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I am doing some background research on sensitivity and elasticity analysis, and I came across the following definitions of elasticity:

$e_{ij}=\frac{a_{ij}\partial \lambda}{\lambda \partial a_{ij}}$

$e_{ij} = \frac{\partial \log \lambda}{\partial \log a_{ij}}$

Why are these equivalent?

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    @Arturo thanks. I had searched for the latex symbol under 'del' instead of 'partial'.2011-02-10

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From the Chain Rule: $\frac{\partial\log\lambda}{\partial \log a_{ij}}\;\frac{\partial\log a_{ij}}{\partial t} = \frac{\partial\log\lambda}{\partial t} = \frac{1}{\lambda}\frac{\partial\lambda}{\partial t}.$ Since $\frac{\partial\log a_{ij}}{\partial t} = \frac{1}{a_{ij}}\frac{\partial a_{ij}}{\partial t}$ "solving" for $\frac{\partial\log\lambda}{\partial \log a_{ij}}$ gives $\frac{\partial\log\lambda}{\partial\log a_{ij}} = \frac{\quad\frac{1}{\lambda}\frac{\partial \lambda}{\partial t}\quad}{\frac{1}{a_{ij}}\frac{\partial a_{ij}}{\partial t}} = \frac{a_{ij}\partial \lambda}{\lambda\partial a_{ij}}.$

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Because $\partial \log x=\frac{\partial x}{x}$, whatever it is that $\partial$ may mean.

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    @David: the rationale i$s$ that it is true... You can justify it using the chain rule, for example. If that is what you want to ask, then ask that :)2011-02-10