The complete supplied teaching material, including deeper learning and source qualifications. Reflection questions here are not automatically graded. The supplied Week 4 application questions do not include an answer key.
Mathematics A — Week 4
HSG Assessment Year · Autumn 2026 · Monastery Masterclass
Derivatives: Understanding How Things Change
12–15 min core lesson
Lecture 4 · 6 October 2026
Exercise Series 3
Every business decision involves change. What happens to costs when production increases? How quickly does revenue grow? When does growth begin to slow down?
A derivative answers these questions by measuring how quickly a function changes at a particular point.
By the end of this lesson, you will be able to calculate derivatives, interpret their meaning, determine whether functions increase or decrease, identify changes in curvature, and use derivatives to solve certain limits.
1. From average change to instantaneous change
Imagine a factory whose total production cost is
\[ C(x)=x^2, \]
where \(x\) is the number of units produced.
At 3 units, the cost is \(C(3)=9\). At 4 units, it is \(C(4)=16\).
The average additional cost per unit between these production levels is
\[ \frac{C(4)-C(3)}{4-3}=\frac{16-9}{1}=7. \]
But what if we want to know how quickly costs are changing exactly at 3 units?
We examine progressively smaller changes in production.
Visual 1 — Zooming in on the slope at \(x=3\)
As the change approaches zero from either direction, the average slope approaches 6.
This is the central idea of differentiation: zoom in far enough, and a smooth curve behaves almost like a straight line.
The derivative definition
For \(y=f(x)\), the absolute change is
\[ \Delta y=f(x_0+\Delta x)-f(x_0). \]
The average rate of change, also called the difference quotient, is
\[ \frac{\Delta y}{\Delta x} =\frac{f(x_0+\Delta x)-f(x_0)}{\Delta x}. \]
If this quotient approaches a finite limit as \(\Delta x\to0\), the derivative exists:
\[ \boxed{ f'(x_0)=\lim_{\Delta x\to0} \frac{f(x_0+\Delta x)-f(x_0)}{\Delta x} } \]
Formulary status: Learn or derive independently. This definition is not printed on the supplied exam formulary.
Worked example — Deriving a marginal cost
For \(C(x)=x^2\), at any production level \(x\),
\[ \begin{aligned} C'(x) &=\lim_{h\to0}\frac{(x+h)^2-x^2}{h}\\ &=\lim_{h\to0}(2x+h)\\ &=\boxed{2x}. \end{aligned} \]
Therefore, \(C'(3)=6\).
The average additional cost from 3 to 4 units was 7, but the instantaneous marginal cost at 3 units is 6.
The distinction matters: a derivative measures a local rate, not necessarily the exact change over a whole interval.
2. Differentiability, continuity and tangent lines
A function is differentiable at \(x_0\) when its derivative at that point exists as a finite number.
Geometrically, this means we can assign the graph one well-defined, finite tangent slope.
Visual 2 — A smooth curve versus a sharp corner
Smooth curve
One tangent slope
Sharp corner
Different slopes from each side
Consider
\[ f(x)=|x|. \]
The function is continuous at zero: you can draw its graph without lifting your pen.
However, its slopes differ:
\[ f'_-(0)=-1,\qquad f'_+(0)=1. \]
Because the left and right derivatives are unequal, \(f'(0)\) does not exist.
This establishes an important logical relationship:
\[ \boxed{ \text{Differentiable}\implies\text{Continuous} } \]
But the reverse implication is false. A continuous function may have a corner, cusp or vertical tangent.
Tangent lines
The derivative \(f'(x_0)\) is the slope of the tangent line at \((x_0,f(x_0))\).
Using the point-slope equation,
\[ \boxed{ y=f(x_0)+f'(x_0)(x-x_0) } \]
For \(f(x)=x^2\) at \(x_0=3\),
\[ y=9+6(x-3)=6x-9. \]
At \(x=3.1\), this tangent predicts \(y=9.6\). The actual function value is \(9.61\).
