In that paper, they consider the following production function:

$Y = F(A_KK,A_LL)$.


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?


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.

  • $\begingroup$ 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 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 at 15:37

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