# why does $\frac{b\partial a}{a \partial b} = \frac{\partial \log a}{\partial \log b}$?

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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According to the linked-to document, you have $\partial$ (\partial), not $\delta$ (\delta). –  Arturo Magidin Feb 10 '11 at 21:22
Notice that the formula in your title is wrong... –  Mariano Suárez-Alvarez Feb 10 '11 at 21:24
@Arturo thanks. I had searched for the latex symbol under 'del' instead of 'partial'. –  David LeBauer Feb 10 '11 at 21:25

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 $$\frac{\partial\log\lambda}{\partial\log a_{ij}} = \frac{\quad\frac{1}{\lambda}\frac{\partial \lambda}{\partial t}\quad}{\frac{1}{a_{ij}}\frac{\partial a_{ij}}{\partial t}} = \frac{a_{ij}\partial \lambda}{\lambda\partial a_{ij}}.$$
Because $\partial \log x=\frac{\partial x}{x}$, whatever it is that $\partial$ may mean.
See the edit I just made; the source has $\partial$, not $\delta$. –  Arturo Magidin Feb 10 '11 at 21:23