Difficulty: Introductory-Intermediate | Prerequisites: Basic calculus (derivatives, chain rule)
Microeconomics leans heavily on calculus for optimisation: firms maximise profit, consumers maximise utility, and the tool for finding maxima and minima is setting derivatives (or partial derivatives) equal to zero. This review covers the mechanical skills you need before the course layers on economic interpretation. If you are comfortable with the power rule, the chain rule, and basic graphing, you are ready.
Partial derivatives work the same as ordinary derivatives, except you treat every other variable as a constant. To find maxima or minima, set each partial derivative equal to zero and solve. When graphing, rewrite the equation to isolate y, find where the function crosses each axis, and note the shape (line, parabola).
Partial derivative (∂f/∂x)
The rate of change of a function f with respect to one variable (x), holding all other variables constant. Notation: ∂f/∂x or f_x.
Think of it as: an ordinary derivative where you pretend the other variables are just numbers.
First-order condition (FOC)
The equation obtained by setting a derivative (or partial derivative) equal to zero. First-order conditions are necessary for a maximum or minimum of a smooth function.
In simple terms, this means: at a peak or a trough, the slope is flat, so the derivative is zero.
Critical point
A point where all first-order conditions are satisfied. Could be a maximum, minimum, or saddle point. The second-order condition (checking the sign of the second derivative) tells you which.
Vertex of a parabola
The highest or lowest point of a quadratic function y = a(x - h)^2 + k. The vertex is at (h, k). If a < 0 the parabola opens downward (vertex is a maximum). If a > 0 it opens upward (vertex is a minimum).
Inverse function (in the graphing context)
Rewriting Q = f(P) as P = g(Q), or y = f(x) in a form ready to graph. Economists graph P on the vertical axis, so they routinely invert demand and supply functions.
∂f/∂x: The exp term has no x in it, so it drops out. Differentiate 2(x - 1)^3 with the chain rule:
∂f/∂x = 6(x - 1)^2
∂f/∂y: The 2(x - 1)^3 term has no y, so it drops out. Differentiate exp((y - 1)^2) with the chain rule:
∂f/∂y = 2(y - 1) · exp((y - 1)^2)
∂f/∂x: Treat y^4 as a constant. Power rule on x^2:
∂f/∂x = 8x y^4
∂f/∂y: Treat 4x^2 as a constant. Power rule on y^4:
∂f/∂y = 16x^2 y^3
Simplify first: ln(x^3) = 3 ln(x). So f = 3y ln(x) + (1/9) y^3 x^(-1).
∂f/∂x: Differentiate each term with respect to x:
d/dx [3y ln(x)] = 3y / x
d/dx [(1/9) y^3 x^(-1)] = -(1/9) y^3 x^(-2) = -y^3 / (9x^2)
∂f/∂x = 3y/x - y^3/(9x^2)
∂f/∂y: Differentiate each term with respect to y:
d/dy [3y ln(x)] = 3 ln(x)
d/dy [(1/9) y^3 x^(-1)] = (3/9) y^2 x^(-1) = y^2 / (3x)
∂f/∂y = 3 ln(x) + y^2/(3x)
This is a downward-opening parabola (the coefficient on the squared term is -3, which is negative).
Vertex: The expression (x + 1)^2 is minimised when x = -1, giving y = 10 - 0 = 10. So the vertex is at (-1, 10).
y-intercept: Set x = 0. y = 10 - 3(1)^2 = 10 - 3 = 7. The y-intercept is (0, 7).
x-intercepts: Set y = 0. 0 = 10 - 3(x + 1)^2, so (x + 1)^2 = 10/3, giving x + 1 = ±√(10/3). Therefore x = -1 ± √(10/3) ≈ -1 ± 1.83. The x-intercepts are approximately (-2.83, 0) and (0.83, 0).
First, solve for y:
2y = x - 5 + 3(x - 1) = x - 5 + 3x - 3 = 4x - 8
y = 2x - 4
This is a straight line with slope 2 and y-intercept -4.
y-intercept: Set x = 0. y = -4. The point is (0, -4).
x-intercept: Set y = 0. 0 = 2x - 4, so x = 2. The point is (2, 0).
Expand: f(y) = 5(y^2 + 4y + 4) - 10y^2 = 5y^2 + 20y + 20 - 10y^2 = -5y^2 + 20y + 20
FOC: f'(y) = -10y + 20 = 0
Solve: y = 2
Note: f''(y) = -10 < 0, confirming this is a maximum.
FOC: f'(y) = 6y^2 - 6y - 12 = 0
Divide by 6: y^2 - y - 2 = 0
Factor: (y - 2)(y + 1) = 0
Solutions: y = 2 or y = -1
To determine which is a max vs min, check the second derivative: f''(y) = 12y - 6.
At y = 2: f''(2) = 24 - 6 = 18 > 0, so y = 2 is a local minimum.
