0

$$\dfrac{\partial\ln E(p_x,p_y,V)}{\partial\ln p_x}=\dfrac1{E(p_x,p_y,V)}\cdot\frac{\partial E(p_x,p_y,V)}{\partial p_x}\cdot\dfrac{\partial p_x}{\partial\ln p_x}$$

How did we get this differentiation?I can't handle $\partial\ln p_x$ in denominator?

Silent
  • 6,520

2 Answers2

1

Hint

$$\frac{\partial \log E(p_x,p_y,V)}{\partial \log p_x}\cdot\frac{\partial \log p_x}{\partial p_x}=\frac{\partial \log E(p_x,p_y,V)}{\partial p_x}$$

Tim Ratigan
  • 7,247
  • Thank you so much,Sir. I wanted to comment this yesterday ASAP, but the site showed that your answer has been deleted! Then I slept. – Silent Dec 16 '13 at 03:27
1

Use the chain rule $$ \frac{\partial y}{\partial x}=\frac{\partial y}{\partial z}\frac{\partial z}{\partial x} $$ and the differentiation of inverse $$ \frac{\partial z}{\partial x}= \frac{1}{\frac{\partial x}{\partial z}}. $$ so that for $y=\ln E, x=\ln p_x, z=p_x$ $$ \dfrac{\partial\ln E}{\partial\ln p_x}=\frac{\partial \ln E}{\partial p_x}\cdot\dfrac{\partial p_x}{\partial\ln p_x}=\dfrac1{E}\cdot\frac{\partial E}{\partial p_x}\cdot\dfrac{\partial p_x}{\partial\ln p_x}=\dfrac1{E}\cdot\frac{\partial E}{\partial p_x}\cdot p_x $$ This is, by definition, the elasticity of $E$ with respect to $p_x$.

alexjo
  • 14,976
  • Thank you soooo much Sir, I expected mathematical answer only, but got the economic intuitions too!! – Silent Dec 16 '13 at 03:30