Marginal Product As A Derivative

From a Production Schedule to a Derivative

A production function describes the maximum output that can be produced from specified quantities of inputs, given the technology and production conditions represented by the function. When inputs can take continuous values, the function can describe output at quantities between the entries in a production schedule. Its slope with respect to one input measures how output responds locally when that input changes and the other inputs are kept fixed.

The earlier material on discrete marginal product measures output changes between observed input levels. A derivative gives the corresponding local measure for a smooth production function. The distinction matters: a derivative is a rate at one input combination, while a finite difference summarizes a change over an interval.

Definition: Suppose output is $Q=f(L,K)$, where $L$ and $K$ are two inputs. The marginal product of $L$ at $(L,K)$ is the partial derivative of output with respect to $L$, holding $K$ constant: $$ MP_L(L,K)=\frac{\partial Q}{\partial L} =\lim_{h\to 0}\frac{f(L+h,K)-f(L,K)}{h}. $$ Similarly, the marginal product of $K$ is $$ MP_K(L,K)=\frac{\partial Q}{\partial K} =\lim_{h\to 0}\frac{f(L,K+h)-f(L,K)}{h}. $$

The limit definition requires the relevant derivative to exist at the input combination being studied. In practice, for the smooth functions used here, this means differentiating the formula with respect to the chosen input and then substituting the stated input quantities. The other input is treated as a constant during that differentiation, even if it is represented by a letter.

If output is measured in units of product and $L$ in units of labor, $MP_L$ is measured in units of product per unit of labor. It describes the local output response to a small change in $L$, not the number of workers or machines themselves. The value is specific to the input combination: the same production function can have different marginal products at different values of $L$ and $K$.

Why the Partial Derivative Is the Right Measure

To measure the effect of changing $L$, compare output at $L$ with output at $L+h$, using exactly the same quantity of $K$ in both cases. The quotient of the output change by the input change is the finite rate of change. Letting $h$ approach zero gives the local rate at the original input combination.

Change one input.
Compare $f(L+h,K)$ with $f(L,K)$. The value $K$ is unchanged in both expressions.

Calculate output change per input change.
The quotient $\frac{f(L+h,K)-f(L,K)}{h}$ measures the average output response over the small change $h$.

Take the local limit.
As $h$ approaches zero, the quotient approaches the slope at the starting input combination. This is $\frac{\partial f}{\partial L}$, the marginal product of $L$.

This derivation also explains why holding the other input fixed is part of the definition. If $K$ changes as $L$ changes, the observed output change can reflect both inputs. It would not isolate the partial effect of $L$. A partial derivative is therefore not automatically the output change when a firm changes several inputs together.

For a small change in $L$, with $K$ fixed, the derivative gives a useful approximation:

\Delta Q \approx MP_L(L,K)\,\Delta L.

This is a local approximation. It is exact for a finite change only in special cases, such as when output is linear in $L$ over the interval. For a nonlinear function, the marginal product can vary as $L$ changes, so the exact finite output change need not equal the starting marginal product multiplied by the entire input change.

Calculating Marginal Product

Worked Example: A Bilinear Production Function

A hypothetical workshop has production function $Q=6L+4K+2LK$, where $Q$ is units of output per shift, $L$ is labor-hours per shift, and $K$ is machine-hours per shift. Find both marginal products at $L=3$ and $K=5$, and interpret the marginal product of labor.

Method. Differentiate with respect to one input at a time, treating the other input as fixed. The common mistake to avoid is treating both $L$ and $K$ as if they changed when taking a partial derivative.

Marginal product of labor. While differentiating with respect to $L$, regard $K$ as a constant. The derivative of $6L$ is 6, the derivative of $4K$ is 0, and the derivative of $2LK$ is $2K$. Thus,

$$ MP_L=6+2K. $$

Substitute $K=5$:

$$ MP_L=6+2(5)=6+10=16. $$

At this input combination, the local rate is 16 units of output per additional labor-hour, with machine-hours held at 5.

Marginal product of machines. Now differentiate with respect to $K$, treating $L$ as constant. The derivative of $6L$ is 0, the derivative of $4K$ is 4, and the derivative of $2LK$ is $2L$. Therefore,

$$ MP_K=4+2L=4+2(3)=10. $$

The marginal product of machines at the same input combination is 10 units of output per additional machine-hour, holding labor-hours at 3.

Check. Output at the stated combination is $6(3)+4(5)+2(3)(5)=18+20+30=68$ units. If labor increases by $0.1$ hour while $K=5$ stays fixed, the labor-dependent terms increase by $6(0.1)+2(0.1)(5)=0.6+1=1.6$ units. This matches $MP_L\Delta L=16(0.1)=1.6$; here the finite change matches exactly because the function is linear in $L$ when $K$ is fixed.

Worked Example: A Square-Root Production Function

A hypothetical packaging process is represented by $Q=5K\sqrt{L}+2L$, where $Q$ is packages per hour, $L$ is labor-hours per hour, and $K$ is machine-hours per hour. Find the marginal product of labor at $L=16$, $K=3$, and state the function’s relevant domain condition.

