e****t 发帖数: 17914 | |
f*******n 发帖数: 12623 | 2 P is a function of what? just L? Why isn't it "dP/dL" (instead of ∂P/&
#8706;L)
You might want to look at implicit differentiation:
http://en.wikipedia.org/wiki/Implicit_function#Implicit_differe |
j*******o 发帖数: 34 | 3 Let P(L+x)=P+y
then [P(L+x)-P(L)]/[(L+x)-L]=y/x
Define (dP/dL)=lim{y/x, x->0}
We have
(1): 1+exp(beta L - lambda P) = lambda (P - c L)
(2): 1+exp[beta (L+x) - lambda (P+y)] = lambda [P+y - c (L+x)]
Let A = exp(beta L - lambda P)
We have the following by (2)-(1)
A[exp(beta x - lambda y)-1] = lambda (y - c x)
A[(beta x - lambda y) + Delta] = lambda ( y - c x)
By taking limit we have
(dP/dL) = (A+1) lambda / (A+c lambda)
=(exp(beta L - lambda P)+1) lambda / (exp(beta L - lambda P)+c lambda) |
f*******n 发帖数: 12623 | 4 F(P,L) = 1 + exp(beta L - lambda P) - lambda (P - c L) = 0
dP/dL = - (partial F/partial L)/(partial F / partialP)
= -( beta exp(beta L - lambda P) + c lambda ) / ( - lambda exp(beta L -
lambda P) - lambda ) |