<p>I am not sure why you expect method two to work, as generally
<span class="math-container">$$
\frac{\partial F(K,N)}{\partial N} \neq \frac{\partial F(K,N)}{\partial K}\frac{\partial K}{\partial N}.
$$</span>
These are <em>partial</em> derivatives. Unless information to the contrary exists
<span class="math-container">$$\frac{\partial K}{\partial N} = 0.$$</span></p>
<p>You can easily see that these formulas are not connected by calculating the area of a rectangle which sides have lengths of <span class="math-container">$a,b$</span>. Thus <span class="math-container">$A(a,b) = a\cdot b$</span> and
<span class="math-container">$$
\frac{\partial A(a,b)}{\partial b} = a \neq b \cdot 0 = \frac{\partial A(a,b)}{\partial a}\frac{\partial a}{\partial b}.
$$</span>
In case you insist on homogeneity of degree one you can take the square root of the area.</p>

CC BY-SA 4.0
2021-05-27T11:30:46.907Z
thank you for your reply. Isn't this can be thought of chain rule?
LJNG
CC BY-SA 4.0
2021-05-27T12:22:55.910Z
@LawrenceGuo Not if you are talking about partial derivatives.
Giskard
CC BY-SA 4.0
2021-05-27T11:16:37.763Z
2021-05-27T11:16:37.763Z
Giskard
<p>Given a homogeneous of degree one function
<span class="math-container">$$Y=F(K,N) \\
y=\frac{Y}{N}
=F(\frac{K}{N},1):=f(k)$$</span></p>
<p>I am looking for the partial derivative for <span class="math-container">$N$</span></p>
<p><strong>Method 1</strong></p>
<p><span class="math-container">$$F_N=\frac{\partial F(K,N)}{\partial N}=\frac{\partial Nf(k)}{\partial N}\\
F_N=f(k)+N\frac{\partial f(k)}{\partial N}\\
F_N=f(k)+N\frac{\partial f(k)}{\partial k}\frac{\partial k}{\partial N}, \quad k=KN^{-1}\\
F_N=f(k)+Nf'(k)(-1)KN^{-2}\\
F_N=f(k)+f'(k)(-1)KN^{-1}=f(k)-f'(k)k
 $$</span></p>
<p><strong>Method 2</strong>
<span class="math-container">$$F_N=\frac{\partial F(K,N)}{\partial K}\frac{\partial K}{\partial N}\\
F_N=\frac{N\partial f(k)}{N\partial k}\frac{\partial K}{\partial N}\\
F_N=f'(k)\frac{\partial Nk}{\partial N}=f'(k)k
$$</span></p>
<p>I have no idea where is the <strong>method 2</strong> gone wrong.  Any help would be appreciated!</p>
<p><strong>Re-edited Solved</strong></p>
<p><span class="math-container">$$\frac{dF(K,N)}{dN}=\frac{\partial F(K,N)}{\partial K}\frac{dK}{dN}+\frac{\partial F(K,N)}{\partial N}\frac{dN}{dN}\\
\frac{dNf(k)}{dN}=f'(k)k+\frac{\partial F(K,N)}{\partial N}\\
f(k)-f'(k)k=\frac{\partial F(K,N)}{\partial N}
$$</span></p>

<wages><marginal>
I don't know why these two method yield different solutions for marginal product of labour
LJNG
# I don't know why these two method yield different solutions for marginal product of labour

(Asked by: username_0 on 2021-05-27)

Given a homogeneous of degree one function
$$Y=F(K,N) \\
y=\frac{Y}{N}
=F(\frac{K}{N},1):=f(k)$$

I am looking for the partial derivative for $N$

**Method 1**

$$F\_N=\frac{\partial F(K,N)}{\partial N}=\frac{\partial Nf(k)}{\partial N}\\
F\_N=f(k)+N\frac{\partial f(k)}{\partial N}\\
F\_N=f(k)+N\frac{\partial f(k)}{\partial k}\frac{\partial k}{\partial N}, \quad k=KN^{-1}\\
F\_N=f(k)+Nf'(k)(-1)KN^{-2}\\
F\_N=f(k)+f'(k)(-1)KN^{-1}=f(k)-f'(k)k
$$

**Method 2**
$$F\_N=\frac{\partial F(K,N)}{\partial K}\frac{\partial K}{\partial N}\\
F\_N=\frac{N\partial f(k)}{N\partial k}\frac{\partial K}{\partial N}\\
F\_N=f'(k)\frac{\partial Nk}{\partial N}=f'(k)k
$$

I have no idea where is the **method 2** gone wrong. Any help would be appreciated!

**Re-edited Solved**

$$\frac{dF(K,N)}{dN}=\frac{\partial F(K,N)}{\partial K}\frac{dK}{dN}+\frac{\partial F(K,N)}{\partial N}\frac{dN}{dN}\\
\frac{dNf(k)}{dN}=f'(k)k+\frac{\partial F(K,N)}{\partial N}\\
f(k)-f'(k)k=\frac{\partial F(K,N)}{\partial N}
$$

**Tags:** wages, marginal

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## 1 answers

### Answer by username_1 on 2021-05-27. Score: 4 | Accepted answer

I am not sure why you expect method two to work, as generally
$$
\frac{\partial F(K,N)}{\partial N} \neq \frac{\partial F(K,N)}{\partial K}\frac{\partial K}{\partial N}.
$$
These are *partial* derivatives. Unless information to the contrary exists
$$\frac{\partial K}{\partial N} = 0.$$

You can easily see that these formulas are not connected by calculating the area of a rectangle which sides have lengths of $a,b$. Thus $A(a,b) = a\cdot b$ and
$$
\frac{\partial A(a,b)}{\partial b} = a \neq b \cdot 0 = \frac{\partial A(a,b)}{\partial a}\frac{\partial a}{\partial b}.
$$
In case you insist on homogeneity of degree one you can take the square root of the area.

> **Comments:**
> **username_0**: thank you for your reply. Isn't this can be thought of chain rule?  
> **username_1**: @LawrenceGuo Not if you are talking about partial derivatives.  

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