Session 04 - Differentiation

Mathematics for Master Students

Dr. Tobias Vlcek

Recap & Your Questions

Session 03 in a Nutshell

A function assigns each input exactly one output
The gallery: linear, quadratic, polynomial, rational, exponential, logarithm
Maximal domain, red flags: division by zero, roots of negatives, logs of non-positives
Composition chains functions (order matters); inverses undo
Convex vs concave: where the chord lies relative to the graph
Today: how fast does the output change, \(f(a+h)\) finally pays off

Common Issues in the Homework

One restriction forgotten: with \(\frac{\sqrt{x+1}}{x-2}\), both the root (\(x \geq -1\)) and the denominator (\(x \neq 2\)) restrict: collect every red flag, then intersect
Composition order: \((f \circ g)(x)\) means \(g\) acts first. Several solutions applied \(f\) first
The log-inequality flip: dividing by \(\ln 0.85 < 0\) flips the inequality. Dividing by a negative number always does

Which tasks gave you trouble? Bring them up now. We take the time to go through them before we move on to today’s topic.

Warm-Up: Compose and Restrict

Question

Let \(f(x) = \sqrt{x}\) and \(g(x) = 2x + 7\). Compute \(\;(f \circ g)(1)\;\) and find the domain of \(f \circ g\).

Inside out: \((f \circ g)(1) = f(g(1)) = f(9) = 3\)
The root needs \(g(x) \geq 0\): \(\;2x + 7 \geq 0\), so \(x \geq -\frac{7}{2}\)
Domain of \(f \circ g\): \(\;[-\frac{7}{2}, \infty)\), the inner function must land where the outer one is defined

Today’s Plan

Rates of change: from secant lines to the tangent and the derivative
Rules of differentiation: power, sum, product, quotient, chain, \(e^x\) and \(\ln x\)
Second derivative: the convex/concave intuition made precise
Marginal analysis: marginal cost, marginal revenue, elasticity of demand

Rates of Change

How Fast Does Revenue React?

Session 3 modelled revenue as a function of price: \(R(p) = 1200p - 20p^2\).

A pricing decision is rarely “pick \(p\) from scratch”, it is nudge the current price a little
Central question: if I move the price from \(p\) to \(p + h\), how fast does revenue react?
The answer depends on where we stand on the curve. Let’s look

The Same Nudge, Very Different Effects

Raising the price by €1 gains exactly €780 of revenue at \(p = 10\), but loses €20 at \(p = 30\). One curve, very different local behaviour.

The Average Rate of Change

How much does \(f\) change per unit of input, between \(a\) and \(a + h\)?

\[\frac{f(a+h) - f(a)}{h} \qquad \text{the [difference quotient]{.highlight}}\]

Change in output divided by change in input, “rise over run”
The numerator is exactly Session 3’s expression \(f(a+h)\) minus \(f(a)\), the raw material, cashed in
For a linear function this is always the slope \(m\). For a curved graph it depends on \(a\) and \(h\)

Geometry: The Secant Line

The difference quotient is the slope of the secant line through \((a, f(a))\) and \((a+h, f(a+h))\): here \(\frac{9 - 1}{2} = 4\) for \(f(x) = x^2\) from \(a = 1\) to \(a + h = 3\).

Your Turn: Compute the Secant Slope

Question

For \(f(x) = x^2\), what is the difference quotient from \(a = 2\) to \(a + h = 4\)? Type the number in the chat.

\[\frac{f(4) - f(2)}{4 - 2} = \frac{16 - 4}{2} = 6\]

The secant from \(1\) to \(3\) had slope \(4\), this one has slope \(6\): the parabola gets steeper as we move right, exactly what the next slides confirm.

From Secant to Tangent

Keep \(a\) fixed and pull the second point closer: the secants tilt toward one limiting line, the tangent at \(a\). Its slope is the rate of change right at \(a\).

Letting \(h\) Shrink

For \(f(x) = x^2\) at \(a = 1\), the secant slope is \(\frac{(1+h)^2 - 1}{h}\):

\(h\)	\(1\)	\(0.5\)	\(0.1\)	\(0.01\)	\(0.001\)
secant slope	\(3\)	\(2.5\)	\(2.1\)	\(2.01\)	\(2.001\)

As \(h\) shrinks toward \(0\), the slopes settle on the value \(2\)
We write \(\lim_{h \to 0}\) for “the value it settles on as \(h\) shrinks toward \(0\)”
That is all the limit machinery we need, no formal theory required in this tutorial

The Derivative

The derivative of \(f\) at \(a\) is the instantaneous rate of change:

\[f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h}\]

Geometric reading: slope of the tangent line at \((a, f(a))\)
Business reading: how fast the output reacts to a tiny nudge at \(a\)
Computed at every point, it defines a new function \(f'\), the derivative of \(f\)
Positive \(f'\): increasing; negative \(f'\): decreasing; the sign is a trend report

Your Turn: Read the Sign

Question

You are told \(f'(a) = -3\). What does that say?

