RS - Ch. 8 - Multivariate Calculus
Chapter 8
Multivariate Calculus
Isaac Barrow (1630-1677)
Augustin Louis Cauchy (1789–1857)
1
8.1 Multivariate Calculus: Partial Differentiation
• Now, y depends on several variables: y = f(x1, x2, …, xn.)
The derivative of y w.r.t. one of the variables –while the other
variables are held constant- is called a partial derivative.
y  f ( x1 , x 2 ,  , x n )
lim
 x1  0
f ( x1   x1 , x 2 ,  , x n )  f ( x1 , x 2 ,  , x n )
y
 lim
 x1 x1  0
 x1

In general,
y
 f1
x1
lim
 xi  0
( partial derivative w.r.t. x 1 )
y
y

 f i , i  1...n
 x i xi
2
1
RS - Ch. 8 - Multivariate Calculus
8.1 Partial Differentiation: Example
Cobb - Douglas production function : Q  AK  L 
 A  96 ; 
MPP K
MPP L
   1
Q  96K 0.3 L0.7
Q

 0 . 3 96 K  .7 L0 .7  28 . 8 K  .7 L0 .7
K
Q

 0 . 7 96 K 0 .3 L  0 .3  67 . 2 K 0 .3 L  0 .3
L
3
8.1 Partial differentiation: Market Model
Qd  a  bP
( a , b  0)
Qs   c  dP
(c, d  0)
1 b  Q   a 
1  d   P     c 

   
  d  b   a  Q * 
1
 
 b  d    1 1    c   P * 
ad  bc
ac
P* 
bd
bd
*
d
Q
Q *
b

0

0
bd
bd
a
c
P *
P *
1
1

0

0
c b  d
a b  d
Q* 
4
2
RS - Ch. 8 - Multivariate Calculus
8.1 Partial differentiation: Market Model
• Using linear algebra, we have:
 d
Q *   b  d 
 *   1
P  
 b  d 
1
*
x A d
b 
b  d   a 
 1   c 
b  d 
x* d  A1
 Q *

x
  a*
d  P
 a
*
Q *   d

c    b  d
*
P   1
c   b  d
b 
bd
1 

bd
Note: We call the matrix of first partial derivatives the Jacobian, J.
5
8.1 Partial Differentiation: Likelihood
In the usual estimation problem in Classical Linear Model (CLM),
the unknowns are the parameters (typical in the CLM, β and σ2). We
treat the data (xt and yt) as (conditionally) known numbers. Assuming
normality for the error term, εt,the (log) likelihood function is:

1
1
T
T
T
T
T
2
log L   ln(2 2 )2  2 t 1 t   ln 2  ln 2  2 t 1 ( yt  x1,t 1  x2,t 2 )2
2
2
2
2
2
1st partial derivatives :
 ln L
1
1 T
T
  2 t 1 2( yt  x1,t 1  x2,t 2 )(x1,t )  2 t 1 ( yt x1,t  x12,t 1  x2,t x1,t 2 )

1
2
 ln L 1 T
  ( yt x2,t  x1,t x2,t 1  x22,t 2 )
2  2 t 1
T
1
 ln L
T
2
  2  4 t 1 t
2 2
 2
1
T
1 T
T
1 T
2
J   2 t 1 ( yt x1,t  x12,t 1  x2,t x1,t 2 )
 ( yt x2,t  x1,t x2,t 1  x22,t 2 )  2 2  2 4 t1t 
 2 t 1

6
3
RS - Ch. 8 - Multivariate Calculus
8.1 The Jacobian

The Jacobian is the matrix of first partial derivatives at the point x:.
J  f x1 (x) ... f xn (x) 
Notation: J or Dfx. For the one equation case, J is a row vector.
Sometimes, J is written as a column vector as f(x) and called the
gradient or gradient vector at x.

A vector is characterized by its length and direction. To emphasize
the direction, the length, h, can be standardized, say ║h║=1. The
direction is studied with directional derivatives.

7
8.1 Directional Derivatives
We can think that the partial derivatives of z = f(x, y, w, …)
represent the rates of changes of z in the x, y, ….

Suppose that we now wish to find the rate of change of z at (x0, y0)
in the direction of an arbitrary unit vector u = <a, b>.

