Testing Diminishing Marginal Utility With Second Derivatives

Introduction

Marginal utility measures the change in utility associated with a small increase in a good, holding other quantities fixed. To test whether that marginal utility declines as the quantity of the good rises, examine how the marginal utility itself changes. For a twice-differentiable utility function, that question is answered by a second partial derivative.

The key is to keep two effects separate. A second partial derivative with respect to the same good tests how that good’s own marginal utility changes. A cross-partial derivative tests how the quantity of a different good changes it. The signs of these derivatives answer different questions.

Definition: For a twice-differentiable utility function \(U(x,y)\), the marginal utilities are \(U_x=\partial U/\partial x\) and \(U_y=\partial U/\partial y\). The own-good second partials are \(U_{xx}=\partial^2U/\partial x^2\) and \(U_{yy}=\partial^2U/\partial y^2\). If \(U_{xx}<0\) at a bundle, marginal utility from good \(x\) is declining locally as \(x\) increases while \(y\) is held fixed. If \(U_{xx}>0\), it is increasing locally; if \(U_{xx}=0\), it is locally constant to first order. The corresponding interpretations apply to \(U_{yy}\) and good \(y\).

Reading the Own-Good Second Partial

The order of differentiation matters for interpretation. First differentiate utility with respect to \(x\) to obtain the marginal utility of \(x\), \(U_x\). Differentiate that result with respect to \(x\) again to obtain \(U_{xx}\). This second derivative measures the rate of change of \(x\)’s marginal utility as \(x\) changes, with \(y\) fixed. It is not the level of marginal utility itself.

A negative value at one bundle gives a local conclusion. To conclude that marginal utility declines throughout an interval of \(x\)-values, with \(y\) held at a specified value, check that \(U_{xx}\) is negative throughout that interval. If the sign changes, a single evaluation is not enough to support a claim about the entire interval.

\frac{\partial U_x}{\partial x}=U_{xx}, \qquad \frac{\partial U_y}{\partial y}=U_{yy}.

The cross-partial \(U_{xy}\), by contrast, differentiates \(U_x\) with respect to \(y\). When it is positive, increasing \(y\) raises the marginal utility of \(x\), locally, holding \(x\) fixed. When it is negative, increasing \(y\) lowers the marginal utility of \(x\), locally. This does not by itself show whether \(x\)’s own marginal utility falls as \(x\) rises.

Key distinction: \(U_{xx}\) tests how \(x\)’s own marginal utility changes with \(x\). \(U_{xy}\) tests how \(x\)’s marginal utility changes with \(y\). A sign for one does not determine the sign of the other.

Why the Test Works

The justification is the ordinary derivative test applied to the marginal utility. Fix \(y=y_0\) and define \(g(x)=U_x(x,y_0)\). The function \(g\) is the marginal utility of \(x\) along that fixed-\(y\) slice, and its derivative is \(g'(x)=U_{xx}(x,y_0)\). If this derivative is negative throughout an interval, \(g\) decreases throughout that interval.

The finite-change version makes the conclusion explicit. If \(U_{xx}\) is continuous, the fundamental theorem of calculus gives the change in marginal utility between \(x_1\) and \(x_2\), with \(y=y_0\):

U_x(x_2,y_0)-U_x(x_1,y_0) = \int_{x_1}^{x_2} U_{xx}(t,y_0)\,dt.

For \(x_2>x_1\), if \(U_{xx}(t,y_0)<0\) throughout the interval, the integral is negative. Therefore \(U_x(x_2,y_0)<U_x(x_1,y_0)\). This is why a sign check over the interval, not merely at an endpoint, supports the full-interval conclusion. A zero at a single point does not necessarily prevent a strict decline over the whole interval; the integral, or the derivative’s sign over the interval, settles that question.

Worked Applications

Worked Example: Negative Own-Good Curvature and a Positive Cross-Effect

Suppose \(U(x,y)=30x-2x^2+3xy+12y-y^2\), where \(x\) and \(y\) are quantities of two goods. At \(y=1\), determine whether the marginal utility of \(x\) declines as \(x\) rises from 2 to 3. Also determine how increasing \(y\) affects \(x\)’s marginal utility at \(x=2\). Use direct differentiation and check the conclusions numerically.

Method. Differentiate first to find \(U_x\), then differentiate with respect to \(x\) for the own-good test and with respect to \(y\) for the cross-effect.

Step 1: Find the marginal utility of \(x\). Differentiating each term with respect to \(x\) gives \(U_x=30-4x+3y\). The terms \(12y-y^2\) do not depend on \(x\), so their derivatives are zero.

Step 2: Test the own-good effect. Differentiating \(U_x\) with respect to \(x\) gives \(U_{xx}=-4\). This is negative for every \(x\) and \(y\), so at fixed \(y=1\), marginal utility of \(x\) declines throughout the interval from 2 to 3.

Step 3: Calculate endpoint marginal utilities. At \((2,1)\), \(U_x=30-4(2)+3(1)=30-8+3=25\). At \((3,1)\), \(U_x=30-4(3)+3(1)=30-12+3=21\). The marginal utility falls by \(21-25=-4\).