(a) \(f\) is decreasing at \(a\), with tangent slope \(-3\)	(b) \(f(a) = -3\)
(c) \(f\) has a minimum at \(a\)	(d) \(f\) drops by exactly \(3\) between \(a\) and \(a + 1\)

(a): the derivative is a slope, and a negative slope means the graph falls at \(a\).

is only approximately right, and only for a nearly straight graph: the tangent slope describes the trend at \(a\), not a whole unit further on. (b) confuses the value with the slope.

Worked Example: Differentiating \(f(x) = x^2\)

Expand (Session 3 did this): \(f(a+h) = (a+h)^2 = a^2 + 2ah + h^2\)
Subtract: \(f(a+h) - f(a) = 2ah + h^2\)
Divide by \(h\): \(\dfrac{2ah + h^2}{h} = 2a + h\)
Shrink \(h\) toward 0: \(\;f'(a) = 2a\)

Check against the table: at \(a = 1\) we get \(f'(1) = 2\), exactly where the secant slopes settled. And at \(a = 3\): slope \(6\), the parabola gets steeper as we move right.

Notation for Derivatives

All of these mean the same thing:

Notation	Read as
\(f'(x)\)	“\(f\) prime of \(x\)”
\(\dfrac{df}{dx}\)	“derivative of \(f\) with respect to \(x\)”
\(\dfrac{d}{dx} f(x)\)	“\(\frac{d}{dx}\) applied to \(f(x)\)”

\(\frac{df}{dx}\) deliberately looks like a difference quotient: it reminds you where the derivative comes from. We mostly write \(f'(x)\); with several variables around (e.g. \(C(q)\)), the \(\frac{dC}{dq}\) form says with respect to what.

The Tangent as a Local Approximation

Near \(a\), the graph and its tangent are almost indistinguishable:

\[f(a + h) \;\approx\; f(a) + f'(a) \cdot h \qquad \text{for small } h\]

The difference quotient rearranged: \(f'(a) \approx \frac{f(a+h) - f(a)}{h}\), multiply by \(h\) and move \(f(a)\) across
Example: \(f(x) = x^2\) at \(a = 3\), \(f'(3) = 6\): \(\;f(3.1) \approx 9 + 6 \cdot 0.1 = 9.6\), exact value \(9.61\)
Business reading: “one more unit changes the output by about \(f'(a)\)”, the idea behind marginal cost, later today

Your Turn: Average vs Instantaneous

Question

A courier van covers 150 km in 2 hours. Halfway through, a speed camera clocks it at 90 km/h. Which number is an average rate of change, which is a derivative, and can both be correct at once?

\(\frac{150}{2} = 75\) km/h is a difference quotient, distance change over time change
The camera reading is the instantaneous rate, the derivative of distance at that moment
Both are correct: the average smooths out faster and slower stretches

Not Every Function Has a Derivative

Kinks, Jumps, and Smoothness

At the kink, secants from the left settle on \(-1\), from the right on \(+1\): no single tangent, no derivative at \(0\)
Jumps are even worse: no tangent can be drawn across a gap
Intuition: differentiable = smooth graph: no kinks, no jumps (tiered shipping tariffs kink at the breakpoints)

Rules of Differentiation

Constants and the Power Rule

Nobody computes difference quotients all day, rules do the work:

Constants: \(f(x) = c\) has \(f'(x) = 0\), a horizontal line has slope zero
Power rule: \(f(x) = x^n\) has \(f'(x) = n x^{n-1}\)
Recipe: exponent down front, then reduce the exponent by one
Example: \((x^5)' = 5x^4\), and our \((x^2)' = 2x\) is the case \(n = 2\)

The Power Rule Takes All Exponents

The rule also covers negative and fractional exponents, rewrite first:

\(\dfrac{1}{x} = x^{-1}\), so \(\left(\dfrac{1}{x}\right)' = -1 \cdot x^{-2} = -\dfrac{1}{x^2}\)
\(\sqrt{x} = x^{1/2}\), so \(\left(\sqrt{x}\right)' = \dfrac{1}{2} x^{-1/2} = \dfrac{1}{2\sqrt{x}}\)

Session 3’s exponent rules pay off: rewrite as a power, differentiate, translate back. Fractions and roots stop being special cases.