8
4
RS - Ch. 8 - Multivariate Calculus
8.1 Directional Derivatives
Consider the surface S with equation z = f(x, y) [the graph of f ]
and we let z0 = f(x0, y0) => The point P(x0, y0, z0) lies on S.

The vertical plane that passes through P in the direction of u
intersects S in a curve C.

The slope of the tangent line
T to C at the point P is the rate
of change of z in the direction
of u.

9
8.1 Directional Derivatives
Now, let Q(x, y, z) be another point on C.
 P’, Q’ be the projections of P, Q on the xy-plane.

 The vector P ' Q ' is parallel to u.

 P ' Q ' = h u = <h a, h b>,
for some scalar h.


Then,
x – x0 = ha => x = x0 + ha
y – y0 = hb => y = y0 + hb
z

h
f ( x o  ha , y o  hb )  f ( x 0 , y 0 )
h
If we take the limit as h → 0, we
get the rate of change of z (w. r. to
distance) in the direction of u.
10
5
RS - Ch. 8 - Multivariate Calculus
8.1 Directional Derivatives
The directional derivative of f at (x0, y0) in the direction of a unit vector
u = <a, b> is:

Du f ( x 0 , y 0 )  lim
h0
f ( x o  ha , y o  hb )  f ( x 0 , y 0 )
h
if the limit exists.
Special cases:
- If u = i = <1, 0>, then Di f = fx.
- If u = j = <0, 1>, then Dj f = fy.

That is, the partial derivatives of f with respect to x and y are just
special cases of the directional derivative.
11
8.1 The Jacobian Determinant
The Jacobian determinant, |J|, at a point x gives information about
the behavior of F near x. For instance, the continuously differentiable
function F is invertible near a point x∈Rn if |J|≠ 0.

Use |J| to test the existence of functional dependence between
functions. If |J| = 0, => functional dependence, that i, a solution to
a system of equations does not exist.
 Not limited to linear functions as |A|
 For the 2x2 case:

J 
 y1  x1
y1 x 2
 y 2  x1
y 2 x 2
Carl Jacobi (1804 – 1851, Germany)
12
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RS - Ch. 8 - Multivariate Calculus
8.1 Cross partial derivatives

The partial derivative is also a function of x: f’(x) = g(x1, x2, …, xn.)
If the n partial derivatives are continuous functions at point x, we
say that f is continuously differentiable at x.

If the n partial derivatives are themselves differentiable on an open
set S ∈Rn , we can compute their partial derivates. For example:

2 f
f f

(
)
xi x j xi x j
 The result of this differentiation is known as the cross partial derivative
of f with respect to xi and xj. It is usually denoted as fij.
When i=j, cross partial derivatives becomes the second-order
derivative, denoted as fii.. The matrix of all second derivatives is

13
8.1 Cross partial derivatives: Hessian
The matrix of all second derivatives is called the Hessian, usually
denoted by H. For example:

  2 f  x12
H  2
  f (  x 2  x1 )
 2 f (  x1  x 2 ) 

 2 f  x 22 
Example: Cobb-Douglas function, Q = AKL
Q K   AK
 1 
L
Q L   A K  L  1
   (1   ) AK   2 L
H  
 1  1
L
  AK
 AK
 1  1
L
  (1   ) A K

L
2



Note: fij..= fji.. This is a general result (Young’s Theorem). Then, H is
a symmetric matrix. H plays a very important role in optimization. 14
7
RS - Ch. 8 - Multivariate Calculus
8.1 Cross partial derivatives: Hessian - Example
We want to calculate H for a function, using econometrics
notation, of β1 and β2 and σ2(we treat xt and yt as constants, along
with σ2). This (log) function is:

log L  f ( y1 , y 2 ,..., yT |  ,  2 )  
1st derivative s :
 ln L
1
fx 

 1
2 2
fy 
 ln L
1
 2

 2


T
t 1
T
t 1
T
T
1
ln 2  ln  2 
2
2
2 2
2( yt  x1,t 1  x2 ,t  2 )(  x1,t ) 

1
2
T
t 1

( yt  x1,t ' 1  x2 ,t '  2 ) 2
T
t 1
( yt x1,t  x12,t 1  x2 ,t x1,t  2 )
( yt x2 ,t  x1,t x2 ,t 1  x22,t  2 )
2nd derivative s and cross derivative s :
f xx 
1
 2 ln L
 2
2