The tangent gives a good local approximation near the contact point. It is not generally identical to the curve.
Formulary status: The tangent-line equation must be learned or reconstructed from the derivative and point-slope form. The formulary prints a tangent-plane formula for two-variable functions, but not this one-variable tangent-line formula.
3. The differentiation toolkit
Calculating every derivative from its limit definition would be painfully slow. Differentiation rules let us calculate slopes efficiently.
Basic derivatives
The power rule applies wherever the real power function is differentiable. Logarithms require positive arguments, and the trigonometric rules use radians.
Five rules that combine functions
Let \(u=u(x)\) and \(v=v(x)\).
The sum, product, quotient and chain rules are not printed on the supplied formulary. You should be able to apply them without assistance.
Worked example — The chain rule
Consider
\[ f(x)=\ln(1+3x^2). \]
Think of the function as two machines connected together:
\(x\)
\(u=1+3x^2\)
\(\ln u\)
Differentiate the outside, then multiply by the derivative of the inside.
The outer derivative is \(1/u\), and the inner derivative is \(6x\).
Therefore,
\[ f'(x)=\frac{1}{1+3x^2}\cdot6x =\boxed{\frac{6x}{1+3x^2}}. \]
If you write only \(1/(1+3x^2)\), you have forgotten the inner derivative.
Worked example — Product and chain rules together
Suppose
\[ g(x)=x^2e^{2x}. \]
Apply the product rule, then the chain rule to the exponential:
\[ \begin{aligned} g'(x) &=2xe^{2x}+x^2(2e^{2x})\\ &=\boxed{2xe^{2x}(1+x)}. \end{aligned} \]
This is the key exam skill: recognizing the structure of a function before differentiating it.
What is actually on your exam formulary?
The supplied HSG Mathematics A formulary is selective. It is not a complete differentiation cheat sheet.
For the printed logarithm formula, \(a>0\), \(a\ne1\), and \(x>0\). For tangent, \(\cos x\ne0\).
For composite expressions such as \((\ln u(x))'\), you still need the independently learned chain rule.
4. Using derivatives to solve difficult limits
Sometimes substituting into a fraction gives an indeterminate form, such as
\[ \frac00\qquad\text{or}\qquad\frac{\infty}{\infty}. \]
Indeterminate means that direct substitution does not determine the limit. It does not mean that the limit is necessarily zero or infinite.
L’Hôpital’s rule can help. Under its differentiability and limit conditions,
\[ \boxed{ \lim_{x\to a}\frac{f(x)}{g(x)} = \lim_{x\to a}\frac{f'(x)}{g'(x)} } \]
for a \(0/0\) or \(\infty/\infty\) form, provided the relevant derivative quotient is defined near the limit point and has an appropriate limit.
The rule also applies to suitable one-sided limits and limits at infinity.
Worked example — Applying L’Hôpital twice
Calculate
\[ \lim_{x\to0}\frac{e^x-1-x}{x^2}. \]
Direct substitution gives \(0/0\), so differentiate numerator and denominator separately:
\[ \lim_{x\to0}\frac{e^x-1}{2x}. \]
This is still \(0/0\). Apply the rule again:
\[ \lim_{x\to0}\frac{e^x}{2} =\boxed{\frac12}. \]
The rule can be repeated as long as its conditions remain satisfied.
An exam shortcut
Before applying L’Hôpital, check the supplied formulary. It already prints
\[ \lim_{x\to0}\frac{\sin x}{x}=1, \qquad \lim_{x\to0}\frac{e^x-1}{x}=1, \]
as well as two special limits involving powers, exponentials and logarithms at infinity.
Recognizing one of these forms can save several steps.
Formulary status: The four special limits are printed on page 2 of the English formulary. L’Hôpital’s rule itself is not printed.
Common mistake: L’Hôpital is not permission to differentiate every fraction. A direct limit of \(2/3\), for example, cannot be replaced by a derivative quotient merely because the functions are differentiable.
