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:


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 $MPL=\frac{\partial F}{\partial L}$?

  • $\begingroup$ 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. $\endgroup$
    – VCG
    Sep 6, 2016 at 1:29
  • $\begingroup$ @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? $\endgroup$
    – Omrane
    Sep 6, 2016 at 1:33
  • $\begingroup$ 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 : en.wikipedia.org/wiki/Finite_difference $\endgroup$
    – VCG
    Sep 6, 2016 at 1:36
  • $\begingroup$ @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 $F(K,L+1)-F(K,L)$? $\endgroup$
    – Omrane
    Sep 6, 2016 at 1:42
  • $\begingroup$ Well if the class is in a discrete context, they probably draw the curve smooth for convenience/display. $\endgroup$
    – VCG
    Sep 6, 2016 at 1:52

1 Answer 1


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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