Why the Second Derivative Helps
A production function describes how much output a firm can produce from its inputs. In the short run, one input may vary while other inputs are held fixed. The first derivative with respect to the variable input measures the change in output from a small additional amount of that input. The second derivative tells us whether that marginal product is increasing or decreasing as the input rises.
This distinction matters because diminishing marginal product does not mean that output is falling. Output can still increase as more input is used, even while each additional unit contributes less than the previous one. To decide whether output itself is falling, we check the first derivative; to decide whether marginal product is falling, we check the second derivative.
For the usual interval-wide test, keep the domain in view: the function must be twice differentiable on the interval being tested. If \(f''(L)>0\) throughout an interval, marginal product is increasing there. If \(f''(L)=0\) throughout an interval, marginal product is constant there. A second derivative equal to zero at just one input level is not enough to classify marginal product on either side of that point; its sign around the point must be checked.
Why a Negative Second Derivative Means Declining Marginal Product
The second derivative is the rate of change of marginal product because \(MP_L=f'(L)\), so the derivative of marginal product with respect to \(L\) is \(f''(L)\). The following argument makes the interval test precise.
Here \(a<b\), and the Mean Value Theorem applied to \(f'\) guarantees some \(c\) between \(a\) and \(b\) for which the equality holds, provided \(f'\) is continuous on the closed interval and differentiable in its interior. If \(f''(L)<0\) throughout that interval, then \(f''(c)<0\), while \(b-a>0\). Therefore \(f'(b)-f'(a)<0\), or \(f'(b)<f'(a)\). Marginal product at the higher input level is lower. This is why the sign of the second derivative tests whether marginal product is declining.
This result concerns the behaviour of marginal product, not the sign of marginal product itself. The first derivative and the second derivative answer different questions:
| Question | Test | Interpretation |
|---|---|---|
| Is output increasing or decreasing as the input rises? | Sign of \(f'(L)\) | Positive means output rises locally; negative means output falls locally. |
| Is marginal product increasing or decreasing as the input rises? | Sign of \(f''(L)\) | Negative means marginal product is declining; positive means it is increasing. |
The signs can occur in combination. If \(f'(L)>0\) and \(f''(L)<0\), output is rising while marginal product is diminishing. If both derivatives are negative, output is falling and marginal product is also declining. If \(f'(L)<0\) while \(f''(L)>0\), output is falling but marginal product is increasing; it may be becoming less negative.
Worked Example: Marginal Product Declines Before It Turns Negative
A hypothetical workshop’s short-run production function is \(Q=f(L)=20L+4L^2-\frac{1}{3}L^3\), where \(L\geq 0\) is labour input and all other inputs are fixed. Find where marginal product is diminishing, where it is positive, and whether diminishing marginal product implies falling output in this example.
Method. Differentiate once to find marginal product, then differentiate again to test how marginal product changes. Solve each sign condition separately.
Step 1: Find marginal product.
Step 2: Find the second derivative.
Marginal product is diminishing when \(8-2L<0\). Solving gives \(-2L<-8\), so \(L>4\). It is increasing for \(0\leq L<4\), and \(f''(4)=0\) marks the change in the sign of its rate of change.
Step 3: Find where marginal product is positive. Factor the first derivative:
On the stated domain \(L\geq0\), \(L+2\) is positive. Thus marginal product is positive for \(0\leq L<10\), zero at \(L=10\), and negative for \(L>10\). Taken together, marginal product is diminishing and positive for \(4<L<10\), and diminishing and negative for \(L>10\).
For instance, at \(L=6\), \(MP_L=20+8(6)-6^2=20+48-36=32\) and \(f''(6)=8-12=-4\). Output is still increasing there, but the marginal product is declining. At \(L=12\), \(MP_L=20+96-144=-28\) and \(f''(12)=8-24=-16\). Output is falling there, and marginal product is also declining.
Check. At \(L=4\), the second derivative is zero, and the sign of \(8-2L\) switches from positive to negative as \(L\) passes 4. At \(L=10\), the first derivative is zero. These are distinct input levels: diminishing marginal product begins before marginal product becomes negative.
Common mistake avoided: concluding that output must be falling as soon as \(f''(L)<0\). At \(L=6\), the second derivative is negative but the first derivative is positive, so output is still increasing.
Test Signs on the Relevant Input Range
A formula can have different second-derivative signs at different input levels. The conclusion should therefore state the interval, not merely report a derivative value at one point. A reliable procedure is to differentiate, solve the inequality \(f''(L)<0\), and then compare that interval with the production function’s relevant domain. If the question also asks whether output is rising or falling, solve \(f'(L)>0\) or \(f'(L)<0\) separately.
Identify the variable input and treat the other inputs as fixed parameters in the production function.
Calculate \(f'(L)\). This is marginal product and determines whether output rises or falls locally.
Calculate \(f''(L)\), then solve its sign condition over the stated input domain to classify marginal product.
Use the sign of \(f'(L)\) for output and the sign of \(f''(L)\) for whether marginal product is changing.
Worked Example: Diminishing Marginal Product With Falling Output
Over the range \(0\leq L\leq8\), a hypothetical packing operation has production function \(Q=f(L)=80+24L-2L^2\). Determine whether marginal product is diminishing throughout the range, and assess output and marginal product at \(L=5\) and \(L=7\).
Method. Use the first derivative to assess output’s response to more labour and the second derivative to assess whether marginal product declines.
Step 1: Differentiate.
Since \(f''(L)=-4<0\) everywhere on \(0\leq L\leq8\), marginal product is diminishing throughout the stated range.
Step 2: Evaluate at \(L=5\).
