Marginal Utility as a Partial Derivative

Why Marginal Utility Uses a Partial Derivative

A utility function assigns a numerical utility level to each consumption bundle. With two goods, write the function as \(U(x,y)\), where \(x\) and \(y\) are the quantities consumed. To ask how utility changes when consumption of \(x\) changes, the quantity of \(y\) must be held fixed. That one-good change is measured by a partial derivative.

This distinction is useful because a consumer's bundle has more than one quantity that could change. A change in utility associated with increasing \(x\) alone is not the same calculation as a change in utility when both \(x\) and \(y\) change. Here the focus is the first calculation: the marginal utility of one good, with the other good held constant.

Definition: If \(U(x,y)\) is differentiable, the marginal utility of good \(x\) at \((x,y)\) is \(MU_x=\frac{\partial U(x,y)}{\partial x}\), the partial derivative of utility with respect to \(x\), holding \(y\) fixed. The marginal utility of good \(y\) is \(MU_y=\frac{\partial U(x,y)}{\partial y}\), holding \(x\) fixed. These derivatives are local rates of change, measured in utility units per unit of the relevant good.

The phrase holding the other good fixed is part of the definition, not an optional simplifying assumption. When differentiating with respect to \(x\), treat \(y\) as a constant. When differentiating with respect to \(y\), treat \(x\) as a constant. If the utility function is differentiable at a bundle, these partial derivatives describe the instantaneous response of its numerical utility representation to a small change in one quantity.

The marginal utility is not itself the utility level at a bundle. The utility level is \(U(x,y)\); the marginal utilities are derivatives evaluated at that bundle. Their units also differ: if utility is measured in utility units and \(x\) is measured in units of a good, \(MU_x\) has units of utility units per unit of \(x\).

Deriving the Partial-Derivative Formula

The definition follows the ordinary derivative idea, but restricts the change to one input. Start at \((x,y)\), increase \(x\) by a small amount \(h\), and leave \(y\) unchanged. The resulting change in utility is \(U(x+h,y)-U(x,y)\). Divide by the change in \(x\), then take the limit as that change approaches zero.

MU_x = \lim_{h\to 0} \frac{U(x+h,y)-U(x,y)}{h} = \frac{\partial U(x,y)}{\partial x}

The corresponding definition for good \(y\) changes \(y\) by a small amount while keeping \(x\) fixed. For a polynomial, the limit can be verified by expanding the expression and cancelling the terms that vanish as \(h\) approaches zero.

Worked Example: Deriving a Polynomial Marginal Utility

Suppose \(U(x,y)=4x^2+3xy+2y^2\), with \(x,y\geq 0\). Derive the marginal utility of \(x\), then calculate it at the bundle \((2,1)\).

Method. Use the one-variable difference quotient, keeping \(y\) fixed. For the change from \(x\) to \(x+h\), the difference in utility is

\begin{aligned} U(x+h,y)-U(x,y) &=4\big((x+h)^2-x^2\big)+3\big((x+h)y-xy\big)\\ &=4(2xh+h^2)+3hy\\ &=8xh+4h^2+3hy. \end{aligned}

Divide by \(h\), for \(h\neq 0\), and take the limit:

\frac{U(x+h,y)-U(x,y)}{h}=8x+4h+3y, \qquad MU_x=\lim_{h\to 0}(8x+4h+3y)=8x+3y.

At \((x,y)=(2,1)\), \(MU_x=8(2)+3(1)=16+3=19\). For completeness, the utility level at this bundle is \(U(2,1)=4(2^2)+3(2)(1)+2(1^2)=16+6+2=24\). Thus the utility level is 24, while the marginal utility of \(x\) is 19 utility units per unit of \(x\).

Check. The \(2y^2\) term does not vary when \(x\) changes, so it contributes zero to \(MU_x\). The result \(8x+3y\) correctly includes the effects of the \(x^2\) and \(xy\) terms only. A common mistake is to differentiate every term as though every quantity were changing.

Polynomial Utility Functions

For polynomial utility functions, use the ordinary rules of differentiation on the variable of interest. A term that does not contain that variable is constant for the partial derivative and contributes zero. A cross term, such as \(xy\), does contribute: its derivative with respect to \(x\) is \(y\), because \(y\) is held fixed; its derivative with respect to \(y\) is \(x\).

For example, if \(U(x,y)=a x^2+bxy+c y^2\), where \(a,b,c\) are constants, then

MU_x=2ax+by, \qquad MU_y=bx+2cy.

These expressions are not utility levels. They are formulas for the local marginal utilities, which must still be evaluated at the specified quantities when a question asks for their values at a particular bundle.

Worked Example: Evaluating Both Marginal Utilities With a Cross Term

Let \(U(x,y)=2x^2y+3xy^2+4y\), where \(x,y\geq 0\). Find both marginal utilities and the utility level at \((1,2)\).

Method. Differentiate each term with respect to one variable at a time. For \(MU_x\), hold \(y\) fixed:

$$ MU_x=4xy+3y^2+0. $$

At \((1,2)\), \(MU_x=4(1)(2)+3(2^2)=8+12=20\). For \(MU_y\), hold \(x\) fixed. The derivative of \(2x^2y\) is \(2x^2\), that of \(3xy^2\) is \(6xy\), and that of \(4y\) is 4:

MU_y=2x^2+6xy+4, \qquad MU_y(1,2)=2(1^2)+6(1)(2)+4=2+12+4=18.

