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!


1 Answer 1


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.

enter image description here

Let me know if you have any questions.

  • $\begingroup$ 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, 2018 at 14:13
  • 1
    $\begingroup$ 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). $\endgroup$ Nov 28, 2018 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, 2018 at 18:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.