Your Turn: Apply the Power Rule

Question

Let \(f(x) = x^4\). What is \(f'(2)\)? Type the number. Bonus: differentiate \(g(x) = \dfrac{1}{x^2}\).

\[f'(x) = 4x^3, \qquad f'(2) = 4 \cdot 8 = 32\]

\[g(x) = x^{-2}, \qquad g'(x) = -2x^{-3} = -\frac{2}{x^3}\]

Common Mistake

\(f'(2) = 4 \cdot 2^4 = 64\) forgets to reduce the exponent. Exponent down front and one less on top.

Sums and Constant Multiples

\[(c \cdot f)'(x) = c \cdot f'(x) \qquad (f + g)'(x) = f'(x) + g'(x)\]

Together: differentiate term by term, constants tag along
Example, a cost function: \(C(q) = 0.5q^2 + 10q + 200\)
\(C'(q) = 0.5 \cdot 2q + 10 + 0 = q + 10\)
The fixed costs \(200\) vanish: they do not react to producing one unit more

Every polynomial is now differentiable in one line. Keep \(C'(q) = q + 10\) in mind, it returns later today.

A Cubic Revenue Function

Revenue from a promotion budget \(x\): \(\;R(x) = -0.5x^3 + 6x^2 + 20x\)

Term by term: \((-0.5x^3)' = -1.5x^2\), \(\;(6x^2)' = 12x\), \(\;(20x)' = 20\)
So \(R'(x) = -1.5x^2 + 12x + 20\)
At \(x = 2\): \(R'(2) = -6 + 24 + 20 = 38 > 0\), revenue still rises with the budget
At \(x = 10\): \(R'(10) = -150 + 120 + 20 = -10 < 0\), the budget has overshot

The sign of \(R'\) is the trend report: positive means “spend more”, negative means “you overshot”.

Your Turn: Differentiate Term by Term

Question

A cost function is \(C(q) = 0.1q^2 + 4q + 50\). What is \(C'(20)\)? Type the number.

\[C'(q) = 0.2q + 4, \qquad C'(20) = 4 + 4 = 8\]

At \(20\) units, cost grows at €8 per unit, while the fixed cost \(50\) has disappeared from \(C'\) entirely: it never reacts to output.

The Product Rule

\[(f \cdot g)'(x) = f'(x)\,g(x) + f(x)\,g'(x)\]

“Derivative of the first times the second, plus the first times derivative of the second”
Example: \(u(x) = (x^2 + 1)(3x - 2)\)
\(u'(x) = 2x \cdot (3x - 2) + (x^2 + 1) \cdot 3 = 9x^2 - 4x + 3\)
Check: expanding first gives \(u(x) = 3x^3 - 2x^2 + 3x - 2\), so \(u'(x) = 9x^2 - 4x + 3\) ✓

Common Mistake

\((f \cdot g)' \neq f' \cdot g'\), try \(f(x) = g(x) = x\): the product \(x^2\) has derivative \(2x\), not \(1\).

The Quotient Rule

\[\left(\frac{f}{g}\right)'(x) = \frac{f'(x)\,g(x) - f(x)\,g'(x)}{\left(g(x)\right)^2}\]

Like the product rule, but with a minus, the numerator’s order matters
Example, Session 3’s average cost per unit: \(A(q) = \frac{200 + 5q}{q}\)
\(A'(q) = \frac{5 \cdot q - (200 + 5q) \cdot 1}{q^2} = -\frac{200}{q^2}\)
Always negative: average cost falls as volume grows: economies of scale, now proven

The Chain Rule

For a composition \((f \circ g)(x) = f(g(x))\), Session 3’s chaining:

\[\left(f(g(x))\right)' = f'(g(x)) \cdot g'(x)\]

Recipe: outer derivative (inner left untouched) times inner derivative
Example: \(v(x) = (2x^2 + 1)^3\), outer \((\cdot)^3\), inner \(2x^2 + 1\)
\(v'(x) = 3(2x^2 + 1)^2 \cdot 4x = 12x\,(2x^2 + 1)^2\)

Common Mistake

Stopping at \(3(2x^2 + 1)^2\), the inner derivative \(4x\) must come along. Forgetting it is the classic chain-rule slip.