Method. Differentiate with respect to $L$ while treating $K$ as a constant. Since the square root is differentiable for $L>0$, the calculation applies at $L=16$. It does not use the ordinary derivative formula at $L=0$, where the derivative of $\sqrt{L}$ is not finite.

Rewrite the square root as a power: $\sqrt{L}=L^{1/2}$. The derivative of $L^{1/2}$ is $\frac{1}{2}L^{-1/2}=\frac{1}{2\sqrt{L}}$. Hence,

MP_L=5K\left(\frac{1}{2\sqrt{L}}\right)+2 =\frac{5K}{2\sqrt{L}}+2.

Substitute $L=16$ and $K=3$. First, $\sqrt{16}=4$, so

MP_L=\frac{5(3)}{2(4)}+2 =\frac{15}{8}+2 =1.875+2 =3.875.

Thus the marginal product is 3.875 packages per additional labor-hour at the stated input combination, with machine-hours fixed at 3.

Check. Output at the input combination is $5(3)(4)+2(16)=60+32=92$ packages per hour. The derivative calculation has used the same labor and machine units as the production function; no conversion of $K$ into $L$ is appropriate.

Worked Example: Comparing a Derivative With a Finite Change

A hypothetical assembly operation has production function $Q=L^2+3LK+2K^2$, with output measured in units per shift and both inputs measured in hours per shift. At $L=4$ and $K=2$, calculate the marginal product of labor. Then calculate the exact output change when labor rises from 4 to 4.5 hours while machine-hours remain fixed.

Method. First find the partial derivative at the starting combination. Then evaluate the production function at both endpoints and subtract. The common mistake to avoid is using the starting derivative as though it were necessarily the exact rate over a finite interval.

Differentiate with respect to $L$, holding $K$ fixed. The three terms have derivatives $2L$, $3K$, and 0, respectively:

$$ MP_L=2L+3K. $$

At $L=4$, $K=2$,

$$ MP_L=2(4)+3(2)=8+6=14. $$

The local marginal product at the starting point is 14 output units per labor-hour. Now calculate the exact finite change. At the starting point,

$$ Q(4,2)=4^2+3(4)(2)+2(2^2)=16+24+8=48. $$

At the endpoint, with labor at 4.5 and machine-hours still at 2,

$$ Q(4.5,2)=(4.5)^2+3(4.5)(2)+2(2^2) =20.25+27+8=55.25. $$

The exact output increase is $55.25-48=7.25$ units. The labor increase is $4.5-4=0.5$ hour, so the finite rate over this interval is $7.25/0.5=14.5$ output units per labor-hour. This differs from the starting marginal product of 14.

Check. The derivative gives the local rate at $L=4$, not a guaranteed exact finite rate through the whole interval from 4 to 4.5. The exact output change, 7.25 units, is consistent with the endpoint calculations.

Common Errors and Interpretation

The most important procedural rule is to identify the input named in the question before differentiating. For example, in the first worked example, differentiating $2LK$ with respect to $L$ gives $2K$, not $2L$. The latter would result from differentiating with respect to $K$. At $L=3$, $K=5$, incorrectly using $2L$ would produce $6+2(3)=12$ for $MP_L$, rather than the correct value $6+2(5)=16$.

A second trap is confusing a partial derivative with an output difference observed when multiple inputs change. The definition of $MP_L$ fixes $K$; if both inputs change, the output response cannot be attributed to labor alone from $MP_L$ without separately accounting for the change in $K$. Always read “holding other inputs constant” as a mathematical condition, not as optional wording.

A third trap is reporting the derivative without units or context. A result such as 14 is incomplete unless it is identified as output units per unit of the input, evaluated at specified input quantities. Also, a marginal product is not itself total output: the production function gives total output, while the partial derivative gives its local rate of response.

Key takeaway: To find an input’s marginal product, take the partial derivative of the production function with respect to that input, hold every other input constant, and then evaluate at the stated input combination. Interpret the result as a local output rate with units.

Check Your Understanding

For each problem, keep the non-target input fixed when differentiating. Answers are provided so that both the derivative and its interpretation can be checked.

A hypothetical production function is $Q=7L+2K+LK$. Find $MP_L$ and $MP_K$ at $L=3$, $K=4$. (Answer: $MP_L=11$ and $MP_K=5$.)
A process has $Q=9\sqrt{L}+2K$, where $L>0$. At $L=9$, $K=5$, calculate output and the marginal product of labor. (Answer: $Q=37$; $MP_L=1.5$ output units per unit of labor.)
For $Q=L^2+2LK+5K$, calculate $MP_L$ at $L=2$, $K=3$. Then calculate the exact output change if $L$ rises by $0.2$ while $K$ is fixed. (Answer: $MP_L=10$; the exact output change is $2.04$ units.)
A hypothetical production function is $Q=4LK+L^2+3K^2$. At $L=2$, $K=1$, find $MP_K$. Then calculate the exact output change when $K$ rises to $1.5$, keeping $L=2$. (Answer: $MP_K=14$; the exact output change is $7.75$ units.)

What You'll Learn