How to apply chain rule to $\log$ function

Question

I'm doing my economics reading and I find this equation for elasticity of substitution:
$$\sigma \equiv \frac{F_K \cdot F_L}{F \cdot F_{KL}}$$ and then $$\frac{d(\log(F_K/F_L))}{d(\log(K/L))}=-\frac{1}{\sigma}.$$
How do I do the later derivative? By chain rule or anything else? Many thanks.

Answer 1

Hint:

$$\frac{df(x)}{dg(x)}=\frac{\frac{df(x)}{dx}}{\frac{dg(x)}{dx}}$$

Answer 2

Take as a reference Chapter 5 in the book "Economists' Mathematical Manual" by Knut Sydsaeter, Arne Strøm, Peter Berck (where unfortunately no proof is given).

As far as I understand $F$ is some production function with inputs $K$ and $L$ (capital and labor?). If $F(K,L)=c$ then $F_KdK+F_L dL=0$ which implies that $dK/dL=-F_L/F_K$. Therefore, with $F_{LK}=F_{KL}$, we get $$\frac{d(\log(F_K/F_L))}{d(\log(K/L))}= \frac{\frac{F_{KK}dK+F_{KL}dL}{F_K}-\frac{F_{LK}dK+F_{LL}dL}{F_L}}{\frac{dK}{K}-\frac{dL}{L}}=\frac{\frac{F_{KK}}{F_K^2}-\frac{2F_{KL}}{F_KF_L}+\frac{F_{LL}}{F_L^2}}{\frac{1}{KF_K}+\frac{1}{LF_L}}\tag{1}.$$ Now if $F(K,L)$ is homogeneous of degree 1 then $KF_K+LF_L=F$ and by taking the derivatives with respect to $K$ and $L$ we find $$F_K+KF_{KK}+LF_{KL}=F_K\quad\text{and}\quad KF_{KL}+F_L+LF_{LL}=F_L.$$ Hence, $KF_{KK}=-LF_{KL}$, $LF_{LL}=-KF_{KL}$, and the right-hand side of (1) simplifies to $$-\frac{FF_{KL}}{F_KF_L}=-\frac{1}{\sigma}.$$

Answer 3

I assume $F(K,L)$ is a production function of capital and labor units.

The elasticity of substitution is defined as: $$\sigma=\frac{\%\Delta \left(\frac{K}{L}\right)}{\%\Delta \left(\frac{F_L}{F_K}\right)} \quad \quad (1)$$ where $\%\Delta$ is the percentage change, $F_L, F_K$ are the marginal product of labor and capital, respectively.

How is $\%\Delta$ found? For example, say the capital (or price, population, etc) increased from $20$ to $28$. It implies that it increased by $40\%$, because: $$\%\Delta K=\frac{\Delta K}{K}=\frac{K_2-K_1}{K_1}=\frac{28-20}{20}=0.4=40\%.$$ So, the formula of the elasticity of substitution $(1)$ can be written as: $$\sigma =\frac{\%\Delta \left(\frac{K}{L}\right)}{\%\Delta \left(\frac{F_L}{F_K}\right)}= \frac{\frac{\Delta \left(\frac{K}{L}\right)}{\frac{K}{L}}}{\frac{\Delta \left(\frac{F_L}{F_K}\right)}{\frac{F_L}{F_K}}}= \frac{\frac{d \left(\frac{K}{L}\right)}{\frac{K}{L}}}{\frac{d \left(\frac{F_L}{F_K}\right)}{\frac{F_L}{F_K}}} \quad \quad (2)$$ where $\Delta=d$ is the change or difference (or differential).

The differential of a function is: $$df(x)=f'(x)\cdot dx$$ and the differential of a composite logarithmic function $f(x)=\ln (g(x))$ is: $$d(\ln (g(x)))=(\ln (g(x))'\cdot dx=\frac1{g(x)}\cdot g'(x)\cdot dx=\frac1{g(x)}\cdot d(g(x))=\frac{d(g(x))}{g(x)} \quad \quad (3)$$

Now with the help of $(3)$ the formula $(2)$ can be written as: $$\sigma =\frac{\%\Delta \left(\frac{K}{L}\right)}{\%\Delta \left(\frac{F_L}{F_K}\right)}= \frac{\frac{\Delta \left(\frac{K}{L}\right)}{\frac{K}{L}}}{\frac{\Delta \left(\frac{F_L}{F_K}\right)}{\frac{F_L}{F_K}}}= \frac{\frac{d \left(\frac{K}{L}\right)}{\frac{K}{L}}}{\frac{d \left(\frac{F_L}{F_K}\right)}{\frac{F_L}{F_K}}}= \frac{d \left(\ln \left(\frac{K}{L}\right)\right)}{d \left(\ln \left(\frac{F_L}{F_K}\right)\right)} \quad \quad (4)$$ Notice that your second formula has a negative sign. Why? Because, there are two similar concepts: MRTS (marginal rate of technical substitution) and MRLS (marginal rate of labor substitution): $$MRTS=\frac{F_L}{F_K}\\ MRLS=\frac{F_K}{F_L}.$$ If in $(4)$ we flip $\frac{F_L}{F_K}$, we get your formula: $$\sigma=\cdots=\frac{d \left(\ln \left(\frac{K}{L}\right)\right)}{d \left(\ln \left(\frac{F_L}{F_K}\right)\right)}=\frac{d \left(\ln \left(\frac{K}{L}\right)\right)}{d \left(\ln \left(\frac{F_K}{F_L}\right)^{-1}\right)}= \frac{d \left(\ln \left(\frac{K}{L}\right)\right)}{d \left(-\ln \left(\frac{F_K}{F_L}\right)\right)}= -\frac{d \left(\ln \left(\frac{K}{L}\right)\right)}{d \left(\ln \left(\frac{F_K}{F_L}\right)\right)}.$$ Now your first formula is a special case of the second formula, namely, when the production function is homogenous of degree $1$, it can be used Euler's homogenous function theorem: $$KF_K+LF_L=F.$$

Maybe I will show its detailed derivation some other time. I will let you digest this. Good luck!