# Second order partial derivative and cross second-order partial derivative

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!

• 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! – Alessandro Nov 28 '18 at 14:13
• 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 :(( – Alessandro Nov 28 '18 at 18:56