Partial Derivatives

One Variable at a Time

When a function depends on several variables, the partial derivative asks: how does the output change when I nudge one input, holding all others constant?

$$f(x,y) = x^2 y + 3xy^2 \qquad \frac{\partial f}{\partial x} = 2xy + 3y^2 \qquad \frac{\partial f}{\partial y} = x^2 + 6xy$$

The $\partial$ symbol replaces $d$ to remind you: other variables are frozen. Mechanically, differentiate with respect to the chosen variable and treat everything else as a constant.

Economic interpretation: in a production function $Q(K,L)$, the partial $\partial Q / \partial L$ is the marginal product of labour — how much extra output one more unit of labour produces, holding capital fixed. Similarly, $\partial Q / \partial K$ is the marginal product of capital.

Cross-partials (second-order mixed partials): $\partial^2 f / \partial x \partial y$ differentiates first with respect to $y$, then $x$. Young's theorem: if both mixed partials are continuous, order doesn't matter — $f_{xy} = f_{yx}$. Economically, the cross-partial tells you whether inputs are complements (the marginal product of one rises when you add more of the other) or substitutes.

The gradient $\nabla f = (f_x, f_y)$ points in the direction of steepest ascent. At an optimum (a peak or trough), $\nabla f = \mathbf{0}$ — both partials are zero. This is the first-order condition for multivariable optimisation, and it says: you can't improve by moving in any direction.

$$\text{First-order conditions}: \quad f_x = 0, \quad f_y = 0$$

Partial derivatives are the calculus of ceteris paribus — the economist's favourite phrase turned into mathematics. Every marginal concept in micro and macro is a partial derivative in disguise.

Partial derivatives in economics

Common pitfall: Forgetting what "holding others constant" costs you. A partial is a laboratory experiment, not a forecast — in the real economy, hiring more labour usually changes capital use too. For total effects, you need the total differential, not one partial.