<p>Let's express everything in terms of <span class="math-container">$x_1$</span> First of all, the elasticity of substitution is determined keeping output fixed, so
<span class="math-container">$$
F(x_1, x_2) = C.
$$</span>
For a constant <span class="math-container">$C$</span>. This gives:
<span class="math-container">$$
F_1 + F_2 \frac{dx_2}{dx_1} = 0 \to \frac{dx_2}{dx_1} = -\frac{F_1}{F_2}.
$$</span></p>
<p>Next, consider the ratio:
<span class="math-container">$$
r = \frac{x_2}{x_1}.
$$</span>
Then:
<span class="math-container">$$
\begin{align*}
\frac{dr}{dx_1} &amp;= \frac{1}{x_1} \frac{dx_2}{dx_1} - \frac{x_2}{(x_1)^2},\\
&amp;= -\frac{1}{x_1}\frac{F_1}{F_2} - \frac{x_2}{(x_1)^2},\\
&amp;= -\frac{x_2 F_1}{x_1}\left(\frac{1}{x_2 F_2}+ \frac{1}{x_1F_1}\right)
\end{align*}
$$</span></p>
<p>Next,
<span class="math-container">$$
MRTS = - \frac{F_1}{F_2}.
$$</span>
So:
<span class="math-container">$$
\begin{align*}
\frac{d MRTS}{dx_1} &amp;= -\left(\frac{F_2\left(F_{11} + F_{12}\frac{dx_2}{dx_1}\right) - F_1\left(F_{21} + F_{22}\frac{dx_2}{dx_1}\right)}{(F_2)^2}\right),\\
&amp;=\frac{-F_2 F_{11}+F_{12} F_1 + F_1 F_{21} - F_1 F_{22}\frac{F_1}{F_2} }{(F_2)^2},\\
&amp;= -\frac{(F_1)^2}{(F_2)}\left(\frac{F_{11}}{(F_1)^2} -2 \frac{F_{12}}{F_1 F_2} + \frac{F_{22}}{(F_2)^2}\right)
\end{align*}
$$</span></p>
<p>So:
<span class="math-container">$$
\begin{align*}
\frac{dr}{dMRTS} &amp;= \left.\left(\frac{dr}{dx_1}\right)\middle/\left(\frac{dMRTS}{dx_1}\right)\right.\\
&amp;= -\frac{\frac{1}{x_2 F_2} + \frac{1}{x_1 F_1}}{\frac{F_{11}}{(F_1)^2} - 2 \frac{F_{12}}{F_1 F_2} + \frac{F_{22}}{(F_2)^2}}\left(-\frac{F_2}{F_1}\frac{x_2}{x_1} \right)
\end{align*} 
$$</span></p>
<p>This gives:
<span class="math-container">$$
\begin{align*}
\frac{d \ln(r)}{d \ln(MRTS)} &amp;= \frac{d r}{d MRTS} \frac{MRTS}{r},\\
&amp;= \frac{dr}{dMRTS} \left(-\frac{F_1}{F_2}\right) \frac{x_1}{x_2},\\
&amp;=-\frac{\frac{1}{x_2 F_2} + \frac{1}{x_1 F_1}}{\frac{F_{11}}{(F_1)^2} - 2 \frac{F_{12}}{F_1 F_2} + \frac{F_{22}}{(F_2)^2}}
\end{align*}
$$</span></p>

CC BY-SA 4.0
2024-11-13T13:45:51.537Z
2024-11-13T13:45:51.537Z
tdm
<p>By definition, ES takes the form
<span class="math-container">$$
\sigma_{12} = \frac{d \ln (x_2/x_1)}{d \ln {MRTS_{12}}}
$$</span>
Given some production function <span class="math-container">$F(x_1,x_2)$</span>, I see a form that
<span class="math-container">$$
\sigma_{12} = -\frac{\frac{1}{x_1 F_1} + \frac{1}{x_2 F_2}}{\frac{F_{11}}{(F_1)^2} - \frac{2 F_{12}}{F_1 F_2} + \frac{F_{22}}{(F_2)^2}}
$$</span>
where <span class="math-container">$F_1,F_2,F_{11},F_{12},F_{22}$</span> are the first and second order partial derivatives, respectively.</p>
<p>However, I don't know how to derive it. It would be great if some hints can be provided.</p>
<p>Thanks!</p>

CC BY-SA 4.0
2024-11-13T13:40:11.600Z
Dear Andy Xu pls. check if this post does not answer your question: https://economics.stackexchange.com/questions/43984/elasticity-of-substitution-of-crs-production-function. If it does we can close your question as a duplicate. If you think it does not pls. clarify why you think it does not.
Jesper Hybel
<producer-theory><elasticity-of-substitution>
General formula of elasticity of substitution
Andy Xu
# General formula of elasticity of substitution

