# Partial Derivatives

Business I · Mathematics II · https://tryals.app/learn/business-i/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?

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

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**: extra output from one more unit of labour, holding capital fixed.

| Partial | Name | Reads as |
|---|---|---|
| $\partial Q/\partial L$ | Marginal product of labour | One more worker, capital frozen |
| $\partial Q/\partial K$ | Marginal product of capital | One more machine, labour frozen |
| $\partial U/\partial x_i$ | Marginal utility of good $i$ | One more unit, basket frozen |

## ## Cross-Partials & Optimization

**Cross-partials** are second-order mixed partials like $\partial^2 f / \partial x \partial y$. By **Young's theorem**, if mixed partials are continuous, order does not matter: $f_{xy} = f_{yx}$.

Economically, cross-partials show if inputs are **complements** (marginal product rises with the other input) or **substitutes**.

**The gradient** $\nabla f = (f_x, f_y)$ points to steepest ascent. At an optimum, the first-order conditions require:

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

> **Common pitfall:** A partial is a laboratory experiment holding things constant. In reality, hiring more labour usually changes capital too. For total effects, use the total differential.

## Practice questions

6 of this lesson's 12 practice questions, with answers. The full set is in the app.

### 1. Match each partial-derivative concept to its economic meaning.

**Answer:**

- $\partial Q / \partial L$ → Marginal product of labour
- $\partial Q / \partial K$ → Marginal product of capital
- $\partial^2 Q / \partial L \partial K$ → Whether inputs are complements
- $\nabla f = 0$ → First-order optimality condition

**Why:** Partials give marginal products; cross-partials reveal complementarity; the zero gradient defines optima.

Page: https://tryals.app/practice/business-i/partial-derivatives/match-each-partial-derivative-concept-to-its-economic-meaning

### 2. Differentiating $f(x,y) = 5x^{3}y^{2}$ with respect to $y$ gives $10x^{3}y$, with the $x^{3}$ carried through untouched. Why does the $x$ term survive unchanged?

A. Because $x^{3}$ and $y^{2}$ are independent variables and neither can be differentiated
B. Because $x^{3}$ would only change if the derivative were taken with respect to $x$ and $y$ together
C. Because the power rule applies only to the variable with the higher exponent
D. Because a partial derivative holds $x$ fixed, so $5x^{3}$ is constant.

**Answer:** D. Because a partial derivative holds $x$ fixed, so $5x^{3}$ is constant.

**Why:** Holding the other variable fixed turns its whole factor into a coefficient. This is exactly why the marginal product of labour still depends on the capital stock.

Page: https://tryals.app/practice/business-i/partial-derivatives/differentiating-f-x-y-5x-y-with-respect-to-y-gives-10x-y-with

### 3. A firm's output depends on capital and labour, and the partial derivative with respect to labour is called the marginal product of labour. What does 'holding capital constant' correspond to in the business?

A. The long run, in which all inputs can be adjusted
B. The assumption that capital and labour are independent inputs
C. The short run, where capital is fixed and labour varies
D. A modelling simplification with no business counterpart

**Answer:** C. The short run, where capital is fixed and labour varies

**Why:** The short run is defined by at least one fixed input, which is the same condition the partial derivative imposes, so the mathematics and the economics agree by construction. Option 3 is a different claim: partials hold other inputs fixed whether or not the inputs interact, and cross-partials measure that interaction.

Page: https://tryals.app/practice/business-i/partial-derivatives/a-firms-output-depends-on-capital-and-labour-and-the-partial

### 4. For $f(x,y) = x^2 y + e^y$, which are correct?

A. $f_x = 2xy$
B. $f_{xx} = 2y$
C. $f_{xy} = 2y$
D. $f_y = x^2 + e^y$

**Answer:** A. $f_x = 2xy$; B. $f_{xx} = 2y$; D. $f_y = x^2 + e^y$

**Why:** $f_x = 2xy$, $f_y = x^2 + e^y$, $f_{xx} = 2y$. But $f_{xy} = \partial(2xy)/\partial y = 2x$, not $2y$.

Page: https://tryals.app/practice/business-i/partial-derivatives/for-f-x-y-x-y-e-y-which-are-correct

### 5. Young's theorem guarantees $f_{xy} = f_{yx}$ for every function of two variables.

**Answer:** False

**Why:** False. Equality holds when the mixed partials are **continuous** at the point. Counterexamples exist, the standard one built from $xy(x^2-y^2)/(x^2+y^2)$ has $f_{xy}(0,0) \neq f_{yx}(0,0)$, so the hypothesis is doing real work, even though it holds for essentially every function met in economics.

Page: https://tryals.app/practice/business-i/partial-derivatives/youngs-theorem-guarantees-fxy-fyx-for-every-function-of-two

### 6. The mixed partials $f_{xy}$ and $f_{yx}$ are equal for the functions used in economics. What does that equality say about the two inputs?

A. That the two inputs contribute equally to output
B. That the production function must be constant returns to scale
C. That the order of differentiation is arbitrary for any function whatever
D. That the complementarity is one number, not two

**Answer:** D. That the complementarity is one number, not two

**Why:** Symmetry means the interaction between two inputs is a single quantity. It is also a modelling constraint, not a mathematical inevitability: it requires the continuity conditions to hold.

Page: https://tryals.app/practice/business-i/partial-derivatives/the-mixed-partials-fxy-and-fyx-are-equal-for-the-functions-used-in