Marginal cost is defined as "the change in the total cost that arises when the quantity produced is incremented by one unit." And given a total cost function $C(q)$ that's differentiable, the marginal cost is the derivative, $C'(q)$. But if I were given $C$ and asked the cost that arises when the quantity produced is increased from 2 to 3, I would simply calculate $C(3)-C(2)$; no need to bring calculus into the picture. In general, $ C(3)-C(2) \neq C'(2)$. For example, if $C(q) = q^2$, then $C(3)-C(2) = 5$, but $C'(2) = 4$.

Thus my question is: Why is the derivative used to represent marginal cost instead of the difference?

Note: I thought this question must've been what's being asked here, but evidently not; there what's being asked is (essentially) why $C'(3) \neq C(3)-C(2)$.


3 Answers 3


The derivative is used in some contexts, but not all, when the cost function is differentiable. In those contexts, it tends to be assumed that supply is continuous, not discrete. This is a matter of convention and of analytic convenience. It has the advantage of being consistent, whether you're approaching the supply point from above or from below.

But in other contexts, given your cost function, assuming that the thing being supplied is discrete and not continuous (that is, it is possible to supply 2 units or 3 units, but not 2.9 or 3.5 or any other fractional unit) then the marginal cost of the third item is indeed 5, not 4.

  • $\begingroup$ The most important concept here is analytic convenience. Using discrete quantities, MC=MR may not have an exact value. Using calculus, you arrive at the exact value. It provides a direct and exact solution. Not an approximate solution. $\endgroup$
    – Jamzy
    Oct 12, 2015 at 1:52
  • $\begingroup$ There are functions which are continuous and differentiable and still may have a supply point where the marginal cost depends on whether you approach the point from above or below. $\endgroup$
    – HRSE
    Oct 12, 2015 at 1:57
  • $\begingroup$ @HREcon can you give a real-world example of such a supply-cost function? $\endgroup$
    – 410 gone
    Oct 12, 2015 at 3:27
  • $\begingroup$ $c(q)=\begin{cases} q, & q\leq 1\\ 2q-1, & q>1 \end{cases}$ is continuous and differentiable, but not continuously differentiable (i.e., the derivative is not a continuous function). $\endgroup$
    – HRSE
    Oct 12, 2015 at 8:29
  • $\begingroup$ @HREcon and that's differentiable everywhere except q=1 $\endgroup$
    – 410 gone
    Oct 12, 2015 at 8:52

To help you discern the two, let's try to explain with words and understand what information are we getting from the derivative and from the difference, respectively:

  1. The derivative gives you information about the change in cost relative to change in quantity produced, in a specific, local, point(quantity)1. In other words you are measuring the change in cost in terms of change in quantity. More mathematically, the derivative of the cost with respect to the quantity gives you the rate of change of the cost over the rate of change in quantity or the slope of the cost curve.

  2. The difference between two points (quantities) on the cost curve: $C(3) − C(2) = 5$ gives you the relative difference in price only of those two points, not accounting for all the intermediate values2. Again more mathematically, the difference just gives you the distance in price between the two points(quantities).

To conclude, the difference between the two is the information they give you, namely:

  • derivative: rate of change of cost in terms of quantity.

  • difference: difference between the total cost for two quantities.

1. In your example, the marginal cost for quantity: $2$, given the total cost function: $C(q) = q^2$ is: $C′(2) = 4$, which means that if you're producing currently 2 items, the next item will increase the cost with $4$ units.

2. The relation $C(3) − C(2) = 5$ means that the total cost for producing 3 items is 5 units more than the total cost of producing 2 items.

  • 1
    $\begingroup$ I agree that the difference between the derivative and the difference is one of instantaneous vs average rate of change (which is essentially what you said, I think). But my question is why is the definition of marginal cost the instantaneous one, when the informal characterization seems to line up better with the average. See what I mean? $\endgroup$ Oct 12, 2015 at 16:05
  • $\begingroup$ I guess my point/problem can also be viewed as this: I don't see the difference between "if you're producing currently 2 items, the next will increase the cost with ___ units" and "the total cost for producing 3 items is ___ units more than the total cost of producing 2 items." Those two phrases seem synonymous, and hence those ___'s should match. See what I mean? $\endgroup$ Oct 12, 2015 at 16:09
  • $\begingroup$ I absolutely get you on this one, it very well may be a simple matter of convention in this case. $\endgroup$
    – Ziezi
    Oct 12, 2015 at 17:49

The function $C(q) = q^2$ is non-linear, therefore the rate of change of $C(q)$ with respect to q is constantly changing.

When you take $\frac{C(3)-C(2)} {3-2}$ you are finding the rate of change over a range of $q$, not the rate of change at $q = 3$.

This is where taking a derivative is needed, because it gives you the rate of change at the point $(q,C)$ as the change in $q$ approaches $0$, rather than an average of the rate of change for every $q$ value from $2 \leq q \leq 3$.

  • $\begingroup$ I don't see what the use of the instantaneous rate of change of the Cost function is, and, in particular, I don't see why it can be called "the total cost that arises when the quantity produced is incremented by one unit", since that's clearly what the difference $C(3)-C(2)$ is. See what I mean? $\endgroup$ Oct 11, 2015 at 16:36
  • $\begingroup$ @QuinnCulver It would be useful in the sense that you could generate a marginal cost curve, then use that curve in a model. For example, modeling a firm by constructing the MC curve along with several others (ATC, AVC, D=MR) and establishing thresholds. denesp: thanks for the edits, I need to learn how to do that! $\endgroup$ Oct 12, 2015 at 22:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.