Your Turn: Spot the Missing Factor

Question

What is the derivative of \(\;(3x + 2)^5\)?

(a) \(5(3x + 2)^4\)	(b) \(15(3x + 2)^4\)
(c) \(5 \cdot 3^4\)	(d) \(15x\,(3x + 2)^4\)

(b): outer derivative \(5(3x+2)^4\), times the inner derivative \(3\).

forgot the inner factor, (c) differentiated the bracket away, (d) invented an \(x\): the inner derivative of \(3x + 2\) is the constant \(3\), nothing more.

Chain Rule in Business: Rates Multiply

A warehouse’s stock grows over time, and storage cost depends on stock:

\[q(t) = 30 + 4t \ \text{ pallets after } t \text{ weeks}, \qquad S(q) = 0.2q^2 \ \text{ cost in €}\]

How fast does the cost grow over time? Differentiate the chain \(S(q(t))\):

\[\frac{dS}{dt} = S'(q) \cdot q'(t) = 0.4q \cdot 4 = 1.6\,(30 + 4t)\]

At \(t = 5\): stock \(q = 50\), and cost rises at \(1.6 \cdot 50 = 80\), €80 per week
The chain rule multiplies the rates along the chain: € per pallet times pallets per week

Your Turn: Pick the Rule

Question

Which rule differentiates each function, and what are the derivatives?

\[\text{(a) } (5x + 1)^4 \qquad \text{(b) } x^3(2x - 7) \qquad \text{(c) } \frac{x}{x^2 + 1}\]

(a) Chain rule, outer \((\cdot)^4\), inner \(5x + 1\): derivative \(4(5x+1)^3 \cdot 5 = 20(5x+1)^3\)
(b) Product rule, or expand to \(2x^4 - 7x^3\) and use the power rule: both give \(8x^3 - 21x^2\)
(c) Quotient rule: \(\left(\frac{x}{x^2+1}\right)' = \frac{1 \cdot (x^2 + 1) - x \cdot 2x}{(x^2 + 1)^2} = \frac{1 - x^2}{(x^2 + 1)^2}\)

Derivatives of \(e^x\) and \(\ln x\)

\(\left(e^x\right)' = e^x\), the exponential is its own derivative: at every point, slope = height
\(\left(\ln x\right)' = \dfrac{1}{x}\), steep near \(0\), ever flatter afterwards: diminishing returns
Other bases pick up a factor: \(\left(a^x\right)' = a^x \ln a\)
\(e\) is exactly the base where that factor is \(\ln e = 1\), the tidy one
With the chain rule: \(\left(e^{3x}\right)' = e^{3x} \cdot 3 = 3e^{3x}\)

Worked Example: Mixing the Rules

Differentiate \(\;u(x) = x \cdot e^{x}\;\) and \(\;v(x) = \ln(x^2 + 1)\).

\(u\) is a product: \(u'(x) = 1 \cdot e^x + x \cdot e^x = (1 + x)\,e^x\)
\(v\) is a chain: outer \(\ln(\cdot)\) with derivative \(\frac{1}{(\cdot)}\), inner \(x^2 + 1\) with derivative \(2x\)
\(v'(x) = \dfrac{1}{x^2 + 1} \cdot 2x = \dfrac{2x}{x^2 + 1}\)

Name the structure first, product or chain, and the rule follows. Two rules can also nest: \(\left(x e^{2x}\right)'\) is a product whose second factor needs the chain rule.

Your Turn: Exponential and Log

Question

Differentiate \(\;f(x) = e^{2x}\;\) and \(\;g(x) = \ln(5x)\). Type both derivatives in the chat.

\(f'(x) = e^{2x} \cdot 2 = 2e^{2x}\): outer \(e^{(\cdot)}\), inner derivative \(2\)
\(g'(x) = \dfrac{1}{5x} \cdot 5 = \dfrac{1}{x}\): the \(5\) cancels
No surprise: \(\ln(5x) = \ln 5 + \ln x\), and the constant \(\ln 5\) differentiates to zero

The Superpower of \(e\)

Session 3’s continuous growth: \(K(t) = K_0 \cdot e^{rt}\). Differentiate (chain rule):

\[K'(t) = K_0 \, e^{rt} \cdot r = r \cdot K(t)\]