 1

f xy 
 2 ln L
1
 2
1 2


f yy 
 2 ln L
1
 2
2

 2

T
2
t 1 1,t
x 0
T
t 1
T
t 1
x2 ,t x1,t
x22,t  0
15
8.1 Cross partial derivatives: Hessian - Example

Then:
f xx  
f xy  
f yy  
1

2
1
2
1
2

T
2
t 1 1,t

t 1
x 0
T

T
t 1
x2,t x1,t
x22,t  0
1
1
T
T


2
 2 t 1 ( x2 ,t x1,t ) 
   2 t 1 ( x1,t )

H 

1
1
T
T
 2 t 1 ( x1,t x2 ,t )
 2 t 1 ( x22,t ) 

 

Note: H plays a very important role in maximum likelihood
estimation. Its inverse is used to calculate standard errors.
16
8
RS - Ch. 8 - Multivariate Calculus
8.2 Differentials
8.1.1 Differentials and derivatives
Problem: What if no explicit reduced-form solution exists because of
the general form of the model?
Example: In the macro model, what is Y / T when
Y = C(Y, T0) + I0 + G0 ?
T0 can affect C direct and indirectly through Y, violating the partial
derivative assumption
Solution: Use differentials!
 Find the derivatives directly from the original equations in the
model.
 Take the total differential
 The partial derivatives become the parameters
17
8.2 Differentials
1)
2)
3)
4)
5)
6)
d
y  f x   lim y x
x 0
dx
y x  f x   r (x)
y x  f  x   r (x)
y  f x x  r (x)x
y  f x x
dy  f x dx
(derivative)
as x  0 then   0
finite as x  0
(differential)
where f x  is the factor of proportionality
A first - order Taylor series approximation of y 1 is :
7)
f x1   f x0   f x0 x1  x0 
8)
y  f x0 x
(difference quotient)
18
9
RS - Ch. 8 - Multivariate Calculus
8.2.1 Differentials and derivatives

What we are going to do:
- From partial differentiation to total differentiation
- From partial derivative to total derivative using total differentials
- Total derivatives measure the total change in y from the direct
and indirect affects of a change in xi

The symbols dy and dx are called the differentials of y and x,
respectively
A differential describes the change in y that results for a specific
and not necessarily small change in x from any starting value of x
in the domain of the function y = f(x).
The derivative (dy/dx) is the quotient of two differentials: dy & dx.
f '(x)dx is a first-order approximation of dy:
19
y  f ( x )  dy  f ' ( x ) dx



8.2.2 Differentials and point elasticity


Let Qd = f(P)
(explicit-function general-form demand equation)
Find the elasticity of demand with respect to price
elastic if  d
dQ d 
 dQ d


 marginal function
dP
 
 
dP 
Qd
average function
P
P
 1, inelastic if  d  1
% Qd
d 

% P
Qd
20
10
RS - Ch. 8 - Multivariate Calculus
8.3 Total Differentials


Extending the concept of differential to smooth continuous
functions w/ two or more variables
Let y = f (x1, x2) Find total differential dy
dy 
y
y
dx 1 
dx
 x1
x 2
dy  f 1 dx
1
 f 2 dx
2
2
21
8.3 Total Differentials - Example




Let U be a utility function: U = U (x1, x2, …, xn)
Differentiation of U with respect to xi
U/ xi is the marginal utility of the good xi
dxi is the change in consumption of good xi
dU 
U
U
U
dx 1 
dx 2   
dx n
 x1
x 2
x n
• dU equals the sum of the marginal changes in the consumption of
each good and service in the consumption function.
• To find total derivative wrt to x1 divide through by the differential
dx1 ( partial total derivative):
dU
U
 U dx 2
 U dx n


 ... 
dx 1
 x1
 x 2 dx 1
 x n dx 1
22
11
RS - Ch. 8 - Multivariate Calculus
8.3 Rules of Differentials (same as derivatives)
Let k is a constant function; u = u(x1); v = v(x2)
 1. dk = 0
(constant-function rule)
n
n-1
 2. d(cu ) = cnu du
(power-function rule)
 3. d(u  v) = du  dv
(sum-difference rule)
 4. d(uv) = v du + u dv
(product rule)
 5.
(quotient rule)
u  vdu  udv

 6. d   
v2
v
 7. d(uvw) = vw du + uw dv + uv dw
d u  v  w   du  dv  dw
23
8.3 Example:
Find the total differential (dz) of the function
x y
1)
z
2)
y