At y = -1: f''(-1) = -12 - 6 = -18 < 0, so y = -1 is a local maximum.
Since the question asks for maximisation, the relevant critical point is y = -1.
This has two variables, so we need two FOCs.
∂f/∂x: Use the chain rule on -3(3x - 15)^2:
∂f/∂x = -3 · 2(3x - 15) · 3 = -18(3x - 15)
Set equal to zero: -18(3x - 15) = 0, so 3x - 15 = 0, giving x = 5.
∂f/∂y: Use the chain rule on -4(6 - 2y)^2:
∂f/∂y = -4 · 2(6 - 2y) · (-2) = 16(6 - 2y)
Set equal to zero: 16(6 - 2y) = 0, so 6 - 2y = 0, giving y = 3.
Solution: x = 5, y = 3
Both squared terms have negative coefficients (-3 and -4), so the function is concave in both variables, confirming this is a maximum. The maximum value is f(5, 3) = 0 - 0 + 10 = 10.
Power rule
d/dx [x^n] = n x^(n-1)
Chain rule
d/dx [f(g(x))] = f'(g(x)) · g'(x)
In words: derivative of the outside times derivative of the inside.
Product rule
d/dx [u · v] = u'v + uv'
Logarithm derivative
d/dx [ln(x)] = 1/x
d/dx [ln(g(x))] = g'(x) / g(x) (chain rule applied)
Exponential derivative
d/dx [e^x] = e^x
d/dx [e^(g(x))] = g'(x) · e^(g(x))
Useful simplification
ln(x^n) = n ln(x). Always simplify logarithms before differentiating.
Treating a constant term as if it depends on the variable. When taking ∂f/∂x, every term without an x in it differentiates to zero. Students sometimes carry constant terms through as if they were part of the derivative.
Forgetting the chain rule. The derivative of (3x - 15)^2 is not 2(3x - 15). You must multiply by the derivative of the inside: 2(3x - 15) · 3 = 6(3x - 15).
Confusing FOC solutions with maxima. A first-order condition gives you critical points. Not all critical points are maxima. Check the second derivative: negative means maximum, positive means minimum.
Mixing up x-intercept and y-intercept. To find the y-intercept, set x = 0. To find the x-intercept, set y = 0. This seems obvious on paper but is a common exam slip under time pressure.
⚠️ Partial derivatives appear in every optimisation problem from here on: profit maximisation, cost minimisation, utility maximisation. The mechanics must be automatic.
⚠️ You will need to graph supply and demand curves from equations. Practice converting Q = f(P) to P = g(Q) quickly.
⚠️ FOC problems are almost guaranteed on the midterm. Expect to differentiate, set equal to zero, solve, and then verify with the second derivative.
⚠️ Show your working. Partial credit on exams comes from demonstrating each step.
Fill in the blank: ∂/∂x [7x^3 y^2] = ______
21x^2 y^2
True or false: The derivative of ln(x^5) with respect to x is 5/x.
True. ln(x^5) = 5 ln(x), and d/dx [5 ln(x)] = 5/x.
Fill in the blank: The y-intercept of y = 3x + 7 is ______.
7 (set x = 0).
True or false: If f''(x) > 0 at a critical point, that point is a maximum.
False. f''(x) > 0 means the function is concave up, so the critical point is a minimum.
Q: Find ∂f/∂x and ∂f/∂y for f(x, y) = 3x^2 y + y^3 ln(x).
A: ∂f/∂x = 6xy + y^3/x. ∂f/∂y = 3x^2 + 3y^2 ln(x).
Q: A function is f(y) = -2y^2 + 8y + 3. Find the value of y that maximises f.
A: f'(y) = -4y + 8 = 0, so y = 2. f''(y) = -4 < 0, confirming a maximum.
Q: Graph y = -(x - 2)^2 + 9. What are the vertex and the x-intercepts?
A: Vertex at (2, 9). For x-intercepts: 0 = -(x - 2)^2 + 9, so (x - 2)^2 = 9, giving x - 2 = ±3. x-intercepts at x = -1 and x = 5.
Q: Find the critical points of f(x, y) = -(x - 4)^2 - 2(y + 1)^2 + 7 and determine whether they are maxima or minima.
A: ∂f/∂x = -2(x - 4) = 0 gives x = 4. ∂f/∂y = -4(y + 1) = 0 gives y = -1. Both squared terms have negative coefficients, so the function is concave and (4, -1) is a maximum. f(4, -1) = 7.
The chain rule and partial derivative skills here are the mechanical foundation for consumer theory (maximising utility subject to a budget constraint) and producer theory (minimising cost, maximising profit). Lagrange multipliers, which you will meet later, are an extension of the FOC approach to constrained optimisation.
Graphing skills carry directly into reading and drawing supply-demand diagrams, indifference curves, isoquants, and cost curves.
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