Partial Derivatives and Optimisation in Two Variables

Teaching resource for Intermediate Microeconomics (Tutorial 1) and Mathematics, Sciences Po Paris.

Almost nothing in microeconomics depends on one variable. Utility depends on several goods, output on several inputs, a voter’s payoff on the tax rate and on her income. Partial derivatives are how you take a derivative when there is more than one variable in the room, and they are the whole engine behind marginal utility, the MRS, marginal products and first-order conditions. This note assumes you can differentiate in one variable; if not, start with the derivatives refresher.

1. What a partial derivative is

Take f(x, y). The partial derivative of f with respect to x asks: if I nudge x a little and hold y exactly fixed, how much does f move? It is written ∂f/∂x or f_x.

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

Geometrically, f(x, y) is a surface. Slicing it with a vertical plane at a fixed y gives a one-variable curve, and f_x is the slope of that curve. There is one partial derivative per variable, and together they form the gradient ∇f = (f_x, f_y).

2. How to compute one

Rule. To differentiate with respect to x, treat every other variable as if it were a number. Then use the ordinary one-variable rules.

The example from the Tutorial 1 slides, g(x₁, x₂) = x₁x₂² − 3x₁:

\frac{\partial g}{\partial x_1} = x_2^2 - 3 \] \[ \frac{\partial g}{\partial x_2} = 2x_1 x_2

For ∂g/∂x₁, the factor x₂² is a constant multiplying x₁, so it survives, and −3x₁ gives −3. For ∂g/∂x₂, the term −3x₁ is a constant and disappears, and x₁x₂² is a constant times x₂².

Three more that you will meet constantly:

U = x^{a} y^{b}: \] \[ U_x = a\,x^{a-1} y^{b} \] \[ U_y = b\,x^{a} y^{b-1}

U = \ln x + \ln y: \] \[ U_x = \frac1x \] \[ U_y = \frac1y

f = e^{2x} \ln y: \] \[ f_x = 2e^{2x} \ln y \] \[ f_y = \frac{e^{2x}}{y}

The chain rule works exactly as before: in the last example the factor 2 comes from differentiating 2x inside the exponential.

3. Reading them as economics

Marginal utility. U_x = ∂U/∂x is the extra utility from one more unit of x, holding y fixed. Diminishing marginal utility is U_xx < 0.
Marginal product. For Q(K, L), Q_L is the marginal product of labour. For Cobb-Douglas Q = A K^α L^β, Q_L = β Q/L: the exponent is the output elasticity.
Marginal rate of substitution. The slope of an indifference curve is −U_x/U_y; the MRS is the ratio U_x/U_y. For U = x^a y^b it is (a/b)(y/x). This ratio is what you set equal to the price ratio in a consumer problem.
Elasticities. (∂Q/∂p)(p/Q) is a partial derivative made unit-free. For Q = A p⁻² m, the price elasticity is −2 and the income elasticity is 1: the exponents.

Why the MRS is a ratio of partials. Along an indifference curve U(x, y) = c, the total change is zero:

dU = U_x\,dx + U_y\,dy = 0 \quad\Longrightarrow\quad \frac{dy}{dx} = -\frac{U_x}{U_y}

That is the total differential, and it gives the slope of any level curve F(x, y) = c as −F_x/F_y without ever solving for y. On the circle x² + y² = 25 at (3, 4) the slope is −3/4.

4. Second-order partials

Differentiate a partial derivative again and you get four second-order partials: f_xx, f_yy, and the two cross partials f_xy (differentiate f_x with respect to y) and f_yx. For any function you will meet in this course the cross partials are equal, f_xy = f_yx (Young’s theorem). Use that as a free check on your algebra.

For f = x³y − 2xy² + y: f_x = 3x²y − 2y², f_y = x³ − 4xy + 1, so

f_{xx} = 6xy \] \[ f_{yy} = -4x \] \[ f_{xy} = f_{yx} = 3x^2 - 4y

5. Unconstrained optimisation in two variables

First-order conditions. At a maximum or minimum the surface is flat in every direction, so every partial derivative is zero. That gives a system of as many equations as variables:

f_x(x^*, y^*) = 0 \] \[ f_y(x^*, y^*) = 0

Second-order conditions. A flat point can be a hilltop, a valley bottom or a saddle (up one way, down the other). Compute

D \;=\; f_{xx}\, f_{yy} - \big(f_{xy}\big)^2

at the critical point, and read:

Condition	Verdict
D > 0 and f_xx > 0	minimum
D > 0 and f_xx < 0	maximum
D < 0	saddle point
D = 0	the test is silent; look at the function directly

The slide example, f(y, z) = y² − 2y + z²: the conditions 2y − 2 = 0 and 2z = 0 give (1, 0); then f_yy = 2, f_zz = 2, f_yz = 0, so D = 4 > 0 with f_yy > 0: a minimum. Completing the square confirms it: f = (y − 1)² + z² − 1.

A saddle for contrast, f(x, y) = x² − y² + 2x: the critical point is (−1, 0), but D = (2)(−2) − 0 = −4 < 0. The first-order conditions found a flat point that is neither a maximum nor a minimum. This is why the second-order check is not optional.

