Derivatives and optimisation

Why this lesson is optional

Bocconi's current test page describes quantitative reasoning as numerical and logical problems using basic mathematics, and it publishes no topic list. A 2024 sample test on Bocconi's website, which the current page does not link to and which may not reflect today's form, included an item on the gradient of a function of two variables. Nothing confirms that calculus appears on the current test. Treat this lesson as extra preparation for that possibility, after the core quantitative reasoning lessons.

What a derivative measures

The derivative $f'(x)$ is the slope of the graph of $f$ at the point $x$: how fast $f$ changes per unit change in $x$. For a straight line the slope is the same everywhere; for a curve it changes from point to point. In economics the derivative of a total is called a marginal quantity. If $C(Q)$ is the total cost of $Q$ units, the marginal cost $C'(Q)$ is approximately the extra cost of one more unit.

The rules you need

Function	Derivative	Example
$x^n$	$n x^{n-1}$	$(4x^3)' = 12x^2$
a constant	$0$	$(500)' = 0$
$e^{kx}$	$k e^{kx}$	$(e^{3x})' = 3e^{3x}$
$\ln x$	$\frac{1}{x}$	$(5\ln x)' = \frac{5}{x}$
$uv$	$u'v + uv'$	$(x e^x)' = e^x + x e^x$
$\frac{u}{v}$	$\frac{u'v - uv'}{v^2}$	$\left(\frac{x}{x+1}\right)' = \frac{1}{(x+1)^2}$
$f(g(x))$	$f'(g(x))\, g'(x)$	$\left(\ln(x^2 + 1)\right)' = \frac{2x}{x^2 + 1}$

Sums are differentiated term by term, and a constant factor stays where it is. The power rule also covers roots and reciprocals once they are written as powers: $\sqrt{x} = x^{1/2}$ has derivative $\frac{1}{2}x^{-1/2} = \frac{1}{2\sqrt{x}}$, and $\frac{1}{x} = x^{-1}$ has derivative $-\frac{1}{x^2}$.

The chain rule says: differentiate the outside function with the inside left alone, then multiply by the derivative of the inside. For $(3x + 1)^4$ the outside is a fourth power and the inside is $3x + 1$, so the derivative is $4(3x + 1)^3 \times 3 = 12(3x + 1)^3$. The rule for $e^{kx}$ in the table is the chain rule at work: the inside, $kx$, has derivative $k$.

Maxima and minima

At a maximum or minimum of a smooth function, away from the ends of its domain, the graph is momentarily flat, so $f'(x) = 0$. Points where this happens are stationary points. To tell which kind you have, look at the second derivative $f''(x)$, the derivative of $f'$:

• $f''(x_0) < 0$: the curve bends downwards there, so $x_0$ is a local maximum;
• $f''(x_0) > 0$: it bends upwards, so $x_0$ is a local minimum;
• $f''(x_0) = 0$: the test is silent; look at how the sign of $f'$ changes on either side.

A point where $f''$ changes sign is a point of inflection, where the curve switches between bending down and bending up. "Local" matters: a local maximum is the highest point nearby, not necessarily the highest value overall. When $x$ is restricted to an interval, compare the stationary values with the values at the ends.

Parabolas without calculus

For $f(x) = ax^2 + bx + c$ the derivative is $2ax + b$, which is zero at

$$x = -\frac{b}{2a}.$$

That is the vertex, and the formula can be used without any calculus. If $a < 0$ the parabola opens downwards and the vertex is its maximum; if $a > 0$ it is its minimum. The vertex also lies midway between the roots, which gives a check: $-x^2 + 10x$ has roots 0 and 10 and its maximum at $x = 5$, and indeed $-\frac{10}{2 \times (-1)} = 5$.

Marginal revenue, marginal cost and profit

Profit is revenue minus cost, $\pi(Q) = R(Q) - C(Q)$. Setting its derivative to zero gives $R'(Q) = C'(Q)$: marginal revenue equals marginal cost. The logic is plain. While one more unit adds more to revenue than to cost, producing it raises profit; the best output is where the two are equal.

$\text{MR} = \text{MC}$ is only the first condition. It can also hold where profit is at a minimum, so check the second derivative: profit is at a maximum when $\pi''(Q) = R''(Q) - C''(Q) < 0$, that is, where marginal cost is rising faster than marginal revenue. When the price $p$ is fixed, $R = pQ$, so $\text{MR} = p$ and the condition becomes $p = \text{MC}$ with marginal cost rising. When the price falls as the firm sells more, marginal revenue is below the price: with demand $P = a - bQ$, revenue is $aQ - bQ^2$ and $\text{MR} = a - 2bQ$.

Fixed costs have derivative zero. They change the level of profit but not the output that maximises it.

Partial derivatives and the gradient

A function of two variables, $f(x, y)$, has two partial derivatives. For $\frac{\partial f}{\partial x}$, treat $y$ as a fixed number and differentiate with respect to $x$; for $\frac{\partial f}{\partial y}$, do the reverse. For $f(x, y) = x^2 y + 3y^2$:

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

The gradient collects them into a vector, in this order: $\nabla f = \left(\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}\right)$. At $(1, 2)$ it is $(4, 13)$. The gradient points in the direction in which $f$ rises fastest, and at an interior maximum or minimum both partial derivatives are zero.

In economics the partial derivatives of a production function are marginal products: $\frac{\partial Q}{\partial L}$ is roughly the extra output from one more unit of labour with capital held fixed. For a Cobb–Douglas function $Q = A K^{\alpha} L^{\beta}$ the exponents also describe returns to scale: if $\alpha + \beta = 1$, doubling both inputs doubles output; if the sum is above 1, output more than doubles; below 1, it less than doubles.