<p>In the Lagrangian function <span class="math-container">$\mathcal{L}(w,r,q,l,k,\lambda)$</span>, the arguments <span class="math-container">$l$</span> and <span class="math-container">$k$</span> are independent variables, so they do not depend on <span class="math-container">$w$</span>.</p>

CC BY-SA 4.0
2022-10-11T09:48:28.257Z
2022-10-11T09:48:28.257Z
VARulle
<p>In the proof of Shepherd's Lemma (production and supply), we have the following step:</p>
<p><span class="math-container">$$\frac{\partial \mathcal{L}}{\partial w} = \frac{\partial \ [wl + rk + \lambda(q - f(l,k))]}{\partial w} = l^c(w,r,q)$$</span></p>
<p>The other variables <span class="math-container">$l, k$</span> are also dependent on <span class="math-container">$w$</span>, why do we treat them as constants here instead of calculating it as <span class="math-container">$\frac{\partial L}{\partial w} = w \frac{\partial l^c}{\partial w} + l^c + r \frac{\partial k^c}{\partial w}$</span>?</p>

CC BY-SA 4.0
2022-10-12T08:22:14.343Z
It is Shephard's lemma, named in honor of Ronald Shephard: https://en.wikipedia.org/wiki/Ronald_Shephard
Bertrand
<microeconomics>
Partial derivatives in Shepherd's lemma
DrLupo
# Partial derivatives in Shepherd's lemma

(Asked by: username_0 on 2022-10-11)

In the proof of Shepherd's Lemma (production and supply), we have the following step:

$$\frac{\partial \mathcal{L}}{\partial w} = \frac{\partial \ [wl + rk + \lambda(q - f(l,k))]}{\partial w} = l^c(w,r,q)$$

The other variables $l, k$ are also dependent on $w$, why do we treat them as constants here instead of calculating it as $\frac{\partial L}{\partial w} = w \frac{\partial l^c}{\partial w} + l^c + r \frac{\partial k^c}{\partial w}$?

**Tags:** microeconomics

> **Comments:**
> **username_1**: It is Shephard's lemma, named in honor of Ronald Shephard: https://en.wikipedia.org/wiki/Ronald_Shephard  

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## 1 answers

### Answer by username_2 on 2022-10-11. Score: 2 | Accepted answer

In the Lagrangian function $\mathcal{L}(w,r,q,l,k,\lambda)$, the arguments $l$ and $k$ are independent variables, so they do not depend on $w$.


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