<blockquote>
  <p>Question: is the following correct ?</p>
  
  <p>$$\frac{dL}{dq}=1/\frac{\partial q}{\partial L},\;\frac{dK}{dq}=1/\frac{\partial q}{\partial K}$$</p>
</blockquote>

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

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

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

<p>But total derivatives are another matter.<br>
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</p>

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

<p>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 </p>

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

CC BY-SA 3.0
2015-06-15T20:31:13.120Z
2015-06-15T20:31:13.120Z
Alecos Papadopoulos
<p>It is valid but only in the short run with the assumption that the capital is fixed.</p>

<p>By assuming that output depends on labor and capital you can write</p>

<p>$$q=q(L,K)$$</p>

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

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

CC BY-SA 3.0
2015-07-08T12:44:50.497Z
2015-07-08T12:44:50.497Z
AnilB
<p>Marginal costs MC is defined as $MC=\frac{dC}{dq}$. Taking into account that $C=wL+rK$,</p>

<p>$$MC=\frac{dC}{dq}=w\frac{dL}{dq}+r\frac{dK}{dq}$$</p>

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

<p>Question: is the following correct </p>

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

<p>which implies</p>

<p>$$MC=w\frac{1}{MP_{L}}+r\frac{1}{MP_{K}}$$</p>

<p>If no, then no need to read further.</p>

<p>If yes, then, consider profit maximization of a firm.</p>

<p>$$\max_{L,K}pq\left(L,K\right)-wL-rK$$</p>

<p>FOC:</p>

<p>$$\begin{cases}pMP_{L}=w\\pMP_{K}=r\end{cases}\Rightarrow\begin{cases}MP_{L}=\frac{w}{p}\\MP_{K}=\frac{r}{p}\end{cases}$$</p>

<p>Therefore,</p>

<p>$$MC=w\frac{1}{MP_{L}}+r\frac{1}{MP_{K}}=w\frac{1}{w/p}+r\frac{1}{r/p}=p+p=2p$$</p>

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

CC BY-SA 3.0
2015-06-15T09:14:11.190Z
What's your basis for thinking that dL/dq might equal 1/(dq/DL)?  dL/dq is a derivative, not a fraction
410 gone
CC BY-SA 3.0
2015-06-15T09:25:58.623Z
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.
ji borrob
CC BY-SA 3.0
2015-06-15T09:27:59.533Z
@EnergyNumbers, btw, the use of what you wrote is valid thanks to inverse function theorem.
ji borrob
CC BY-SA 3.0
2015-06-15T11:11:27.967Z
Can you show the conditions for that, hold? (in particular, can you show that there are no zeros where there shouldn't be)
410 gone
CC BY-SA 3.0
2015-06-15T11:19:55.897Z
(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)
410 gone
CC BY-SA 3.0
2015-06-15T14:49:47.517Z
This post may be of some help, http://economics.stackexchange.com/a/1700/61
Alecos Papadopoulos
<microeconomics><production-function><producer-theory>
Is it true that $\frac{dL}{dq}=1/\frac{\partial q}{\partial L}$?
ji borrob
# Is it true that $\frac{dL}{dq}=1/\frac{\partial q}{\partial L}$?

(Asked by: username_0 on 2015-06-15)

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

$$MC=\frac{dC}{dq}=w\frac{dL}{dq}+r\frac{dK}{dq}$$

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

$$MC=w\frac{1}{MP\_{L}}+r\frac{1}{MP\_{K}}$$

If no, then no need to read further.

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

$$\max\_{L,K}pq\left(L,K\right)-wL-rK$$

FOC:

$$\begin{cases}pMP\_{L}=w\\pMP\_{K}=r\end{cases}\Rightarrow\begin{cases}MP\_{L}=\frac{w}{p}\\MP\_{K}=\frac{r}{p}\end{cases}$$

Therefore,

$$MC=w\frac{1}{MP\_{L}}+r\frac{1}{MP\_{K}}=w\frac{1}{w/p}+r\frac{1}{r/p}=p+p=2p$$

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

**Tags:** microeconomics, production-function, producer-theory
> **Comments:**
> **username_1**: What's your basis for thinking that dL/dq might equal 1/(dq/DL)?  dL/dq is a derivative, not a fraction  
> **username_0**: 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.  
> **username_0**: @EnergyNumbers, btw, the use of what you wrote is valid thanks to inverse function theorem.  
> **username_1**: Can you show the conditions for that, hold? (in particular, can you show that there are no zeros where there shouldn't be)  
> **username_1**: (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)  
> **username_2**: This post may be of some help, http://economics.stackexchange.com/a/1700/61  

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## 2 answers

### Answer by username_2 on 2015-06-15. Score: 7

> 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

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


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### Answer by username_3 on 2015-07-08. Score: 6

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

$$q=q(L,K)$$

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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