<p>Okay, so the key is that there are nuances in the notation, they are calculating a <em>full</em> derivative, not a <em>partial</em> derivative. They write between 9.19 and 9.20 that</p>
<blockquote>
<p>Because <span class="math-container">$f_l$</span> and <span class="math-container">$f_k$</span> are functions of both <span class="math-container">$k$</span> and <span class="math-container">$l$</span>, we must be careful in taking the derivative of this expression</p>
</blockquote>
<p>You write</p>
<blockquote>
<p><span class="math-container">$\frac{d|MRTS|}{dl}=\frac{-K}{L^2}$</span></p>
</blockquote>
<p>but actually if you are doing movement on an isoquant then <span class="math-container">$k$</span> is a function of <span class="math-container">$l$</span>, and hence</p>
<p><span class="math-container">$$
\frac{\partial|MRTS(k,l)|}{\partial l} = \frac{k}{l^2} 
$$</span>
while
<span class="math-container">$$
\frac{d|MRTS(k,l)|}{dl} = 
\frac{\partial|MRTS(k,l)|}{\partial l} + \frac{\partial|MRTS(k,l)|}{\partial k}
\frac{\text{d}k}{\text{d}l}.
$$</span></p>

CC BY-SA 4.0
2023-12-31T18:08:43.943Z
Thank you, this makes sense. As a corollary, if we want to show convexity of isoquants, we should show total derivative of MRTS (absolute value) to be negative right?
AcadEconInd
CC BY-SA 4.0
2024-01-01T12:32:10.273Z
Yes. If you are looking at the total derivative, you are measuring change in MRTS while moving on the isoquant. If you are looking at the partial derivative, you are measuring change while moving in the positive $l$ direction (with no movement along the $k$ axis) from a point of the isoquant, so essentially you would be moving off the isoquant and would not really get any info about its slope/convexity.
Giskard
CC BY-SA 4.0
2024-01-01T14:40:19.707Z
Thank you. On a side note, I love your profile description: "leaving economics for something 'tangible'" resonates with me!
AcadEconInd
CC BY-SA 4.0
2024-01-01T14:48:39.233Z
@AcadEconInd Thank you & good luck!
Giskard
CC BY-SA 4.0
2023-12-31T11:17:27.837Z
2024-01-01T14:54:08.087Z
2024-01-01T14:54:08.087Z
Giskard
<p>Consider a production function <span class="math-container">$f(L,K)=\sqrt{KL}$</span>.</p>
<p>The |MRTS|=<span class="math-container">$K/L$</span>, and <span class="math-container">$\frac{d|MRTS|}{dl}=\frac{-K}{L^2}$</span></p>
<p>However, if I use the expression given in Nicholson and Snyder (Microeconomic Theory, Chapter 9 [Production Functions], equation 9.21):</p>
<p><a href="https://i.sstatic.net/GZpor.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/GZpor.png" alt="enter image description here" /></a></p>
<p>and plug in values of the partial derivatives, then I get <span class="math-container">$-2K/L^2$</span>.</p>
<p>I have checked the calculation twice so I don't think that's the issue. I have also checked with other functional forms and I get an extra constant in most cases. Am I doing something wrong?</p>

CC BY-SA 4.0
2023-12-31T10:57:04.067Z
Hi! Can you please refine your Nicholson and Snyder reference, so that it includes a subchapter name or something similarly identifiable.
Giskard
CC BY-SA 4.0
2023-12-31T11:02:23.550Z
I did the calculations for $f(K,L) = KL$ because the derivations are easier to do and I also get 
$$
\frac{f_k^2 \cdot 0 - 2 \cdot L \cdot K \cdot 1 + f_l^2 \cdot 0}{L^3} = -2\frac{K}{L^2}
\neq
-\frac{K}{L^2}
$$
Giskard
CC BY-SA 4.0
2023-12-31T11:08:53.577Z
@Giskard Updated the reference, it is equation 9.21 in the 10th edition
AcadEconInd
<production-function><producer-theory>
Slope of isoquants
AcadEconInd
# Slope of isoquants

(Asked by: username_0 on 2023-12-31)

Consider a production function $f(L,K)=\sqrt{KL}$.

The |MRTS|=$K/L$, and $\frac{d|MRTS|}{dl}=\frac{-K}{L^2}$

However, if I use the expression given in Nicholson and Snyder (Microeconomic Theory, Chapter 9 [Production Functions], equation 9.21):

[![enter image description here](https://i.sstatic.net/GZpor.png)](https://i.sstatic.net/GZpor.png)

and plug in values of the partial derivatives, then I get $-2K/L^2$.

I have checked the calculation twice so I don't think that's the issue. I have also checked with other functional forms and I get an extra constant in most cases. Am I doing something wrong?

**Tags:** production-function, producer-theory
> **Comments:**
> **username_1**: Hi! Can you please refine your Nicholson and Snyder reference, so that it includes a subchapter name or something similarly identifiable.  
> **username_1**: I did the calculations for $f(K,L) = KL$ because the derivations are easier to do and I also get  $$ \frac{f_k^2 \cdot 0 - 2 \cdot L \cdot K \cdot 1 + f_l^2 \cdot 0}{L^3} = -2\frac{K}{L^2} \neq -\frac{K}{L^2} $$  
> **username_0**: @Giskard Updated the reference, it is equation 9.21 in the 10th edition  

---

## 1 answers

### Answer by username_1 on 2023-12-31. Score: 0 | Accepted answer

Okay, so the key is that there are nuances in the notation, they are calculating a *full* derivative, not a *partial* derivative. They write between 9.19 and 9.20 that

> Because $f\_l$ and $f\_k$ are functions of both $k$ and $l$, we must be careful in taking the derivative of this expression

You write

> $\frac{d|MRTS|}{dl}=\frac{-K}{L^2}$

but actually if you are doing movement on an isoquant then $k$ is a function of $l$, and hence

$$
\frac{\partial|MRTS(k,l)|}{\partial l} = \frac{k}{l^2}
$$
while
$$
\frac{d|MRTS(k,l)|}{dl} =
\frac{\partial|MRTS(k,l)|}{\partial l} + \frac{\partial|MRTS(k,l)|}{\partial k}
\frac{\text{d}k}{\text{d}l}.
$$

> **Comments:**
> **username_0**: Thank you, this makes sense. As a corollary, if we want to show convexity of isoquants, we should show total derivative of MRTS (absolute value) to be negative right?  
> **username_1**: Yes. If you are looking at the total derivative, you are measuring change in MRTS while moving on the isoquant. If you are looking at the partial derivative, you are measuring change while moving in the positive $l$ direction (with no movement along the $k$ axis) from a point of the isoquant, so essentially you would be moving off the isoquant and would not really get any info about its slope/convexity.  
> **username_0**: Thank you. On a side note, I love your profile description: "leaving economics for something 'tangible'" resonates with me!  
> **username_1**: @AcadEconInd Thank you & good luck!  

---