The growth rate is proportional to the current level: at every instant, capital grows at \(r\) times itself
That is the defining property of exponential growth: interest earns interest
This is why \(e\) rules finance and growth models, the promised superpower

The Rule Table

Building blocks

\(f(x)\)	\(f'(x)\)
\(c\)	\(0\)
\(x^n\)	\(n x^{n-1}\)
\(e^x\)	\(e^x\)
\(a^x\)	\(a^x \ln a\)
\(\ln x\)	\(\frac{1}{x}\)

Combination rules

Rule	Formula
Constant multiple	\((c f)' = c f'\)
Sum	\((f + g)' = f' + g'\)
Product	\((f g)' = f'g + f g'\)
Quotient	\(\left(\frac{f}{g}\right)' = \frac{f'g - f g'}{g^2}\)
Chain	\((f(g))' = f'(g) \cdot g'\)

These ten lines differentiate every function in Session 3’s gallery: practice comes with the tasks, the table lives on the cheatsheet.

Higher-Order Derivatives & Curvature

Differentiating Twice

\(f'\) is again a function, so differentiate again:

\[f''(x) = \left(f'\right)'(x) \qquad \text{also written } \frac{d^2 f}{dx^2}\]

\(f'\) reports how fast \(f\) changes; \(f''\) reports how fast the slope changes
Example: \(f(x) = x^3 - 3x\), Session 3’s cubic
First: \(f'(x) = 3x^2 - 3\); then: \(f''(x) = 6x\)
Third, fourth, \(\dots\) derivatives exist too: for us, two is all we need

Your Turn: Differentiate Twice

Question

Let \(f(x) = 2x^3 - 5x\). What is \(f''(3)\)? Type the number.

\[f'(x) = 6x^2 - 5, \qquad f''(x) = 12x, \qquad f''(3) = 36\]

Positive: at \(x = 3\) the slope is increasing, the graph bends upward, convex there. The next slide makes that reading official.

Curvature: The Second-Derivative Test

Session 3’s chord intuition, now with a formal test on an interval:

\(f''(x) \geq 0\) everywhere \(\;\Rightarrow\;\) convex: slope increasing, graph bends upward, chords above
\(f''(x) \leq 0\) everywhere \(\;\Rightarrow\;\) concave: slope decreasing, graph bends downward, chords below
\(f(x) = x^2\): \(f''(x) = 2 > 0\): convex everywhere, as the chords showed
\(f(x) = \ln x\): \(f''(x) = -\frac{1}{x^2} < 0\), concave everywhere: diminishing returns, certified

Curvature in Business: Diminishing Returns

Sales from an advertising budget \(a\) (in €1000): \(\;S(a) = 100\sqrt{a}\)

\(S'(a) = \dfrac{50}{\sqrt{a}} > 0\): more budget, more sales, always
\(S''(a) = -\dfrac{25}{a^{3/2}} < 0\): but each extra €1000 buys less than the previous one
Concave everywhere: the classic shape of diminishing returns
The first euro of advertising works hardest, the millionth barely moves the needle

Reading \(f\), \(f'\) and \(f''\) Together

For \(x < 0\): \(f'' < 0\): slope falling, \(f\) bends downward (concave)
For \(x > 0\): \(f'' > 0\): slope rising, \(f\) bends upward (convex)
At \(x = 0\) the bend switches, an inflection point; Session 5 turns these signs into an optimisation tool

Your Turn: What Does \(C''(q) > 0\) Mean Here?

Question

A warehouse’s handling cost \(C(q)\) satisfies \(C'(q) > 0\) and \(C''(q) > 0\) for all realistic \(q\). What does each sign say about the next pallet?

\(C'(q) > 0\): handling more pallets costs more, the cost curve rises
\(C''(q) > 0\): each additional pallet costs more than the previous one, the rise steepens
Business reading: the warehouse strains against capacity, the cost curve is convex

Marginal Analysis

Marginal Cost

Economists call \(C'(q)\) the marginal cost at output level \(q\):

Exact meaning: the instantaneous rate at which cost grows at \(q\)
Practical meaning: approximately the cost of one more unit
Why: with \(h = 1\) in the difference quotient, \(C'(q) \approx \dfrac{C(q+1) - C(q)}{1}\)
One unit is a small nudge next to hundreds produced, so the tangent slope is an excellent stand-in

Worked Reading: What Does \(C'(50)\) Say?

Take today’s cost function \(C(q) = 0.5q^2 + 10q + 200\), so \(C'(q) = q + 10\).

