Marginal costs MC is defined as $MC=\frac{dC}{dq}$. Taking into account that $C=wL+rK$,


Recall that marginal product of labor $MP_{L}=\frac{\partial q}{\partial L}$ and marginal product of capital $MP_{K}=\frac{\partial q}{\partial K}$.

Question: is the following correct

$$\frac{dL}{dq}=1/\frac{\partial q}{\partial L},\;\frac{dK}{dq}=1/\frac{\partial q}{\partial K}$$

which implies


If no, then no need to read further.

If yes, then, consider profit maximization of a firm.






The result is wrong for sure. I wonder, at what step of derivation I made a mistake?

  • $\begingroup$ What's your basis for thinking that dL/dq might equal 1/(dq/DL)? dL/dq is a derivative, not a fraction $\endgroup$
    – 410 gone
    Jun 15, 2015 at 9:14
  • $\begingroup$ Perloff in his textbook on micro, uses what you just wrote when discussing short run. For unknown reason I voluntarily wrote in my notes more general version for long run, i.e. using partial derivatives. But after reading the chapter on demand on factors of production, I updated my notes and realized that there must be a mistake somewhere. $\endgroup$
    – ji borrob
    Jun 15, 2015 at 9:25
  • 3
    $\begingroup$ @EnergyNumbers, btw, the use of what you wrote is valid thanks to inverse function theorem. $\endgroup$
    – ji borrob
    Jun 15, 2015 at 9:27
  • $\begingroup$ Can you show the conditions for that, hold? (in particular, can you show that there are no zeros where there shouldn't be) $\endgroup$
    – 410 gone
    Jun 15, 2015 at 11:11
  • $\begingroup$ (and be careful with partial derivatives - I think you want to take the inverse of the matrix, rather than of just an individual element in the matrix, don't you - sorry if this is a red-herring, it's been a while since I worked through this) $\endgroup$
    – 410 gone
    Jun 15, 2015 at 11:19

2 Answers 2


Question: is the following correct ?

$$\frac{dL}{dq}=1/\frac{\partial q}{\partial L},\;\frac{dK}{dq}=1/\frac{\partial q}{\partial K}$$

In general, no. Since $q= f(L,K)$ is a multivariable, single-valued function, then by the implicit function theorem applied on the implied equation $H = f(L,K)-q=0$, what we can say is only that

$$\frac {\partial L}{\partial q} = -\frac {\partial H/\partial q}{\partial H /\partial L} = -\frac {-1}{MP_L} = \frac {1}{\partial q /\partial L}$$

Essentially, here we treat $q$ as a univariate function (by taking partial derivatives, we keep all other variables fixed).

But total derivatives are another matter.
For $q = f(L,K)$, we have the implicit relation $L = L(q,K)$ and so by taking the total differential and divide throughout by $dq$ we get

$$\frac {d L(q,K)}{d q} = \frac {\partial L(q,K)}{\partial q} + \frac {\partial L(q,K)}{\partial K}\frac {dK}{dq}$$

But $\frac {\partial L}{\partial K} \neq 0$, because we implicitly keep $q$ constant in order to calculate this partial derivative, and so by changing $K$ we have also to change $L$. The other possibility would be that $dK/dq$ is zero. If we do not have such a production function (or we are not in a range where such a thing holds), we obtain

$$\frac {dL}{dq} \neq \frac {\partial L}{\partial q} = \frac {1}{\partial q/\partial L}$$


It is valid but only in the short run with the assumption that the capital is fixed.

By assuming that output depends on labor and capital you can write


Now taking the total derivative $$dq=\frac{\partial q}{\partial L}dL+\frac{\partial q}{\partial K}dK$$

In the short run capital is fixed such as $dK=0$ $$\longrightarrow\quad dq=\frac{\partial q}{\partial L}dL\longrightarrow \frac{dL}{dq}=1/\frac{\partial q}{\partial L}$$


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