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I'm following a paper (Full text available here!) where at some point (pag.17 and 20) the author get the following derivative:

$$\frac{\partial V}{\partial L}=Y-X\frac{dY}{dX}=\alpha X^{-\frac{c}{b}}Y^{\frac{1}{b}}$$

where: $Y=\frac{V}{L}$ and $X=\frac{K}{L}$

Then, starting from this he calculates the partial derivative with respect to L and the cross second-order partial derivative (the partial derivative with respect to K), whose results are shown below:

$$\frac{\partial^2 V}{L^{2}}=-\frac{\alpha }{bL} X^{-\frac{c}{b}}Y^{\frac{1}{b}-1}\left ( X\frac{dY}{dX}-cY \right )$$

$$\frac{\partial^2 V}{dKdL}=\frac{\alpha }{bL} X^{-\frac{c}{b}-1}Y^{\frac{1}{b}-1}\left ( X\frac{dY}{dX}-cY \right )$$

I was stuck trying to derive these latter derivatives. Is there anyone who can help me with this? Thank you so much!

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Is there anyone who can help me with this?

Here it is. Equations 1-3, and 5-6 are obtained in preparation for the 2nd derivatives of V with respect to L and K.

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Let me know if you have any questions.

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  • $\begingroup$ First of all thank you very much for your extremely clear answer! :) Then, if you have a minute I would like to ask your help with another passage of the same paper that I did not fully understand. Further on (page 20), the author calculates the elasticity of substitution (σ) which is equal to (3.24): $$\sigma =\frac{b}{1-\frac{c}{X}\frac{f}{f'}}$$ Up to here everything is clear. Then the author rewrite (3.24) in the following way, obtaining (3.25): $$\sigma =\frac{b}{1-c\left ( 1+\frac{R}{X} \right )}$$ It is this last result that I cannot get. Thank you so much, I owe you a beer! $\endgroup$ – Alessandro Nov 28 '18 at 14:13
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    $\begingroup$ I wish I could help on that one. Yesterday I just computed the derivatives without looking at the paper at issue, but I'll need to delve into it to get a sense of what R means. If I get the chance to do so, I will be happy to address (3.25). $\endgroup$ – Iñaki Viggers Nov 28 '18 at 18:43
  • $\begingroup$ Sorry, I forgot to write that R is the marginal rate of substitution of L for K, so basically just: $$R=-\frac{dK}{dL}=\frac{MPL}{MPK}$$ I tried to make this calculation, which according to the formulas of the previous comment if divided by X and then adding 1 should be exactly equal to $$\frac{f}{Xf'}$$ i.e. equal to $$\frac{Y}{X\frac{dY}{dX}}$$ But I can't get this latter equality :(( $\endgroup$ – Alessandro Nov 28 '18 at 18:56

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