Step 4: Check the cross-effect. Differentiating \(U_x\) with respect to \(y\) gives \(U_{xy}=3\). Thus, at \(x=2\), increasing \(y\) by one unit raises \(U_x\) by 3; for example, \(U_x(2,1)=25\) and \(U_x(2,2)=30-8+6=28\).

Check and common mistake avoided. The endpoint change in \(U_x\) is \(-4\), matching the constant \(U_{xx}=-4\) over a one-unit increase in \(x\). The positive cross-partial does not contradict diminishing marginal utility in \(x\): it describes a change in \(y\), not in \(x\).

Worked Example: Testing Both Goods and Their Cross-Effect

Let \(U(x,y)=40x-x^2+4xy+18y-3y^2\). At the bundle \((3,2)\), determine whether each good has diminishing marginal utility in its own quantity. Then calculate how \(x\)’s marginal utility changes when \(y\) increases from 1 to 2 while \(x=3\).

Method. Find both first partial derivatives, then each own-good second partial and the cross-partial. Use the relevant derivative for each question.

Step 1: Find both marginal utilities. Differentiating with respect to \(x\) gives \(U_x=40-2x+4y\). Differentiating with respect to \(y\) gives \(U_y=4x+18-6y\).

Step 2: Find the own-good second partials. Differentiating \(U_x\) with respect to \(x\) gives \(U_{xx}=-2\). Differentiating \(U_y\) with respect to \(y\) gives \(U_{yy}=-6\). Both are negative everywhere, so each good has declining marginal utility as its own quantity rises, holding the other quantity fixed.

Step 3: Evaluate marginal utilities at the stated bundle. At \((3,2)\), \(U_x=40-2(3)+4(2)=40-6+8=42\), and \(U_y=4(3)+18-6(2)=12+18-12=18\). These are the marginal utility levels; the negative second partials describe how those levels change with their respective own quantities.

Step 4: Calculate the cross-effect on \(U_x\). Differentiating \(U_x\) with respect to \(y\) gives \(U_{xy}=4\). Directly, at \((3,1)\), \(U_x=40-6+4=38\), and at \((3,2)\), \(U_x=40-6+8=42\). The change is \(42-38=4\), as predicted.

Check and common mistake avoided. The requested change in \(x\)’s marginal utility as \(y\) changes is governed by \(U_{xy}\), not \(U_{xx}\). Using \(U_{xx}=-2\) for that calculation would answer a different question: how \(U_x\) changes when \(x\) changes.

Worked Example: Checking a Sign That Depends on Quantity

Suppose \(U(x,y)=12x-\frac{2}{3}x^3+5y+2xy^2\), on the range \(0\leq x\leq2\) and \(0\leq y\leq1\). Does \(x\)’s marginal utility decline throughout the stated range of \(x\), for every fixed \(y\) in the stated range? Compare the marginal utilities at \(x=0.5\) and \(x=1.5\) when \(y=0.5\).

Method. Differentiate to obtain \(U_{xx}\), inspect its sign over the whole range, and verify the direction with two marginal-utility evaluations.

Step 1: Find \(U_x\). Differentiating with respect to \(x\) gives \(U_x=12-2x^2+2y^2\). In particular, the derivative of \(-\frac{2}{3}x^3\) is \(-2x^2\).

Step 2: Find and inspect \(U_{xx}\). Differentiating again with respect to \(x\) gives \(U_{xx}=-4x\). On \(0\leq x\leq2\), this is nonpositive, and it is strictly negative whenever \(x>0\). Therefore \(U_x\) is nonincreasing over the full range and strictly decreases between any two distinct points in that range. The value \(U_{xx}=0\) at the single endpoint \(x=0\) does not reverse the conclusion.

Step 3: Evaluate at the requested points. With \(y=0.5\), \(y^2=0.25\), so \(2y^2=0.5\). At \(x=0.5\), \(U_x=12-2(0.5)^2+0.5=12-0.5+0.5=12\). At \(x=1.5\), \(U_x=12-2(1.5)^2+0.5=12-4.5+0.5=8\). The marginal utility decreases by \(8-12=-4\).

Step 4: Separate the cross-effect. Differentiating \(U_x\) with respect to \(y\) gives \(U_{xy}=4y\). At \(y=0.5\), \(U_{xy}=4(0.5)=2\), so increasing \(y\) raises \(U_x\) locally at that value of \(y\). This cross-effect does not change the own-good test, which uses \(U_{xx}\).

Check and common mistake avoided. The own-good sign varies with \(x\), so it was necessary to inspect the full range. It would be incorrect to call \(U_x\) constant just because \(U_{xx}=0\) at \(x=0\); the second derivative is negative at every positive \(x\) in the range.

Worked Example: A Marginal Utility That Falls and Then Rises

Consider \(U(x,y)=2x^3-3x^2+10x+y\), for \(0\leq x\leq1\). Determine whether \(x\)’s marginal utility diminishes throughout this interval. Find the turning point and compare marginal utilities at \(x=0\), \(x=0.5\), and \(x=1\).