Marginal product is positive at \(L=5\), so output is locally increasing as labour rises. The second derivative is negative, so marginal product is diminishing.
Step 3: Evaluate at \(L=7\).
At \(L=7\), marginal product is negative, so output is locally decreasing as labour rises. Marginal product is still diminishing because \(f''(7)=-4\). Indeed, \(f'(6)=24-24=0\): output reaches a stationary point there, then \(f'(L)<0\) for \(L>6\) within the stated range.
Check. The function’s values at \(L=5\) and \(L=7\) happen to be equal, but that does not make the marginal products at those points equal: they are \(4\) and \(-4\). The derivative describes the local response at each input level.
Common mistake avoided: treating negative marginal product and diminishing marginal product as synonyms. At \(L=5\), marginal product is diminishing but positive; at \(L=7\), it is diminishing and negative.
A Zero Second Derivative Needs Context
If \(f''(L_0)=0\) at a single input level \(L_0\), that equality alone does not establish that marginal product is constant, nor does it tell us whether marginal product is diminishing nearby. We must inspect the sign of \(f''\) on either side. If its sign changes from negative to positive, marginal product changes from diminishing to increasing. If its sign does not change, the point may simply be a zero within a region of the same behaviour.
Worked Example: A Change From Diminishing to Increasing Marginal Product
A hypothetical process has production function \(Q=f(L)=6L-L^2+\frac{1}{6}L^3\) for \(0\leq L\leq4\). Identify the ranges where marginal product is diminishing or increasing. Then check whether marginal product is negative at \(L=1\), \(L=2\), and \(L=3\).
Method. Find both derivatives, solve the second-derivative sign condition, and evaluate the first derivative at the specified input levels.
Step 1: Differentiate.
Step 2: Classify the change in marginal product. The second derivative is negative when \(L<2\), zero at \(L=2\), and positive when \(L>2\). On the stated domain, marginal product is diminishing for \(0\leq L<2\), and increasing for \(2<L\leq4\). The value \(f''(2)=0\) is a change point, not evidence that marginal product is constant over an interval.
Step 3: Check whether marginal product is negative.
All three marginal products are positive. Marginal product falls from \(4.5\) at \(L=1\) to \(4\) at \(L=2\), then rises to \(4.5\) at \(L=3\). Thus it diminishes on the lower-input side and increases on the higher-input side, without being negative at any of these checked points.
Check. The sign of \(f''(L)=L-2\) changes at 2, matching the switch in whether marginal product decreases or increases. The first derivative values separately confirm that output is locally increasing at all three input levels.
Common mistake avoided: reporting “diminishing at \(L=2\)” just because \(f''(2)=0\), or inferring negative marginal product from diminishing marginal product. Neither conclusion follows from that equality.
Common Errors and Practical Checks
- Using the wrong derivative. \(f'(L)\) is marginal product; \(f''(L)\) tests whether marginal product changes. A question about diminishing marginal product calls for the second derivative.
- Confusing a falling marginal product with falling output. A negative second derivative can occur while the first derivative remains positive. Check both signs if the question asks about output.
- Ignoring the input domain. A mathematical inequality may produce input levels outside the feasible range. State conclusions only on the domain relevant to the production function.
- Classifying from one point alone. A negative second derivative at a point gives local information there; an interval-wide claim requires checking its sign throughout the interval.
- Treating \(f''(L)=0\) at one point as constant marginal product. Constant marginal product on an interval requires the second derivative to be zero throughout that interval.
The derivative test is useful because it turns a verbal claim about successive contributions into a precise condition. But the interpretation remains economic: \(f'(L)\) concerns the effect on output of a small increase in the input, while \(f''(L)\) concerns how that effect itself changes. Keeping those questions separate prevents a sign calculation from being given the wrong meaning.
Check Your Understanding
For each problem, use the first derivative to assess marginal product and the second derivative to test whether it is diminishing. Respect the stated input domain.
- A hypothetical process has \(Q=30L+2L^2-\frac{1}{3}L^3\) for \(L\geq0\). Find \(MP_L\) and \(f''(L)\), then classify both at \(L=3\). (Answer: \(MP_L=33>0\) and \(f''(3)=-2<0\); output is increasing and marginal product is diminishing.)
- For \(Q=50+18L-\frac{3}{2}L^2\) on \(0\leq L\leq10\), determine whether marginal product is diminishing throughout, and whether output is rising or falling at \(L=8\). (Answer: \(f''(L)=-3\) throughout; \(MP_L(8)=-6\), so marginal product is diminishing and output is falling at \(L=8\).)
- A production function is \(Q=4L+L^2-\frac{1}{3}L^3\), with \(L\geq0\). Find the input range on which marginal product is both positive and diminishing. (Answer: \(1<L<1+\sqrt{5}\), approximately \(1<L<3.236\).)
- For \(Q=10L-\frac{1}{2}L^2+\frac{1}{6}L^3\) on \(0\leq L\leq3\), find where marginal product is diminishing and determine whether \(f''(1)=0\) alone implies constant marginal product. (Answer: marginal product is diminishing for \(0\leq L<1\) and increasing for \(1<L\leq3\); no, it is not constant on an interval.)
- A hypothetical operation has \(Q=24L-2L^2\) on \(0\leq L\leq12\). Find \(MP_L\), locate where it changes sign, and compare output at \(L=5\) with marginal product there. (Answer: \(MP_L=24-4L\), which is zero at \(L=6\), positive below 6 and negative above 6; \(Q(5)=70\) and \(MP_L(5)=4>0\), while marginal product is diminishing throughout because \(f''(L)=-4\).)