

for all points (x, y ) within some disk centered at (x0 , y0 ).

If the preceding inequality holds for every point (x, y ) in the do-
main of f , then f has a global maximum at (x0 , y0 ).
Similarly, we can define the local minimum and the global minimum

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Maxima and Minima for functions of Two Variables

Saddle points
The point (x0 , y0 ) is a saddle point of z = f (x, y ) if it is a critical
point and, for every disk D containing (x0 , y0 ), there are points (x1 , y1 ),
(x2 , y2 ) in D such that

f (x1 , y1 ) < f (x0 , y0 ) < f (x2 , y2 )

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Maxima and Minima for functions of Two Variables

Find extrema of the function z = f (x, y )

Rule 1: If the function
z = f (x, y )
has a local extrema at (x0 , y0 ) then (x0 , y0 ) is a critical point of f . It
means that (x0 , y0 ) is a solution of following system
(
fx (x, y ) = 0
fy (x, y ) = 0

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Maxima and Minima for functions of Two Variables

Find extrema of the function z = f (x, y )

Rule 2: (Second Derivative Test for Local Extrema)
For every critical point P (x0 , y0 ), we define
2
D (x0 , y0 ) = fxx (x0 , y0 )fyy (x0 , y0 ) − (fxy (x0 , y0 ))

If D > 0 and fxx (x0 , y0 ) > 0 then f has a local minimum at P

If D > 0 and fxx (x0 , y0 ) < 0 then f has a local maximum at P
If D < 0 then P is a saddle point of f
If D = 0 then the test is inconclusive.

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Maxima and Minima for functions of Two Variables

Find extrema of the function z = f (x, y )

Rule 3: (Second Derivative Test for Global Extrema)
Define
D (x, y ) = fxx (x, y )fyy (x, y ) − (fxy (x, y ))2

If D (x, y ) > 0 and fxx (x, y ) > 0, for all points (x, y ) in the domain
of f then f has a global minimum at P
If D (x, y ) > 0 and fxx (x, y ) < 0, for all points (x, y ) in the domain
of f then f has a global maximum at P

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Maxima and Minima for functions of Two Variables

Examples
1) Find the local extrema of

f (x, y ) = x 3 + y 3 − 12xy

2) A monopolist sells two competitive products, A and B, for which

the demand function are

qA = 16 − pA + pB , qB = 24 + 2pA − 4pB ,

where pA , pB are the prices of the products. The joint-cost is

C (qA , qB ) = 2qA + 4qB .

How many units of A and B should be sold to maximize the

monopolist’s profit?

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Lagrange Multipliers

Problem
Find the extrema of the function z = f (x, y ) subject to the constraint

ϕ (x, y ) = M

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Lagrange Multipliers

First solution

Using the constraint ϕ (x, y ) = 0, we can express one of the vari-

ables in term of the other variable.
Then we transform z, which is a function of two variables, into a
function of one variable
Find the extrema of the last function.

Example: Find the extrema of following function

z = f (x, y ) == x 2 − 3xy + 12x

subject to the constraint 2x + 3y = 6.

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Lagrange Multipliers

Method of Lagrange multipliers

We define a new function which is called to be Lagrange function

L(x, y , λ ) = f (x, y ) + λ (ϕ (x, y ) − M )

Rule 1: Suppose that f , when restricted to points on the curve

ϕ (x, y ) = M,

has a local extremum at the point (x0 , y0 ) and

(ϕx (x0 , y0 ))2 + (ϕy (x0 , y0 ))2 6= 0

Then there is a number λ0 called a Lagrange mutiplier, for which

Lx (x0 , y0 , λ0 ) = Ly (x0 , y0 , λ0 ) = Lλ (x0 , y0 , λ0 ) = 0.

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Lagrange Multipliers

Method of Lagrange multipliers

Rule 2: (Second Derivative Test)
Assume that M (x0 , y0 ) is a sationary point of f with respect to
the Lagrange multiplier λ0 . This means

Lx (x0 , y0 , λ0 ) = Ly (x0 , y0 , λ0 ) = Lλ (x0 , y0 , λ0 ) = 0.

We define
 
0 ϕx (x0 , y0 ) ϕy (x0 , y0 )
H (x0 , y0 , λ0 ) =  ϕx (x0 , y0 ) Lxx (x0 , y0 , λ0 ) Lyx (x0 , y0 , λ0 ) 
ϕy (x0 , y0 ) Lxy (x0 , y0 , λ0 ) Lyy (x0 , y0 , λ0 )

If det(H (x0 , y0 , λ0 ))> 0 then f has a local maximum at M

If det(H (x0 , y0 , λ0 ))< 0 then f has a local minimum at M
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Lagrange Mulipliers

Examples

1) Find the extrema of the function f (x, y ) = x + y subject to

x2 + y2 = 1

2) A firm want to produce 100 units of output in the cheapest pos-

sible manner. If there are two input factors ` and k, and their
prices per unit are fixed at p` = 2 and pk = 1, respectively, then
the production function is

q = 3`2/3 k 1/3 .

Find the value of ` and k to obtain firm’s goal.

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