Method. Locate where the second derivative changes sign, then evaluate the first derivative to confirm what happens on either side.

Step 1: Differentiate with respect to \(x\). The marginal utility is \(U_x=6x^2-6x+10\).

Step 2: Find the second derivative and its sign change. Differentiating again gives \(U_{xx}=12x-6\). Setting this equal to zero gives \(12x-6=0\), so \(x=0.5\). For \(0\leq x<0.5\), \(12x-6<0\); for \(0.5<x\leq1\), \(12x-6>0\). Thus marginal utility falls before \(x=0.5\) and rises after it; it does not diminish throughout the interval.

Step 3: Check marginal-utility values. At \(x=0\), \(U_x=6(0)^2-6(0)+10=10\). At \(x=0.5\), \(U_x=6(0.5)^2-6(0.5)+10=1.5-3+10=8.5\). At \(x=1\), \(U_x=6(1)^2-6(1)+10=6-6+10=10\). It falls from 10 to 8.5, then rises from 8.5 to 10.

Check and common mistake avoided. The negative second derivative near \(x=0\) does not establish diminishing marginal utility on the entire interval. The sign change at \(0.5\), confirmed by the evaluated marginal utilities, rules out that full-interval conclusion.

Common Interpretation Errors

Confusing a marginal utility with its rate of change. \(U_x\) is the marginal utility of \(x\); \(U_{xx}\) tells how \(U_x\) changes as \(x\) changes.
Using a cross-partial for an own-good claim. \(U_{xy}\) concerns the effect of \(y\) on \(U_x\), not the effect of \(x\) on \(U_x\).
Checking only one bundle for an interval-wide claim. A sign at one point gives local information. Inspect \(U_{xx}\) throughout the interval before asserting that marginal utility declines throughout it.
Assuming a negative second partial means marginal utility is negative. It means marginal utility is falling, not that its level is below zero.
Ignoring what is held fixed. The own-good test changes one good’s quantity while holding the other good fixed. If both quantities change, the total change in marginal utility can involve both own and cross-partial effects.

For a twice continuously differentiable utility function, the mixed partials satisfy \(U_{xy}=U_{yx}\). This equality does not make a cross-partial an own-good second partial: both mixed partials still describe how one good’s marginal utility responds to the other good. The own-good tests remain \(U_{xx}\) and \(U_{yy}\).

Second partial derivatives provide a precise way to test diminishing marginal utility within a specified utility representation and range. They do not, by themselves, establish how a consumer chooses among affordable bundles or describe the curvature of indifference curves. This treatment is aligned with the SOA VEE Economics (Validation by Educational Experience) syllabus.

Key takeaway: To test diminishing marginal utility of good \(x\), compute \(U_{xx}\) and inspect its sign as \(x\) varies with \(y\) fixed. To test good \(y\), use \(U_{yy}\). Use a cross-partial such as \(U_{xy}\) only to describe how one good’s marginal utility responds to the other good.

Check Your Understanding

For each problem, identify the relevant derivative and state the conclusion in the specified context.

Let \(U(x,y)=18x-x^2+2xy+6y\). Find \(U_{xx}\) and \(U_{xy}\). Does \(x\)’s marginal utility decline as \(x\) rises with \(y\) fixed, and what is the local effect of increasing \(y\) on \(U_x\)? (Answer: \(U_{xx}=-2\), so \(x\)’s marginal utility declines; \(U_{xy}=2\), so increasing \(y\) raises \(U_x\).)
Let \(U(x,y)=24x-3x^2+5y-y^2\). At \(y=2\), calculate \(U_x\) at \(x=1\) and \(x=3\), and verify the own-good second-derivative test. (Answer: \(U_{xx}=-6\); \(U_x(1,2)=24-6+10=28\) and \(U_x(3,2)=24-18+10=16\), a decline of 12.)
Let \(U(x,y)=8x-x^3+2y+xy\), with \(0\leq x\leq1\). Find \(U_{xx}\) and determine whether \(x\)’s marginal utility declines throughout this interval for fixed \(y\). (Answer: \(U_{xx}=-6x\), which is nonpositive throughout and negative for \(x>0\); \(U_x\) declines over the interval.)
Let \(U(x,y)=x^3-3x^2+9x+4y\), with \(0\leq x\leq3\). Find where \(U_{xx}\) changes sign and state whether marginal utility diminishes throughout the interval. (Answer: \(U_{xx}=6x-6\), changing sign at \(x=1\); marginal utility falls for \(x<1\) and rises for \(x>1\), so it does not diminish throughout.)
Let \(U(x,y)=20x-x^2+3xy-2y^2\). At \((x,y)=(4,2)\), calculate \(U_x\), \(U_{xx}\), and \(U_{xy}\). Explain which derivative answers each of these two questions: whether \(x\)’s own marginal utility falls as \(x\) rises, and whether increasing \(y\) raises \(U_x\). (Answer: \(U_x=20-8+6=18\), \(U_{xx}=-2\), and \(U_{xy}=3\); \(U_{xx}\) shows the own marginal utility falls, while \(U_{xy}\) shows increasing \(y\) raises \(U_x\).)

What You'll Learn