2x2 2x2
z
z
dz 
dx 
dy
x
y
3)
4)
z
2x2
x
z
  x
 4 xy
y  2x2  4x2


 2  2
2
2 2
x  x  2 x
2x 
2x
2x2
 
6)
 
2
x  2x  2 y

4x4
2x3
  x
y    y 
z
1






y y  2 x 2 2 x 2   y  2 x 2  2 x 2
1
 x  2 y 
dz 
dx  2 dy
3
2x
2x

5)
2
2 x  4 x  4 xy
24
12
RS - Ch. 8 - Multivariate Calculus
8.3.1 Finding Total Derivatives from Differentials
Given
1)
y  f x1,x2 , ,xn 
Total differential dy is equal to the sum of the partial changesin y :
y
y
y
2)
dy 
dx1 
dx2   
dxn
x1
x2
xn
3)
dy  f1dx1  f 2 dx2  ...  f n dxn
The partial total derivativeof y wrt x1 , for example,is found by
dividing both sides by dx1
4)
dx
dx
dy
 f1  f 2 2    f n n
dx1
dx1
dx1
25
8.4 Multivariate Taylor Series

Recall Taylor’s series formula
f ( x )  T ( x , c )  f c  

f
/
c  x  c 1
1!

f
//
c  x  c 2
2!
f (n) c 
x  c n
n!
We want to generalize the Taylor polynomial to multivariate
functions. A similar logic to the univariate case gives us:
f ( x )  T ( x , a )  f a   Df ( a ) x  a  
1


1
x  a T H (a )( x  a )  
2!
Using abbreviated notation:
T(x,a) = Σj=0 to n (1/j!)(x-a)j Djf(a).
26
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RS - Ch. 8 - Multivariate Calculus
8.4 Multivariate Taylor Series
Example: 1-st order Taylor series, around a=(d,c)=(0,0) of
f(x,y ) = [(1+x)/(1+y)] -1
f(x,y )= [(1+x)/(1+y)] -1
=> f(c=0,d=0) = [(1+0)/(1+0)] - 1 = 0
=> fx(c=0,d=0) = 1
fx = 1/(1+y)
fy = (-1)(1+x)/(1+y)2
=> fy(c=0,d=0) = -1
Then, 1st-order Taylor’s series formula:
f(x,y) ≈ T(x,y; 0) = 0 + 1 (x-0) + (-1) (y-0) = x – y
• Application to Relative PPP:
ef,TPPP = [(1 + Id)/(1 + If)] - 1 ≈ (Id - If),
where ef,T = (St+T/St) -1
27
8.5 Homogeneous Functions
Definition:
A function f(x1, ..., xn) is homogeneous of degree r if multiplication of each
of its independent variables by a constant j will alter the value of the
function by the proportion jr, that is;
if f (jx1, ..., jxn) = jrf(x1, ... xn), for all f (jx1, ... jxn) in the domain of f

Special cases:
- If r = 0, j0 = 1,, the function is homogeneous of degree zero
- If r = 1, j1 = j, the function is homogeneous of degree one,
sometimes called linearly homogeneous.
Note: Technically if j>0, we say positive homogenous.
28
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RS - Ch. 8 - Multivariate Calculus
8.5 Homogeneous Functions: Examples
Examples:
- In applied work, it is common to see homogenous production
functions. For example, a firm increases inputs by k, then output
increases by k. Then, f(.) is homogenous of degree (r= 1), we say, f(.)
shows constant returns to scale. If r>1 (r<1), shows increasing (decreasing)
returns to scale.
- Demand functions are homogeneous. If all prices and income
change by the same amount (the budget constraint does not change),
the demands remain unchanged. That is,
D(jp1, ..., jpn, jI) =D(x1, ... xn,I) => homegenous of degree 0.
Since individual demands have r=0, the aggregate demand (sum of
29
individual demands) also has r=0.
8.5 Homogeneous Functions: Cobb-Douglas
A popular production function is the Cobb-Douglas:
Q = AKL. .
The Cobb-Douglas function is homogeneous of degree  + :

