# Show that the elasticity of substitution is σ [closed]

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?

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

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