5. First and second derivatives tell different stories
So far, the derivative has measured the slope of a function. What if the slope itself changes?
That is what the second derivative measures.
\[ f'(x)=\text{first derivative}, \qquad f''(x)=(f'(x))'=\text{second derivative}. \]
Higher derivatives \(f'''(x),f^{(4)}(x),\ldots\) are obtained by continuing this process.
The first derivative: direction
On an interval where \(f\) is differentiable:
When \(f'\ge0\) throughout an interval, \(f\) is nondecreasing. The analogous rule applies when \(f'\le0\).
A point where \(f'(x)=0\) is called a stationary point. It need not be a maximum or minimum.
The second derivative: curvature
Convex
\(f''(x)>0\)
Slope increases
Concave
\(f''(x)<0\)
Slope decreases
More generally, on an interval where the second derivative exists:
\[ f''(x)\ge0\implies f\text{ is convex}, \]
\[ f''(x)\le0\implies f\text{ is concave}. \]
Strict inequalities are sufficient for strict convexity or concavity, respectively, but are not necessary.
An inflection point occurs where the graph changes its curvature from convex to concave or vice versa.
For a twice-differentiable function, a common method is to solve \(f''(x)=0\) and then check that \(f''\) changes sign. A zero second derivative alone is not sufficient.
For example, \(f(x)=x^4\) has \(f''(0)=0\), but \(f''(x)=12x^2\ge0\) on both sides of zero. There is no inflection point.
6. Worked business application — When marginal costs change
Consider a firm's total cost function for \(x\ge0\):
\[ C(x)=x^3-6x^2+15x+10. \]
Step 1 — Are total costs increasing?
Differentiate:
\[ \begin{aligned} C'(x) &=3x^2-12x+15\\ &=3\big((x-2)^2+1\big)>0. \end{aligned} \]
Since marginal cost is positive at every production level, total cost is strictly increasing.
Step 2 — Are marginal costs increasing?
Differentiate again:
\[ C''(x)=6x-12=6(x-2). \]
This changes sign at \(x=2\).
Visual 3 — The two derivatives tell different stories
The cost curve has an inflection point at
\[ (2,C(2))=(2,24). \]
Conceptual sketch, not to scale: total cost rises throughout, while marginal cost changes from decreasing to increasing. The dashed line represents the local tangent.
The economic interpretation is important: falling marginal cost does not imply falling total cost.
Before \(x=2\), every additional unit still increases total cost, but by progressively less. After \(x=2\), additional units become progressively more expensive.
7. The mean-value theorem: why slope signs work
Suppose you travel 120 km in 2 hours. Your average speed is 60 km/h.
Under suitable continuity and differentiability assumptions, at some instant your speed must have been exactly 60 km/h.
The same principle applies to functions.
If \(f\) is continuous on \([a,b]\) and differentiable on \((a,b)\), the mean-value theorem guarantees a point \(c\in(a,b)\) such that
\[ \boxed{ f'(c)=\frac{f(b)-f(a)}{b-a} } \]
This connects instantaneous slopes to average slopes.
It also explains the monotonicity criterion: if every derivative is positive on an interval, every secant between distinct points has positive slope, so the function is strictly increasing.
Formulary status: Neither the mean-value theorem nor its monotonicity consequences are printed. Understand the statement and when its assumptions are satisfied.
8. Common exam mistakes
Your Week 4 mental model
\(f(x)\) tells you the current level.
\(f'(x)\) tells you how fast the level changes.
\(f''(x)\) tells you how the rate itself changes.
Optional deeper learning
For students who want stronger mathematical foundations. These sections are not required for the first guided reading.
A. Why differentiability guarantees continuity
Assume \(f'(x_0)\) exists. Rewrite the change in the function as
\[ f(x_0+h)-f(x_0) =h\cdot\frac{f(x_0+h)-f(x_0)}{h}. \]
As \(h\to0\), the first factor tends to zero and the second tends to the finite value \(f'(x_0)\).