A  jK
 Aj
  jL 
 
K

j K

L
L   j    AK

L
 Aj

 j  Q
Cases:  +  > 1 increasing returns (paid < share)
 +  < 1 decreasing returns (paid > share)
 +  =1, constant returns (function is linearly homogeneous)
Note: In empirical work it is usually found that + are close to 1.
Assuming linear homogeneity is common.
30
15
RS - Ch. 8 - Multivariate Calculus
8.5 Homogeneous Functions: Cobb-Douglas
More characteristics Q = AKL
- Isoquants are negatively sloped and strictly convex (K, L > 0)
- Function is quasi-concave (K, L > 0)
- If  + =1, the Cobb-Douglas function is linearly homogeneous.
Let j = 1/L, then the average physical product of labor (APPL) and of
capital (APPK) can be expressed as the capital-labor ratio, k  K/L:

Q
  ( k )  Ak 
L
Q
APP L 
  ( k )  Ak 
L
Q L  (k )
APP K 

 Ak  1
k
L K
jQ 
Note: This result applies to linearly homogeneous functions
Q = f(K, L) => jQ  Q/ L  f (K / L, L / L)  f (k,1)  (k)
31
8.5 Homogeneous Functions: Properties

Given a linearly homogeneous production function Q = f(K, L),
the marginal physical products MPPL and MPPK can be expressed
as functions of k alone:
MPP K   ' ( k )
MPP L   ( k )  k  ' ( k )
32
16
RS - Ch. 8 - Multivariate Calculus
8.5 Homogeneous Functions: Euler’s Theorem
Euler’s Theorem
Let f : Rn+ →R be continuous, and differentiable on Rn+. Then, f is
homogeneous of degree r if and only if for all x ∈ Rn+:
r f (x1, ..., xn) = Σi fi xi
Example: Suppose Q = f(K, L), is homogeneous of degree 1, then,
Q  K
Q
Q
 K MPK
 L
L
K
  L MPL 
Then, if each input is paid the amount of its marginal product the
total product will be exactly exhausted by the distributive shares for all
the inputs -i.e., no residual is left.
33
8.5 Homogeneous Functions: Euler’s Theorem
• Euler’s theorem has a useful corollary:
Suppose that ƒ: Rn → R is differentiable and homogeneous of degree
k. Then its first-order partial derivatives fi are homogeneous of degree
k − 1.
Application: Suppose Q = f(K, L), is homogeneous of degree 1, then,
the MPL and MPK are homogeneous of degree 0. This implies:
 2Q
Q Q
 2Q
Q
Q Q
(
) K
(
)  L
 K
 L
2
K L
K L
L
L L
L
L  2Q
 2Q
 
K  L2
K L
0 
which is positive since fii <0. That is, the marginal productivity of one
factor increases when the other factor also increases (Wicksell’s law).
34
17
RS - Ch. 8 - Multivariate Calculus
8.6 Implicit Function Theorem



So far, if we were given F(y, x)=0  y = f(x).
 dy/dx easy to calculate (not always realistic situation.)
Suppose F(y, x) = x3 – 2x2y + 3xy2 - 22 = 0,
 not easy to solve for y=f(x)
=> dy/dx=?
Implicit Function Theorem: given F(y, x1 …, xm) = 0
a) if F has continuous partial derivatives Fy, F1, …, Fm and Fy  0
b) if at point (y0, x10, …, xm0), we can construct a neighborhood
(N) of (x1 …, xm), e.g., by limiting the range of y, y = f(x1 …,
xm), i.e., each vector of x’s  unique y
Then i) y is an implicitly defined function y = f(x1 …, xm) and
ii) still satisfies F(y, x1 … xm) for every m-tuple in the N such
35
that F  0.
8.6.1 Implicit Function Rule

If the function F(y, x1, x2, . . ., xn) = k is an implicit function
of y = f(x1, x2, . . ., xn), then
F y dy  F x 1 dx
1
 F x 2 dx
where Fy = F/y;
2
 ...  F x n d
xn
 0
Fx1 = F/x1
 Implicit function rule
 F(y, x) = 0; F(y, x1, x2 … xn) = 0, set dx2 to n = 0
F y dy   F x 1 dx
1
 F x 2 dx
2
 ...  F x n d
F x1
y
dy
| dx x . dx n  0 
 
 x1
dx 1
Fy
xn
36
18
RS - Ch. 8 - Multivariate Calculus
8.6.1 Deriving the implicit function rule
1) F ( y , x1 , x 2 )  0
(implicit function)
2) y  f ( x1 , x 2 )
(explicit function)
Total differenti ation of (1) :
F y dy  F1dx1  F2 dx 2  0
Total differenti ation of (2) :
dy  f1dx1  f 2 dx 2

