Problem Set 4 From: Mash-Erdene Ganbold (1154248) 1. (a) y = A·L·f(k) = A L f I K AL ,1M = A L f Hk,1L r = MP K = ΔY ΔK = Δy ΔK r = ΔIAL f I K AL ,1MM ΔK = A L f ' I K AL ,1M 1 AL = f ' Hk,1L = f ' HkL Using chain rule: f(g(x)) = f’(g(x))·g(x), where f HgH xLL = f I K AL ,1M gH xL = K AL (b) y = A·L·f(k) = A L f I K AL ,1M w = MP L = ΔY ΔL = Δy ΔL w = ΔIAL f I K AL ,1MM ΔL = A f ' I K AL ,1M - K AL 2 L + A f I K AL ,1M = A· f ' I K AL ,1M - K AL + A f I K AL ,1M = AA f ' I K AL ,1M - K AL + f I K AL ,1ME = A[f(k)-kf’(k)] Using product & chain rule: (f·g)’ = f’·g+f·g’ where f H xL = A f I K AL ,1M gH xL = L
Welcome message from author
This document is posted to help you gain knowledge. Please leave a comment to let me know what you think about it! Share it to your friends and learn new things together.
Transcript
-
Problem Set 4
From: Mash-Erdene Ganbold (1154248)
1.
(a)
y = ALf(k) = A L fI KAL
, 1M = A L fHk, 1L
r = MPK =Y
K=
y
K
r =IALf I K
AL,1MM
K= A L f ' I K
AL, 1M 1
AL
= f ' Hk, 1L = f ' HkL
Using chain rule:
f(g(x)) = f(g(x))g(x), where
fHgHxLL = fI KAL
, 1MgHxL = K
AL
(b)
y = ALf(k) = A L fI KAL
, 1M
w = MPL =Y
L=
y
L
w =IALf I K
AL,1MM
L= A f ' I K
AL, 1M - K
AL2 L+A fI KAL , 1M
= Af ' I KAL
, 1M - KAL+A fI K
AL, 1M
= AA f ' I KAL
, 1M - KAL+ fI K
AL, 1ME
= A[f(k)-kf(k)]
Using product & chain rule:
(fg) = fg+fg where fHxL = A fI K
AL, 1M
gHxL = L
Related Documents
