<p>The (partial) derivative of a continuous function is defined as
\begin{align}
\frac{\partial F(\overline K, L)}{\partial L} := \lim_{\Delta L \to 0}\frac{F(\overline K, L + \Delta L) - F(\overline K, L)}{\Delta L}.
\end{align}
Now if $L \in \mathbb{N}$, then you have a lower bound for the increment $\mathbb{N} \ni \Delta L \geq 1$. Otherwise the definition above is not well defined for $\Delta L = 0$.
Such that we finally arrive at the approximation
 \begin{align}
\frac{\partial F(\overline K, L)}{\partial L} \approx \frac{F(\overline K, L + 1) - F(\overline K, L)}{1}.
\end{align}</p>

CC BY-SA 3.0
2016-09-06T05:29:16.183Z
2016-09-06T05:29:16.183Z
clueless
<p>Let $F(K,L)$ be a production function with variables $K$ for capital and $L$ for labor.</p>

<p>The slope of the $F(\overline K,L)$ ($K$ taken constant) is defined as the marginal product of labor ($MPL$) such that:</p>

<p>$$MPL=F(K,L+1)-F(K,L)$$</p>

<p>Most production functions have a positive decreasing slope due to diminishing marginal product and therefore are not straight lines. How is the $MPL$ formula still valid? Isn't that the same formula to find a gradient $m$ of a line (i.e. $\frac{y_{2}-y_{1}}{x_{2}-x{1}}$)? If we take any function whose representation isn't a line, its slope at a certain point is equal to the slope of the tangent at the point and not the formula for the gradient $m$.</p>

<p>Also, supposing the given formula for $MPL$ is true, shouldn't we also have $MPL=\frac{\partial F}{\partial L}$?</p>

CC BY-SA 3.0
2016-09-06T01:29:43.600Z
So where did you get that initial MPL formula? That would work if we are in discrete scenario with no $\Delta$. You are right that the MPL is the partial derivative.
VCG
CC BY-SA 3.0
2016-09-06T01:33:34.957Z
@VCG, it's written in my intermediate macroeconomics book. Alright so basically the $L+1$ indicates that we are working in a discrete scenario. And in a continuous one, my last line would hold true?
Omrane
CC BY-SA 3.0
2016-09-06T01:36:56.387Z
You should check the exact context in the book, but I think they are dealing with discrete. Because the formula you gave is the difference quotient, the discrete equivalent to derivative : https://en.wikipedia.org/wiki/Finite_difference
VCG
CC BY-SA 3.0
2016-09-06T01:42:23.457Z
@VCG, thanks for the link. It's weird because they draw a production function graph (positive decreasing slope) and label its slope at different points as $MPL$. Isn't drawing a graph implying a continuous scenario? If so, shouldn't I refuse that the slope be $F(K,L+1)-F(K,L)$?
Omrane
CC BY-SA 3.0
2016-09-06T01:52:54.040Z
Well if the class is in a discrete context, they probably draw the curve smooth for convenience/display.
VCG
CC BY-SA 3.0
2016-09-06T01:58:17.227Z
I'm not sure about that refuse slope part - if it's in the context of a HW problem or something, then doing what the book wants is likely the correct approach.
VCG
CC BY-SA 3.0
2016-09-06T05:38:34.483Z
@Sadem you're right though. the graph of the production function $\mathbb{N}_+ \times \mathbb{N}_+ \ni (K,L) \mapsto F(K,L)$ should rather consinst of points $\{(K_i, L_j, F_{ij}) : i,j = 0,1,\ldots\}$ than a line.
clueless
<macroeconomics><production-function>
Slope of a production function
Omrane
# Slope of a production function

(Asked by: username_0 on 2016-09-06)

Let $F(K,L)$ be a production function with variables $K$ for capital and $L$ for labor.

The slope of the $F(\overline K,L)$ ($K$ taken constant) is defined as the marginal product of labor ($MPL$) such that:

$$MPL=F(K,L+1)-F(K,L)$$

Most production functions have a positive decreasing slope due to diminishing marginal product and therefore are not straight lines. How is the $MPL$ formula still valid? Isn't that the same formula to find a gradient $m$ of a line (i.e. $\frac{y\_{2}-y\_{1}}{x\_{2}-x{1}}$)? If we take any function whose representation isn't a line, its slope at a certain point is equal to the slope of the tangent at the point and not the formula for the gradient $m$.

Also, supposing the given formula for $MPL$ is true, shouldn't we also have $MPL=\frac{\partial F}{\partial L}$?

**Tags:** macroeconomics, production-function

> **Comments:**
> **username_1**: So where did you get that initial MPL formula? That would work if we are in discrete scenario with no $\Delta$. You are right that the MPL is the partial derivative.  
> **username_0**: @username_1, it's written in my intermediate macroeconomics book. Alright so basically the $L+1$ indicates that we are working in a discrete scenario. And in a continuous one, my last line would hold true?  
> **username_1**: You should check the exact context in the book, but I think they are dealing with discrete. Because the formula you gave is the difference quotient, the discrete equivalent to derivative : https://en.wikipedia.org/wiki/Finite_difference  
> **username_0**: @username_1, thanks for the link. It's weird because they draw a production function graph (positive decreasing slope) and label its slope at different points as $MPL$. Isn't drawing a graph implying a continuous scenario? If so, shouldn't I refuse that the slope be $F(K,L+1)-F(K,L)$?  
> **username_1**: Well if the class is in a discrete context, they probably draw the curve smooth for convenience/display.  
> **username_1**: I'm not sure about that refuse slope part - if it's in the context of a HW problem or something, then doing what the book wants is likely the correct approach.  
> **username_2**: @Sadem you're right though. the graph of the production function $\mathbb{N}_+ \times \mathbb{N}_+ \ni (K,L) \mapsto F(K,L)$ should rather consinst of points $\{(K_i, L_j, F_{ij}) : i,j = 0,1,\ldots\}$ than a line.  

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## 1 answers

### Answer by username_2 on 2016-09-06. Score: 3 | Accepted answer

The (partial) derivative of a continuous function is defined as
\begin{align}
\frac{\partial F(\overline K, L)}{\partial L} := \lim\_{\Delta L \to 0}\frac{F(\overline K, L + \Delta L) - F(\overline K, L)}{\Delta L}.
\end{align}
Now if $L \in \mathbb{N}$, then you have a lower bound for the increment $\mathbb{N} \ni \Delta L \geq 1$. Otherwise the definition above is not well defined for $\Delta L = 0$.
Such that we finally arrive at the approximation
\begin{align}
\frac{\partial F(\overline K, L)}{\partial L} \approx \frac{F(\overline K, L + 1) - F(\overline K, L)}{1}.
\end{align}


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