 
5) F y f1  F1 dx1  0

3) F y  f1dx1  f 2 dx 2   F1dx1  F2 dx 2  0
4) F y f1  F1 dx1  Fy f 2  F2 dx 2  0


6) F y f 2  F2 dx 2  0
( dx 2  0)  f1 
F
y
 1
Fy
x1
( dx1  0)  f 2 
F
y
 2
Fy
x 2
37
8.6.1 Implicit function problem - Examples







Given the equation F(y, x) = x3 – 2x2y + 3xy2 - 22 = 0,
Q1: Find the implicit function y = f(x) defined at (y=3, x=1)
The function F has continuous partial derivatives Fy, F1, …, Fm :
∂F/∂x =3x2-4xy+3y2
∂F/∂y =-2x2+6xy
At (y0, x10, …, xm0) satisfying F (y, x1 …, xm) = 0, Fy ≠ 0:
F(y = 3, x = 1) = 13 – 2 12 3 + 3 1 32 - 22 = 0;
Fy = -2x2+6xy = -2 12+6 1 3 = 16.
Yes! We have a continuous function f with continuous partial
derivatives.
Q2: Find dy/dx by the implicit-function rule. Evaluate it at (y=3, x=1)
dy/dx = - Fx/Fy =- (3x2-4xy+3y2 )/-2x2+6xy
38
dy/dx = -(3*12-4*1*3+3*32 )/(-2*12+6*1*3)=-18/16=-9/8
19
RS - Ch. 8 - Multivariate Calculus
8.6.2 Derivatives of implicit functions - Examples

Example 1:
If F(z, x, y) = x2z2 + xy2 - z3 + 4yz = 0, then
y
F
 
z
F
z
y
 
2 x
2
z  3z2  4 y
2 xy  4 z
Example 2: Implicit Production function: F (Q, K, L)
F/J = FJ
J = Q, K, L
Applying the implicit function rule:
Q/L = -(FL/FQ)
- MPPL
Q/K = -(FK/FQ)
- MPPK
K/L = -(FL/FK)
- MRTS: Slope of the isoquant

39
8.6.3 Extension to the simultaneous-equation
case




We have a set of m implicit equations. We are interested in the
effect of the exogenous variables (x) on the endogenous variables
(y). That is, dyi/dxj.
Find total differential of each implicit function.
Let all the differentials dxi = 0 except dx1
and divide each term by dx1 (note: dx1 is a choice)
Rewrite the system of partial total derivatives of the implicit
functions in matrix notation
40
20
RS - Ch. 8 - Multivariate Calculus
8.6.3 Extension to the simultaneous-equation
case
Example
: 2x2 System
1 ) F1 ( y 1 , y 2 , x 1 )  0
2 ) F 2 ( y1 , y 2 , x 2 )  0
3)
 F1
 F1
 F1
dy 1 
dy 2 
dx 1  0
 y1
y 2
 x1
4)
 F2
 F2
 F2
dy 1 
dy 2 
dx 2  0
 y1
y2
x2
5)
 F1
 F1
 F1
dy 1 
dy 2  
dx 1  0 dx 2
 y1
y2
 x1
6)
 F2
 F2
dy 1 
dy 2 
 y1
y2
0 dx 1 
 F2
dx 2
x2
41
8.6.3 Extension to the simultaneous-equation
case

Rewrite the system of partial total derivatives of the
implicit functions in matrix notation (Ax=d)
7)
F
F1 dy1 F1 dy 2

 1
x1
y1 dx1 y 2 dx1
10 )
F1 dy1 F1 dy 2


y1 dx 2 y 2 dx 2
8)
F2 dy1 F2 dy 2


y1 dx1 y 2 dx1
11)
F2 dy1 F2 dy 2
F

 2
y1 dx 2 y 2 dx 2
x 2
 F1
 y
1
9) 
 F2
 y
 1
0
F1   dy1 
F
y 2   dx1    1 

   x1 

F2   dy 2  
 0 



dx
y 2   1 
 F1
 y
1
12 ) 
 F2
 y
 1
0
F1   dy1 
y 2   dx 2   0 

   F2 
F2   dy 2    x 
2 

y 2   dx 2 
42
21
RS - Ch. 8 - Multivariate Calculus
8.6.3 Extension to the simultaneous-equation
case

