Given a homogeneous of degree one function $$Y=F(K,N) \\ y=\frac{Y}{N} =F(\frac{K}{N},1):=f(k)$$

I am looking for the partial derivative for $N$

Method 1

$$F_N=\frac{\partial F(K,N)}{\partial N}=\frac{\partial Nf(k)}{\partial N}\\ F_N=f(k)+N\frac{\partial f(k)}{\partial N}\\ F_N=f(k)+N\frac{\partial f(k)}{\partial k}\frac{\partial k}{\partial N}, \quad k=KN^{-1}\\ F_N=f(k)+Nf'(k)(-1)KN^{-2}\\ F_N=f(k)+f'(k)(-1)KN^{-1}=f(k)-f'(k)k $$

Method 2 $$F_N=\frac{\partial F(K,N)}{\partial K}\frac{\partial K}{\partial N}\\ F_N=\frac{N\partial f(k)}{N\partial k}\frac{\partial K}{\partial N}\\ F_N=f'(k)\frac{\partial Nk}{\partial N}=f'(k)k $$

I have no idea where is the method 2 gone wrong. Any help would be appreciated!

Re-edited Solved

$$\frac{dF(K,N)}{dN}=\frac{\partial F(K,N)}{\partial K}\frac{dK}{dN}+\frac{\partial F(K,N)}{\partial N}\frac{dN}{dN}\\ \frac{dNf(k)}{dN}=f'(k)k+\frac{\partial F(K,N)}{\partial N}\\ f(k)-f'(k)k=\frac{\partial F(K,N)}{\partial N} $$


1 Answer 1


I am not sure why you expect method two to work, as generally $$ \frac{\partial F(K,N)}{\partial N} \neq \frac{\partial F(K,N)}{\partial K}\frac{\partial K}{\partial N}. $$ These are partial derivatives. Unless information to the contrary exists $$\frac{\partial K}{\partial N} = 0.$$

You can easily see that these formulas are not connected by calculating the area of a rectangle which sides have lengths of $a,b$. Thus $A(a,b) = a\cdot b$ and $$ \frac{\partial A(a,b)}{\partial b} = a \neq b \cdot 0 = \frac{\partial A(a,b)}{\partial a}\frac{\partial a}{\partial b}. $$ In case you insist on homogeneity of degree one you can take the square root of the area.

  • $\begingroup$ thank you for your reply. Isn't this can be thought of chain rule? $\endgroup$
    – LJNG
    Commented May 27, 2021 at 11:30
  • $\begingroup$ @LawrenceGuo Not if you are talking about partial derivatives. $\endgroup$
    – Giskard
    Commented May 27, 2021 at 12:22

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