(Asked by: username_0 on 2024-11-13)

By definition, ES takes the form
$$
\sigma\_{12} = \frac{d \ln (x\_2/x\_1)}{d \ln {MRTS\_{12}}}
$$
Given some production function $F(x\_1,x\_2)$, I see a form that
$$
\sigma\_{12} = -\frac{\frac{1}{x\_1 F\_1} + \frac{1}{x\_2 F\_2}}{\frac{F\_{11}}{(F\_1)^2} - \frac{2 F\_{12}}{F\_1 F\_2} + \frac{F\_{22}}{(F\_2)^2}}
$$
where $F\_1,F\_2,F\_{11},F\_{12},F\_{22}$ are the first and second order partial derivatives, respectively.

However, I don't know how to derive it. It would be great if some hints can be provided.

Thanks!

**Tags:** producer-theory, elasticity-of-substitution

> **Comments:**
> **username_1**: Dear username_0 pls. check if this post does not answer your question: https://economics.stackexchange.com/questions/43984/elasticity-of-substitution-of-crs-production-function. If it does we can close your question as a duplicate. If you think it does not pls. clarify why you think it does not.  

---

## 1 answers

### Answer by username_2 on 2024-11-13. Score: 1 | Accepted answer

Let's express everything in terms of $x\_1$ First of all, the elasticity of substitution is determined keeping output fixed, so
$$
F(x\_1, x\_2) = C.
$$
For a constant $C$. This gives:
$$
F\_1 + F\_2 \frac{dx\_2}{dx\_1} = 0 \to \frac{dx\_2}{dx\_1} = -\frac{F\_1}{F\_2}.
$$

Next, consider the ratio:
$$
r = \frac{x\_2}{x\_1}.
$$
Then:
$$
\begin{align\*}
\frac{dr}{dx\_1} &= \frac{1}{x\_1} \frac{dx\_2}{dx\_1} - \frac{x\_2}{(x\_1)^2},\\
&= -\frac{1}{x\_1}\frac{F\_1}{F\_2} - \frac{x\_2}{(x\_1)^2},\\
&= -\frac{x\_2 F\_1}{x\_1}\left(\frac{1}{x\_2 F\_2}+ \frac{1}{x\_1F\_1}\right)
\end{align\*}
$$

Next,
$$
MRTS = - \frac{F\_1}{F\_2}.
$$
So:
$$
\begin{align\*}
\frac{d MRTS}{dx\_1} &= -\left(\frac{F\_2\left(F\_{11} + F\_{12}\frac{dx\_2}{dx\_1}\right) - F\_1\left(F\_{21} + F\_{22}\frac{dx\_2}{dx\_1}\right)}{(F\_2)^2}\right),\\
&=\frac{-F\_2 F\_{11}+F\_{12} F\_1 + F\_1 F\_{21} - F\_1 F\_{22}\frac{F\_1}{F\_2} }{(F\_2)^2},\\
&= -\frac{(F\_1)^2}{(F\_2)}\left(\frac{F\_{11}}{(F\_1)^2} -2 \frac{F\_{12}}{F\_1 F\_2} + \frac{F\_{22}}{(F\_2)^2}\right)
\end{align\*}
$$

So:
$$
\begin{align\*}
\frac{dr}{dMRTS} &= \left.\left(\frac{dr}{dx\_1}\right)\middle/\left(\frac{dMRTS}{dx\_1}\right)\right.\\
&= -\frac{\frac{1}{x\_2 F\_2} + \frac{1}{x\_1 F\_1}}{\frac{F\_{11}}{(F\_1)^2} - 2 \frac{F\_{12}}{F\_1 F\_2} + \frac{F\_{22}}{(F\_2)^2}}\left(-\frac{F\_2}{F\_1}\frac{x\_2}{x\_1} \right)
\end{align\*}
$$

This gives:
$$
\begin{align\*}
\frac{d \ln(r)}{d \ln(MRTS)} &= \frac{d r}{d MRTS} \frac{MRTS}{r},\\
&= \frac{dr}{dMRTS} \left(-\frac{F\_1}{F\_2}\right) \frac{x\_1}{x\_2},\\
&=-\frac{\frac{1}{x\_2 F\_2} + \frac{1}{x\_1 F\_1}}{\frac{F\_{11}}{(F\_1)^2} - 2 \frac{F\_{12}}{F\_1 F\_2} + \frac{F\_{22}}{(F\_2)^2}}
\end{align\*}
$$


---