Now evaluate the utility level separately:

$$ U(1,2)=2(1^2)(2)+3(1)(2^2)+4(2)=4+12+8=24. $$

Check. The cross term contributes \(3y^2=12\) to \(MU_x\) and \(6xy=12\) to \(MU_y\) at the bundle. The answers are \(MU_x=20\), \(MU_y=18\), and \(U=24\). The utility level is not interchangeable with either derivative.

Product Utility Functions

A product utility function can be differentiated with the product rule and, when necessary, the chain rule. If one factor depends only on \(x\) and another only on \(y\), then the factor that depends on the unchanged good is constant during the partial differentiation. For instance, if \(U(x,y)=f(x)g(y)\), then differentiating with respect to \(x\) gives \(f'(x)g(y)\), because \(g(y)\) is fixed with respect to \(x\). Differentiating with respect to \(y\) gives \(f(x)g'(y)\).

For a factor raised to a power, apply the chain rule to differentiate the inner expression as well. The next example shows how to keep track of both the factor and its exponent.

Worked Example: Differentiating a Product With Powers

Suppose \(U(x,y)=(x+2)^2(2y+1)^3\), with \(x,y\geq 0\). Calculate \(MU_x\), \(MU_y\), and \(U\) at \((1,0)\).

Method. For \(MU_x\), the factor \((2y+1)^3\) is constant because \(y\) is held fixed. Differentiate \((x+2)^2\):

MU_x=2(x+2)(2y+1)^3, \qquad MU_x(1,0)=2(3)(1^3)=6.

For \(MU_y\), the factor \((x+2)^2\) is constant. The chain rule differentiates \((2y+1)^3\): its derivative is \(3(2y+1)^2\cdot 2\), where 2 is the derivative of the inner expression \(2y+1\). Therefore,

MU_y=(x+2)^2\cdot 3(2y+1)^2\cdot 2, \qquad MU_y(1,0)=3^2\cdot 3\cdot 1^2\cdot 2=9\cdot 3\cdot 1\cdot 2=54.

Finally, evaluate the function itself: \(U(1,0)=(1+2)^2(2(0)+1)^3=3^2\cdot 1^3=9\). The results are \(MU_x=6\), \(MU_y=54\), and \(U=9\). Substituting these into the derivative formulas gives the same values, which checks the calculations.

Common mistake avoided. When finding \(MU_x\), adding the derivative of the \(y\)-factor would incorrectly treat \(y\) as changing. For example, that error would add \(54\) to \(MU_x\) here and produce 60. The partial derivative with respect to \(x\) is 6 because \(y\) is fixed; the value 54 is the partial derivative with respect to \(y\).

Interpretation, Scope, and Common Traps

A marginal utility is a local rate of change in a specified numerical utility representation. If \(MU_x\) is positive at a bundle, the representation's utility level increases locally when \(x\) increases slightly while \(y\) remains fixed. If it is negative, the represented utility level decreases locally under that one-good change. A zero derivative means the local first-order rate of change in that direction is zero; by itself, it does not establish that utility is constant for all changes in \(x\).

The numerical magnitude of marginal utility depends on the utility representation. A strictly increasing transformation can preserve the ranking of bundles while changing derivative values. In particular, if \(\widetilde U=g(U)\) and \(g\) is differentiable, the chain rule gives \(\frac{\partial \widetilde U}{\partial x}=g'(U)\frac{\partial U}{\partial x}\). Thus, the derivative value is not automatically a representation-independent measure of satisfaction. This is consistent with the distinction between ordinal and cardinal utility.

The derivative also assumes differentiability at the point being evaluated. At a point where a function has a kink or is otherwise not differentiable, the partial derivative may not exist, so the definition cannot be applied there in the usual way. And a partial derivative is not a finite change in utility: it is the limiting rate for an increasingly small change in one quantity.

Do not differentiate with respect to both goods at once. A partial derivative changes one input and fixes the other.
Do not confuse a function value with its derivative. Calculate \(U(x,y)\) separately if the utility level is also requested.
Keep cross terms. The term \(xy\) contributes \(y\) to \(MU_x\) and \(x\) to \(MU_y\).
Apply the chain rule to inner expressions. The derivative of \((2y+1)^3\) includes the factor 2.
State the bundle. A marginal-utility formula becomes a numerical marginal utility only after evaluating it at the specified values of \(x\) and \(y\).

Calculating derivatives carefully provides a precise way to describe how a utility representation responds locally to a change in one good. This partial-derivative technique is aligned with the SOA VEE Economics (Validation by Educational Experience) syllabus.

Check Your Understanding

For each problem, calculate the requested derivatives at the stated bundle, and distinguish those values from the utility level where it is requested.

Let \(U(x,y)=5x^2+2xy+y^2\). Find \(MU_x\), \(MU_y\), and \(U\) at \((1,3)\). (Answer: \(MU_x=16,\ MU_y=8,\ U=20\).)
Let \(U(x,y)=(2x+1)(y+2)^2\). Find both marginal utilities and the utility level at \((1,1)\). (Answer: \(MU_x=18,\ MU_y=18,\ U=27\).)
Let \(U(x,y)=x^2y+4xy^2+3y\). At \((2,1)\), find \(MU_x\), \(MU_y\), and \(U\). (Answer: \(MU_x=8,\ MU_y=23,\ U=15\).)
Let \(U(x,y)=(x+1)^3(2y+1)^2\). At \((1,0)\), calculate both partial derivatives and the utility level. Include the chain-rule factor when differentiating with respect to \(y\). (Answer: \(MU_x=12,\ MU_y=32,\ U=8\).)
Let \(U(x,y)=6x^2+3xy+2y^2\). At \((0.5,2)\), find both marginal utilities and the utility level. (Answer: \(MU_x=12,\ MU_y=9.5,\ U=12.5\).)

What You'll Learn