2
$\begingroup$

In that paper, they consider the following production function:

$Y = F(A_KK,A_LL)$.

Where:

K denotes capital, L is labor, and $A_K$ and $A_L$ denote capital-augmenting and labor-augmenting technology, respectively. We assume throughout that F is continuously differentiable, concave, and exhibits constant returns to scale. Let $F_K$ and $F_L$ denote the derivatives of F with respect to capital and labor. We focus on competitive labor markets, which implies that the equilibrium wages is equal to the marginal product of labor:

$W = A_L F_L(A_KK,A_LL)$.

Why? Shouldn't simply be:

$W = F_L(A_KK,A_LL)$.

Without the $A_L$ multiplying the derivative?

Improve this question
$\endgroup$

1 Answer 1

Reset to default
3
$\begingroup$

Chain rule! To make it simple call $\tilde{L} = A_L L$ and $\tilde{K} = A_K K$, so that

$$ Y = F(\tilde{K}, \tilde{L}) $$

so that

$$ \frac{\partial Y}{\partial L} = \frac{\partial \tilde{L}}{\partial L}\frac{\partial F}{\partial \tilde{L}} = A_L F_L(A_K K, A_L L) $$

the trick here is that you should read $F_L$ as the derivative of $F$ with respect to its second argument.

Improve this answer
$\endgroup$
2
  • $\begingroup$ Thanks! I understand. But don't you think it's confusing to call $F_L$ what is actually $F_{(A_LL)}$? Is there a reason for doing so? $\endgroup$
    – Tecon
    Jan 22, 2019 at 15:34
  • 1
    $\begingroup$ @Tecon Yes it is :) But labor is the result of applying $A_L$ to $L$, so in that sense the marginal product of labor is $F_{(A_L L)}$ $\endgroup$
    – caverac
    Jan 22, 2019 at 15:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.