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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?

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    $\begingroup$ I'm voting to close this question as off-topic because this is a question of mathematical notation. $\endgroup$ – Giskard Dec 8 '16 at 10:10
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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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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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