Maths for Intermediate Microeconomics

Tutorial 1 companion, Intermediate Microeconomics, Sciences Po Paris.

Several of you wrote after the first tutorial asking for material to go over the maths and for exercises to practise. This page is that answer. It follows the order of the Tutorial 1 slides (which are on Moodle) and, for each topic, gives you the essentials in a few lines, a note that explains it properly, and exercises with solutions you can reveal one by one. You do not need all of it at once. Work through the parts you found hardest, and email me when something does not click.

If what slows you down is not the calculus but the algebra around it (adding fractions, dividing by a fraction, negative or fractional exponents, solving for x when it sits in a denominator or under a root), start with Fractions and Exponents and its three practice sheets. Every topic below leans on it.

1. Derivatives in one variable

The core of everything else. You need the rules by heart and the chain rule until it is automatic.

(u+v)' = u' + v' \] \[ (uv)' = u'v + uv' \] \[ \left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2} \] \[ (x^{\alpha})' = \alpha x^{\alpha - 1} \] \[ (e^{u})' = u' e^{u} \] \[ (\ln u)' = \frac{u'}{u} \] \[ \big(f(g(x))\big)' = f'(g(x))\,g'(x)

Read: Derivatives: A Refresher, especially section 3 on the chain rule.
Practise: Starter (24, the basic rules), Intermediate (100, product and quotient rules), Upper intermediate (20, chain rule and second derivatives).

The slide exercise f(x) = 3(ln x)² is a chain rule: outer 3u², inner ln x, so f′ = 6 ln x · (1/x).

2. Logarithms and exponentials

The rules you will use in every problem set:

\ln(uv) = \ln u + \ln v \] \[ \ln\frac{u}{v} = \ln u - \ln v \] \[ \ln(u^{\alpha}) = \alpha \ln u \] \[ e^{\ln u} = u

u^{\alpha} u^{\beta} = u^{\alpha+\beta} \] \[ \frac{u^{\alpha}}{u^{\beta}} = u^{\alpha-\beta} \] \[ (u^{\alpha})^{\beta} = u^{\alpha\beta} \] \[ u^{-\alpha} = \frac{1}{u^{\alpha}}

Two traps from the slides: (ln u)^α is not α ln u, and u^α + u^β does not simplify. Also, ln is only defined for positive numbers, and ln 1 = 0.

Why we care: taking logs turns a Cobb-Douglas utility x^a y^b into a ln x + b ln y, which is easier to differentiate and describes the same preferences, since ln is increasing. And a coefficient in a log-log regression is an elasticity.

3. Partial derivatives

When a function has several variables, differentiate with respect to one while treating the others as constants. Marginal utility, marginal product and the MRS are all partial derivatives:

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

Read: Partial Derivatives and Optimisation in Two Variables.
Practise: Partial Derivatives Practice, 22 exercises in three levels, the last ones built on Problem Set 1.

4. Unconstrained optimisation

One variable: set f′(x) = 0 and solve; f″ < 0 means a maximum, f″ > 0 a minimum. Two variables: set both partials to zero and solve the system; then compute

D = f_{xx} f_{yy} - f_{xy}^2

and read: D > 0 with f_xx < 0 is a maximum, D > 0 with f_xx > 0 a minimum, D < 0 a saddle. The slide example f(y, z) = y² − 2y + z² has its critical point at (1, 0) with D = 4 > 0 and f_yy = 2 > 0: a minimum.

Problem 3 of Problem Set 1 is this in one variable with parameters: V(τ, y_i) is concave in τ because ∂²V/∂τ² = −2ȳ < 0, so the first-order condition gives each agent’s preferred tax rate. Sections 5 and 6 of the partial derivatives note work through it.

5. Constrained optimisation: the Lagrangian

The consumer problem, max U(x₁, x₂) subject to p₁x₁ + p₂x₂ = R. Form the Lagrangian, take the partials, and the first two conditions give MRS = price ratio:

\mathcal{L} = U(x_1, x_2) + \lambda\,\big(R - p_1 x_1 - p_2 x_2\big) \] \[ \Longrightarrow\quad \frac{U_{x_1}}{U_{x_2}} = \frac{p_1}{p_2}

Read: Lagrange Multipliers: Optimisation with Constraints. It also explains what λ means, which the exam likes to ask.
Practise: Lagrange Multipliers: Practice, 20 problems from numbers-only to Kuhn-Tucker. The slide problem, max x₁x₂ subject to 2x₁ + 3x₂ = 6, is the same structure as the first ones.

A note on the sign of λ: the slides write +λ(R − g), my notes write −λ(g − c). They are the same thing; keep whichever you like and be consistent.

6. Probability and expectation

A random variable takes values with probabilities. Its expectation is the probability-weighted average:

E[X] = \sum_i p_i\, x_i \] \[ X \sim \mathcal{U}[a, b] \text{ (continuous)}: \quad E[X] = \frac{a+b}{2}

The slide exercises: a fair 5-sided die has E[X] = (1+2+3+4+5)/5 = 3; the lottery ticket has E[X] = −10 · 0.99 + 1000 · 0.01 = 0.1 > 0; a uniform variable on [0, 1] has mean 1/2, which is exactly why average income ȳ = 1/2 in Problem Set 1.

7. Integrals

Integration reverses differentiation, and a definite integral is the area under the curve:

\int x^{\alpha}\,dx = \frac{x^{\alpha+1}}{\alpha+1} \quad (\alpha \neq -1) \] \[ \int \frac{u'}{u}\,dx = \ln u \] \[ \int u' e^{u}\,dx = e^{u}

\int_2^3 x^3\,dx = \left[\frac{x^4}{4}\right]_2^3 = \frac{81 - 16}{4} = 16.25 \] \[ \int_0^5 e^{3x}\,dx = \left[\frac{e^{3x}}{3}\right]_0^5 = \frac{e^{15} - 1}{3}

You will meet integrals when a model has a continuum of agents, as in Problem Set 1 where income is spread over [0, 1]: average income is ∫₀¹ y dy = 1/2.

How to use this before the midterm

If fractions or exponents trip you up, do the fractions and exponents Starter and Intermediate sheets first. They take an evening and make everything after faster.
Do the derivatives Starter sheet in one sitting. If you get fewer than 20 of 24, read the refresher first.
Do the Intermediate sheet in pieces, ten a day. It is about speed and confidence, not difficulty.
Read the partial derivatives note and do its practice sheet through the Intermediate level.
Read the Lagrange note and do the Starter and Intermediate problems.
Redo Problem Set 1, Problem 3, with the book closed.

All notes and practice sheets are on the resources page.

If you get stuck anywhere, email me with the exercise number and what you tried. That is much more useful to both of us than “I don’t understand derivatives” :’)