When the two first-order conditions involve both variables, solve them as a system. A firm with profit π = 20q₁ + 30q₂ − q₁² − q₂² − q₁q₂ has π₁ = 20 − 2q₁ − q₂ and π₂ = 30 − 2q₂ − q₁; solving gives q₁ = 10/3, q₂ = 40/3, and D = 4 − 1 = 3 > 0 with π₁₁ < 0 confirms a maximum.

6. One variable, but with parameters: the Problem Set 1 model

Often you differentiate with respect to one variable while others sit in the formula as parameters. That is a partial derivative too, and it is exactly what Problem 3 of Problem Set 1 asks for. An agent with income y_i gets

V(\tau, y_i) = (1 - \tau)\,y_i + (\tau - \tau^2)\,\bar y

from a tax rate τ. Here y_i and ȳ are parameters; τ is the variable.

\frac{\partial V}{\partial \tau} = -y_i + (1 - 2\tau)\,\bar y \] \[ \frac{\partial^2 V}{\partial \tau^2} = -2\bar y < 0

The second derivative is negative, so V is strictly concave in τ: single-peaked preferences, one best tax rate per agent. Setting the first derivative to zero:

\tau_i^* = \frac12\left(1 - \frac{y_i}{\bar y}\right)

for y_i ≤ ȳ, and zero for anyone richer than average. Now differentiate the answer with respect to the parameter: ∂τ_i*/∂y_i = −1/(2ȳ) < 0. Richer agents prefer lower taxes. That last step, differentiating a solution with respect to a parameter to see how the solution moves, is called comparative statics, and you will do it in every part of this course.

7. Common mistakes

Forgetting that the other variable is a constant. In ∂(x²y)/∂x the answer is 2xy, not 2x: the y stays.
Dropping a term that does contain the variable. In ∂(x₁x₂² − 3x₁)/∂x₁ both terms contain x₁, so both contribute.
Confusing f_xy with f_xx f_yy. The cross partial is a single second derivative, taken in x then in y.
Stopping at the first-order conditions. Without D you do not know whether you found a maximum, a minimum or a saddle.
Evaluating D in general when it depends on the point. Plug in each critical point separately; the sign can differ from one to the next.
Treating the parameter as the variable in Problem Set 1. You differentiate V with respect to τ, not y_i; y_i only comes back when you ask how τ_i* changes with it.

Practice

Five to check yourself. The full practice sheet has twenty-two more, in three levels.

1.
Compute both partials of \( f(x, y) = x^2 y^3 + e^{y} - 4x \).
Solution
\[ f_x = 2xy^3 - 4 \] \[ f_y = 3x^2 y^2 + e^{y} \]
For \( f_x \), \( e^y \) is a constant and vanishes; for \( f_y \), \( -4x \) vanishes.
2.
For \( U(x, y) = x^{1/4} y^{3/4} \), find the MRS and evaluate it at \( (1, 3) \).
Solution
\[ MRS = \frac{U_x}{U_y} = \frac{\tfrac14 x^{-3/4} y^{3/4}}{\tfrac34 x^{1/4} y^{-1/4}} = \frac{y}{3x} \]
At \( (1, 3) \) the MRS is \( 1 \). General pattern for \( x^a y^b \): \( MRS = \frac{a}{b}\cdot\frac{y}{x} \).
3.
Find the slope of the level curve \( xy = 12 \) at \( (3, 4) \) using partial derivatives.
Solution
\[ \frac{dy}{dx} = -\frac{F_x}{F_y} = -\frac{y}{x} = -\frac43 \]
Check by solving explicitly: \( y = 12/x \), \( y' = -12/x^2 = -12/9 = -4/3 \). Same answer, but the partial-derivative route works even when you cannot solve for \( y \).
4.
Find and classify the critical point of \( f(x, y) = -2x^2 - y^2 + 8x + 2y \).
Solution
\[ f_x = -4x + 8 = 0 \Rightarrow x = 2 \] \[ f_y = -2y + 2 = 0 \Rightarrow y = 1 \]
\[ f_{xx} = -4, \quad f_{yy} = -2, \quad f_{xy} = 0 \] \[ D = 8 > 0 \]
\( D > 0 \) and \( f_{xx} < 0 \): a maximum at \( (2, 1) \), where \( f = -8 - 1 + 16 + 2 = 9 \).
5.
In the Problem Set 1 model, suppose average income is \( \bar y = 0.5 \). Which agents prefer a tax rate of at least \( 0.25 \)? What does the median voter choose if median income is \( 0.4 \)?
Solution
\[ \tau_i^* = \frac12\left(1 - \frac{y_i}{0.5}\right) = \frac12 - y_i \geq 0.25 \quad\Longleftrightarrow\quad y_i \leq 0.25 \]
Agents with income up to \( 0.25 \), half the average, want \( \tau \geq 0.25 \). The median voter with \( y_m = 0.4 \) chooses \( \tau^* = 0.5 - 0.4 = 0.1 \): a positive tax, because the median is below the mean. In the problem set the distribution was uniform, median equal to mean, and the tax was zero.

Constrained problems, where a budget or a technology limits the choice, are the next step: Lagrange multipliers. All notes and practice sheets are on the resources page.

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