Transcript math12.-DIFERENSIAL-FUNGSI-MAJEMUK

Diferensial Fungsi
Majemuk
-Diferensial Parsial
- Diferensial Total
- Chain rule
- dll
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1. y  f ( x , z )
y

a
)
f
(
x
,
z
)

 x
x
y' 
y
b ) f x ( x , z ) 
z

y
y
dy  dx  dz
x
z
2. p  f ( q , r , s )
p

a
)
f
(
q
,
r
,
s
)

 q
q

p

p ' b) f r ( q, r , s ) 
r

 c ) f ( q, r , s )  p
 s
s
p
p
p
dp 
dq  dr  ds
q
r
s
Diferensial Parsial
Diferensial Total
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High Order Partial Derivatives
Fungsi dengan lebih dari satu variabel
bebas juga dapat diturunkan lebih dari
satu kali
 Turunan parsial z = f (x,y)  kalau
kontinyu dapat mempunyai turunannya
sendiri.  empat turunan parsial :

 f  f  f
 f
,
,
, dan 2
2
x xy yx
y
2
2
2
2
• Dapat dilambangkan fxx, fxy, fyx, dan fyy
• fxy = fyx
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Partial derivatives
Cobb-Douglas production function (+=1)
 Q = 96K0.3 L0.7

Q
0.7
 0.7
0.7
 0.7
MPP K 
 0.396 L K
 28.8L K
K
Q
0.3  0.3
0.3  0.3
MPP L 
 0.7 96 K L  67.2 K L
L
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Techniques of partial differentiation

Market model
Qd  a  bP a, b  0 

Qs  c  dP c, d  0 
ac
P
bd
ad  bc
Q
bd
Qd  bP  a
Q  dP  C
 s
1  b  Q  a 
 




1  d   P   c 

1 a


Ap 1  c
 c  a  a  c 
P 



1

b
A
 d  b  b  d 


1 d

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Geometric interpretation of partial derivatives
• Market model
a, b  0
Qd  a  bP

Qs  c  dP c, d  0
ac
P
bd
P
a
P
c
P
b
1

0
bd
1

0
bd
 a  c 

0
2
b  d 
 a  c 

0
2
d b  d 
P
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• Market model
Qd  a  bP a, b  0

Qs  c  dP c, d  0
ad  bc
Q
bd
Q
a
Q
c
Q
b
d

0
bd
b

0
bd
 d a  c 

0
2
b  d 
ba  c 

0
2
d b  d 
Q
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Qd  a  bP

Qs  c  dP
Q
S0
a, b  0
c, d  0
Q
S
S1
D1
D
P
Q
b

0
c b  d
D
P
P
1

0
a b  d
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Market model
Qd  a  bP

Qs  c  dP
Q
a, b  0
c, d  0
S1
Q
S0
S0
D
P
Q0
Q1
D1
D0
P
 Q  d a  c 

0
2
b
b  d 
P  a  c 

0
2
d b  d 
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National-income model
Y = C + I0 + G 0
C = a + b(Y-T);
T=d+tY;
b = MPC
t = MPT
(a > 0; 0 < b < 1)
(d > 0; 0 < t < 1)
Y=( a-bd+I+G)/(1-b+tb)
C=(b(1-t)(I+G)+a-bd)/ (1-b+tb)
T=(t(I+G)+ta+d(1-b))/ (1-b+tb)
Y
1

Go 1  b  bt
C
b(1  t )

Go 1  b  bt
T
t

Go 1  b  bt
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Input-output model
 x1  b11 b12   d1 
 x   b
 d 
 2   21 22   2 
 x1   b11 d 1
 x   b d
 2   21 2

b12 d 1 
b22 d 2 
∂x1/∂d1 = b11
 x1   x1 d1 x1 d 2   d1 
 x   x d x d  d 
1
2
2  2 
 2  2
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Note on Jacobian Determinants
Use Jacobian determinants to test the existence of
functional dependence between the functions /J/
 Not limited to linear functions as /A/ (special case
of /J/
 If /J/ = 0 then the non-linear or linear
functions are dependent and a solution
does not exist.

J 
y1 x1
y1 x2
y 2 x1
y 2 x2
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Total Differentials
y
y
dy 
dx1 
dx2
x1
x2
dy  f1dx1  f 2 dx2
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Diferensial Total
Diferensial T otaldy dari y  f(x,z) didefinisikan sebagai :
dy  lim  f ( x  x, z  z )  f ( x, z )
x 0
y 0
ditambahkan f (x,z  z )  ( x, z  z )  0
f ( x  x, z  z )  f ( x, z  z )
 dy  lim
x 
x
x 0
f ( x, z  z )  f ( x, z )
y
lim
y
y 0
 df 
 df 
  dx   dz
 dx 
 dz 
f
f
y
y

dx  dz atau dy  dx  dz
x
z
x
z
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Let Utility function U = U (x1, x2, …, xn)
 Differentiation of U wrt x1..n
 U/ xi is the marginal utility of the good xi
 dxi is the change in consumption of good xi

U
U
U
dU 
dx1 
dx2   
dxn
x1
x2
xn
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Finding the total derivative from the differential
• Given a function y = f (x1, x2, …, xn)
• Total differential dy is:
y
y
y
dy 
dx1 
dx2   
dxn
x1
x2
xn
dy  f1dx1  f 2 dx2  ...  f n dxn
• Total derivative of y with respect to x2 found by
dividing both sides by dx2 (partial total derivative)
dxn
dy
dx1
 f1
 f2   fn
dx2
dx2
dx2
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Chain rule (kaidah rantai)
dz dz dy

 f  y g x 
dx dy dx

This is a case of two or more
differentiable functions, in which each
has a distinct independent variable.
where z = f(g(x)), i.e.,
z = f(y), i.e., z is a function of variable y
and
y = g(x), i.e., y is a function of variable x
• If R = f(Q) and if Q = g(L)
dR dR dQ


 f Q   g L   MR  MPPL  MRPL
dL dQ dL
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Kaidah Rantai
dy dy du

dx du dx
 y  f (u ) dan  u  f ( x),  dy / dx diperolehdengan
mendiferensialkan y ke u dan u ke x serta mengalikanhasilnya.
Kalau z  z ( x, y ) dan x  x(t ) dan y  y(t)
z
z
 dz  dx  dy
x
y
sehingga
z
dz z dx z dy


x
y
dt x dt y dt
t
Pohon
rantai
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Kaidah Rantai

Kalau w = w(x,y,z) dan x = x(u,v), y =
y(u,v), dan z = z(u,v), maka pohon rantai
:
w
x
y
u
z
sehingga:
w w x w y w z



u x u y u z u
dengan rumus serupa untuk w / v
v
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
Kalau z = z(x,y), dan x = x(s), y = y(s),
dan s = s(u,v), maka pohon rantai
menjadi z:
z  z dx z dy  s

 

u  x ds y ds  u
y
x
dengan rumus serupa unt uk z / v
s
u
v
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