Solve the comparative statics of endogenous variables
in terms of exogenous variables using Cramer’s rule
 dy1   F1
 dx   y1
13 )  1   
 dy 2   F2
 dx1   y
 1
F1 
y 2 

F2 
y 2 

dy1

dx1
1
F1
x1
 F1 
  x 
1


 0 
F1
y 2
F2
y 2
0
J
 dy1   F1
 dx   y1
14 )  2   
 dy 2   F2
 dx 2   y
 1
F1 
y 2 

F2 
y 2 
1
 0 
 F2 
  x 
2 

F1 F2
x
x1 y 2

F1 F2 F1 F2
x

x
y1 y 2 y 2 y1

43
8.7 Application: The Market Model

Assume the demand and supply functions for a commodity are
general form explicit functions
Qd = D(P, Y0)
(Dp < 0; DY0 > 0)
Qs = S(P, T0)
(Sp > 0; ST0 < 0)

Q is quantity,
P is price,
(endogenous variables)
Y0 is income,
T0 is the tax (exogenous variables)
no parameters, all derivatives are continuous

Find

Solution:
- Either take total differential or apply implicit function rule
- Use the partial derivatives as parameters
- Set up structural form equations as Ax = d,
- Invert A matrix or use Cramer’s rule to solve for x/d
P/Y0,
Q/Y0,
P/T0
Q/T0
44
22
RS - Ch. 8 - Multivariate Calculus
8.7 Application: The Market Model
1) D ( P , Y0 )  S ( P , T0 )  Q
2 ) F 1(P, Q; Y 0 , T 0 )  D( P , Y 0 ) – Q  0
3) F 2(P, Q; Y 0 , T 0 )  S( P , T 0 ) – Q  0
Suppose we are interested in finding d Q dY0 .
Take the total differenti al of equations (2) & (3) and organize;
4 ) D P/ d P  d Q   DY/ 0 dY 0
5) S P/ d P  d Q   S T/ 0 dT 0
Put equations (4) & (5) in matrix format (Ax  d) ;
D /
6 )  /P
 S P
/
 1  d P    DY0




 1  d Q   0
  dY 0    DY/ dY 0 
0



 S T/ 0   dT 0    S T/ 0 dT 0 
0
45
8.7 Application: The Market Model
 D /  1  d P 
6 )  /P


 S P  1  d Q 
Take the partial total
 dP 
 dY   D /
7 )  0    /P
 dQ   S P
 dY 
 0
  D Y/ 0 dY 0 


/
  S T 0 dT 0 
derivative of equation (6) wrt to dY 0 .
1
 1   D Y/ 0 
 

 1  0 
We want to calculate :
dQ
| J2 |

dY 0
|J |
Calculate the Jacobian determinan t, | J|, and | J 2 | .
J 
D P/
1
S P/
1
 S P/  D P/  0
S P/ D Y/ 0
dQ
 /
 0.
dY 0
S P  D P/
;
J2 
D P/
 D Y/ 0
S P/
0
 S P/ D Y/ 0  0
46
23
RS - Ch. 8 - Multivariate Calculus
8.8 Limitations of Comparative Statics




Comparative statics answers the question: how does the
equilibrium change with a change in a parameter.
The adjustment process is ignored
New equilibrium may be unstable
Before dynamic, optimization
47
Overview of Taxonomy Equations: forms and functions
Form
Primitive
Function
Specific
(parameters)
General
(no parameters)
Explicit
(causation)
y = a+bx
y = f(x)
Implicit
(no causation)
y3+x3-2xy = 0
F(y, x) = 0
48
24
RS - Ch. 8 - Multivariate Calculus
Overview of Taxonomy –
1st Derivatives & Total Differentials
Form
Differentiation
Function
Specific
(parameters)
Explicit
(causation)
d
yb
dx
Implicit
(no causation)
3 y
2
 
General
(no parameters)
dy  f ( x) dx
dy
 f ( x )
dx

 2 x dy  2 x 2  2 y dx  0


2
dy
2x  2 y

dx
3y 2  2x


dy  
Fx
dx
Fy
F
dy
 x
dx
Fy
49
25