\(C'(50) = 60\), at 50 units, cost grows at €60 per unit
Reading: the 51st unit costs approximately €60 to produce
Exact check: \(C(51) - C(50) = 2010.5 - 1950 = 60.5\), the approximation is off by 50 cents
And \(C''(q) = 1 > 0\): marginal cost is increasing, every further unit is dearer

Mind the units: \(C'(q)\) is € per unit, not €: a rate, not a cost.

Your Turn: The 101st Unit

Question

Same cost function, \(C(q) = 0.5q^2 + 10q + 200\). Approximately what does the 101st unit cost? Type the number.

\[C'(100) = 100 + 10 = 110 \qquad \text{so about €110}\]

Exact check: \(C(101) - C(100) = 6310.5 - 6200 = 110.5\). Again the tangent slope misses by 50 cents, and it is far quicker than the subtraction.

Marginal Revenue

The same idea on the revenue side, sell \(q\) units, earn \(R(q)\):

Marginal revenue \(R'(q)\): approximately the revenue from one more unit sold
It usually falls with \(q\), selling more requires lowering the price
Producing one more unit pays off while \(R'(q) > C'(q)\), the extra revenue beats the extra cost
Where the two rates meet lies the profit-maximal output, Session 5 makes this precise

Elasticity of Demand

Rates in units depend on the units, pricing needs percentages. For a demand function \(D(p)\):

\[\varepsilon = \frac{p}{D(p)} \cdot D'(p)\]

Reading: a 1% price increase changes demand by about \(\varepsilon\) percent
Demand falls in price, so \(\varepsilon\) is negative
\(|\varepsilon| > 1\): elastic, demand overreacts to price
\(|\varepsilon| < 1\): inelastic, demand barely reacts

Worked Example: Pricing a Delivery Service

Weekly demand for a same-day delivery service: \(D(p) = 500 - 10p\), so \(D'(p) = -10\).

At \(p = 20\): \(D(20) = 300\), so \(\varepsilon = \frac{20}{300} \cdot (-10) \approx -0.67\)
Inelastic: raising the price 1% loses only 0.67% of demand, revenue rises
At \(p = 30\): \(D(30) = 200\), so \(\varepsilon = -1.5\), elastic: a price increase now backfires
The switch sits at \(\varepsilon = -1\) (here \(p = 25\)), exactly where revenue \(p \cdot D(p)\) peaks

Your Turn: Raise the Price or Not?

Question

At the current price, a freight forwarder estimates the elasticity of demand at \(\varepsilon = -0.4\). Elastic or inelastic, and what does a small price increase do to revenue?

\(|\varepsilon| = 0.4 < 1\): inelastic, customers barely react
A 1% price increase loses only about 0.4% of demand
Price up, demand almost unchanged: revenue increases, typical for services customers depend on

Closing

Key Takeaways

The derivative \(f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h}\) is the instantaneous rate of change, the tangent’s slope
Kinks and jumps break differentiability, smooth graphs are safe
Power, sum, product, quotient and chain rule handle every gallery function
\(e^x\) is its own derivative, growth proportional to the current level
\(f''\) measures curvature: \(f'' \geq 0\) convex, \(f'' \leq 0\) concave, the chord intuition, formalised
\(C'(q) \approx\) cost of one more unit; elasticity \(\varepsilon\) compares percentage reactions

That’s it for today.

Any Questions?

Until the Next Session

Work through the Tasks: the drilling happens there, with worked solutions
Check yourself with the Self-Test quiz
Keep the Cheatsheet next to you, the rule table lives on it
Note down anything unclear, we start next session with your questions

When differentiating, name the structure first: sum, product, quotient or chain? The right rule follows automatically.

Preview: Session 05: Single Variable Optimization

Central question: where is a function largest or smallest?
The recipe: set \(f'\) to zero (flat tangent) and ask \(f''\) whether it is a peak or a valley
Best output, best price, best order quantity
Today’s rules become the toolkit, differentiation was the warm-up

See you there, and bring your questions!

Literature & Further Reading

These sessions cover the essentials: textbooks offer more depth and practice
Start with Sydsæter et al. (2012) or Jacques (2015); full recommendations on the tutorial’s literature page

Jacques, Ian. 2015. Mathematics for Economics and Business. 8th ed. Always Learning. Pearson.

Sydsæter, Knut, Peter J. Hammond, and Arne Strøm. 2012. Essential Mathematics for Economic Analysis. 4th ed. Pearson.