<p>You cannot find $\partial \ln\frac{q(\omega)}{q(\omega')}$, it has no well defined meaning here. However, by the first relation you mention, it follows that $$\ln(q(\omega)/q(\omega'))=-\sigma\ln(p(\omega)/p(\omega')).$$ Taking the partial derivative on both sides <em>with respect to</em> $\ln(p(\omega)/p(\omega'))$ gives $$\frac{\partial \ln\frac{q(\omega)}{q(\omega ')}}{\partial \ln\frac{p(\omega)}{p(\omega ')}}=-\sigma.$$</p>

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2016-12-07T15:42:48.437Z
2016-12-07T15:42:48.437Z
Elias
<p>You might be overthinking this.</p>

<p>To make things clearer, try the following substitution. Let $$y=\ln\frac{q(\omega)}{q(\omega^\prime)}$$ and  $$x=\ln\frac{p(\omega)}{p(\omega^\prime)}$$</p>

<p>The first equation you show then becomes $y=-\sigma x $, and all you're looking for is $\frac{\partial y}{\partial x }$.</p>

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2016-12-07T15:45:04.653Z
2016-12-07T15:45:04.653Z
Theoretical Economist
<p>How do I show that elasticity of substitution is equal to σ from a CES utility function. I have derived the following:</p>

<p>$\frac{q(\omega)}{q(\omega ')}=\left(\frac{p(\omega)}{p(\omega ')}\right)^{-\sigma}$</p>

<p>And then I can use the following to show that the elasticity of substitution is equal to σ:</p>

<p>$\frac{\partial \ln\frac{q(\omega)}{q(\omega ')}}{\partial \ln\frac{p(\omega)}{p(\omega ')}}=-\sigma $</p>

<p>But how do I find: $\partial \ln\frac{q(\omega)}{q(\omega ')}$? What should I differentiate with respect to?</p>

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2016-12-08T10:10:53.643Z
I'm voting to close this question as off-topic because this is a question of mathematical notation.
Giskard
<mathematical-economics>
Show that the elasticity of substitution is σ
Simon
# Show that the elasticity of substitution is σ

(Asked by: username_0 on 2016-12-07)

How do I show that elasticity of substitution is equal to σ from a CES utility function. I have derived the following:

$\frac{q(\omega)}{q(\omega ')}=\left(\frac{p(\omega)}{p(\omega ')}\right)^{-\sigma}$

And then I can use the following to show that the elasticity of substitution is equal to σ:

$\frac{\partial \ln\frac{q(\omega)}{q(\omega ')}}{\partial \ln\frac{p(\omega)}{p(\omega ')}}=-\sigma $

But how do I find: $\partial \ln\frac{q(\omega)}{q(\omega ')}$? What should I differentiate with respect to?

**Tags:** mathematical-economics
> **Comments:**
> **username_1**: I'm voting to close this question as off-topic because this is a question of mathematical notation.  

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## 2 answers

### Answer by username_2 on 2016-12-07. Score: 0 | Accepted answer

You cannot find $\partial \ln\frac{q(\omega)}{q(\omega')}$, it has no well defined meaning here. However, by the first relation you mention, it follows that $$\ln(q(\omega)/q(\omega'))=-\sigma\ln(p(\omega)/p(\omega')).$$ Taking the partial derivative on both sides *with respect to* $\ln(p(\omega)/p(\omega'))$ gives $$\frac{\partial \ln\frac{q(\omega)}{q(\omega ')}}{\partial \ln\frac{p(\omega)}{p(\omega ')}}=-\sigma.$$


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### Answer by username_3 on 2016-12-07. Score: 0

You might be overthinking this.

To make things clearer, try the following substitution. Let $$y=\ln\frac{q(\omega)}{q(\omega^\prime)}$$ and $$x=\ln\frac{p(\omega)}{p(\omega^\prime)}$$

The first equation you show then becomes $y=-\sigma x $, and all you're looking for is $\frac{\partial y}{\partial x }$.


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