Therefore,
\[ \lim_{h\to0}\big(f(x_0+h)-f(x_0)\big)=0. \]
This is exactly continuity at \(x_0\).
B. Differentiating inverse functions
When an invertible differentiable function has nonzero derivative at the relevant point,
\[ \boxed{ (f^{-1})'(y)=\frac1{f'(f^{-1}(y))} } \]
For example, \(f(x)=e^x\) has inverse \(f^{-1}(y)=\ln y\). Thus,
\[ (f^{-1})'(y) =\frac1{e^{\ln y}} =\frac1y,\qquad y>0. \]
The inverse-function derivative formula is not printed. This example also recovers the logarithm derivative that can be obtained directly from the printed formulary.
C. Detecting inflections with higher derivatives
If \(f''(x_0)=0\) and \(f'''(x_0)\ne0\), with the relevant derivatives continuous near \(x_0\), then \(f''\) changes sign and \(x_0\) is an inflection point.
This is a convenient sufficient test. More generally, directly examining the sign of \(f''\) on either side is the most reliable method.
The third-derivative shortcut is not supplied on the formulary.
Check your understanding
Six original application questions · Suggested time: 7–10 minutes. Try them before revealing the solutions.
1
A function is defined by \(f(x)=x^2\) for \(x\le1\) and \(f(x)=ax+b\) for \(x>1\). Find \(a,b\) so that it is differentiable at \(x=1\). Explain the two conditions you need.
2
For \(h(x)=xe^{x^2}\), calculate \(h'(1)\) and identify the differentiation rules involved.
3
Evaluate \(\displaystyle\lim_{x\to0}\frac{1-\cos x}{x^2}\). Explain why L’Hôpital is permissible, and how the formulary helps.
4
Suppose \(f'(x)=(x+1)(x-2)\) on \(\mathbb R\). Determine where \(f\) increases and decreases. What happens when \(x\) crosses \(-1\) and \(2\)?
5
A function is \(g(x)=(x-2)^3+5\). At \(x=2\), is there a stationary point, an inflection point, both, or neither? Is there an extremum?
6
A manufacturer's cost function is \(C(x)=100+5x+0.2x^2\). At \(x=10\), what is the marginal cost? Does producing one entire additional unit cost exactly that amount? Interpret the difference.
Before you unlock the exercises
You are ready to begin Exercise Series 3 when you can explain the derivative as a limit, select the correct differentiation rules, distinguish continuity from differentiability, justify the use of L’Hôpital, and interpret the signs of \(f'\) and \(f''\).
You do not need to memorize long proofs before practising. Focus on recognizing mathematical structure and explaining why a method works.
Official sources and coverage
This lesson is aligned with the HS2026 Mathematics with Statistics A schedule, which identifies 6 October as the differentiability lecture and associates it with Exercise Series 3.
04_CURATED_STUDY_GUIDES_001.md
The mathematical content is grounded in the current German HS2026 Lecture 4 — Differentiability slides by Dr. Dominik Kachel (preserved in SUPPLEMENTARY_ORIGINALS.pdf, merged PDF pp. 158–253), the English Mathematics A Midterm Review (PDF pp. 29–32 for the central derivative rules and limits), and the official Mathematics A Formulary (PDF p. 2).
00_START_HERE.md
04_CURATED_STUDY_GUIDES_001.md
02_CANVAS_LECTURES_AND_READINGS_001.md
Coverage qualification: The English Lecture 4 assignment and its official link are identifiable in Canvas, but the English slide contents were not independently verified from the available Project snapshot. The Week 4 coverage above is corroborated by the actual HS2026 German Lecture 4 slides and the English review material, not by an assumed English slide transcription. Local/global extremum optimization is mainly developed in the following teaching sequence; only its immediate derivative-based intuition is introduced here. The worked examples, diagrams and comprehension questions are original Monastery teaching material, not reproductions of official exam questions.