<p>Chain rule! To make it simple call <span class="math-container">$\tilde{L} = A_L L$</span> and <span class="math-container">$\tilde{K} = A_K K$</span>, so that</p>

<p><span class="math-container">$$
Y = F(\tilde{K}, \tilde{L})
$$</span></p>

<p>so that</p>

<p><span class="math-container">$$
\frac{\partial Y}{\partial L} = \frac{\partial \tilde{L}}{\partial L}\frac{\partial F}{\partial \tilde{L}} = A_L F_L(A_K K, A_L L)
$$</span></p>

<p>the trick here is that you should read <span class="math-container">$F_L$</span> as the derivative of <span class="math-container">$F$</span> with respect to its second argument.</p>

CC BY-SA 4.0
2019-01-22T15:34:25.633Z
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?
Tecon
CC BY-SA 4.0
2019-01-22T15:37:10.433Z
@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 $F_{(A_L L)}$
caverac
CC BY-SA 4.0
2019-01-22T15:04:29.407Z
2019-01-22T15:04:29.407Z
caverac
<p>In that paper, they consider the following production function:</p>

<p><span class="math-container">$Y = F(A_KK,A_LL)$</span>.</p>

<p>Where: </p>

<blockquote>
  <p>K denotes capital, L is labor, and <span class="math-container">$A_K$</span> and <span class="math-container">$A_L$</span> denote capital-augmenting
  and labor-augmenting technology, respectively. We assume throughout
  that F is continuously differentiable, concave, and exhibits constant
  returns to scale. Let <span class="math-container">$F_K$</span> and <span class="math-container">$F_L$</span> 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:</p>
</blockquote>

<p><span class="math-container">$W = A_L F_L(A_KK,A_LL)$</span>.</p>

<p>Why? Shouldn't simply be:</p>

<p><span class="math-container">$W = F_L(A_KK,A_LL)$</span>.</p>

<p>Without the <span class="math-container">$A_L$</span> multiplying the derivative?</p>

<macroeconomics><labor-economics><theory><technology>
Factor-augmenting technologies (Acemoglu and Restrepo 2018c)
Tecon
# Factor-augmenting technologies (Acemoglu and Restrepo 2018c)

(Asked by: username_0 on 2019-01-22)

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?

**Tags:** macroeconomics, labor-economics, theory, technology

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## 1 answers

### Answer by username_1 on 2019-01-22. Score: 3 | Accepted answer

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.

> **Comments:**
> **username_0**: 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?  
> **username_1**: @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 $F_{(A_L L)}$  

---

