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Week 5 — Differentials, Rates of Change & Elasticities
Mathematics with Statistics A · HS2026
Lecture 5 · Exercise Series 3
10–15 min guided lesson · HSG Assessment Year
The big idea: how sensitive is a function?
Imagine you're managing a company. You increase production, change your prices, or spend more on advertising.
Three questions immediately matter:
- How much will the result change?
- How large is that change relative to the current result?
- How sensitive is the result to a percentage change in the input?
This week, you learn three mathematical tools for answering those questions.
\(df\)
Differential
Absolute change
\(\rho_f\)
Rate of change
Relative growth per unit
\(\varepsilon_f\)
Elasticity
% response to 1% input
By the end, you should be able to choose the appropriate tool, calculate it, interpret its units and limitations, and avoid common exam traps.
Exam-formulary guide. Throughout this lesson:
- [F] Printed on formulary — directly available on the supplied Mathematics A formulary, page 2.
- [D] Derived from formulary — follows from a printed formula, but is not itself printed.
- [I] Independently needed — a definition, derivative rule, reasoning step or method to know or derive yourself.
The three headline formulas are all [F]. Knowing when to use them is your responsibility.
1. Differential — predicting an actual change
Concept 1 · Absolute changes
Intuition: replace a curve with a straight line
Remember Week 4: the derivative \(f'(x_0)\) measures a function's instantaneous slope.
If the input moves a small distance, the function behaves approximately like its tangent line.
That's the entire idea behind the differential.
Actual curve
Tangent approximation
Original conceptual sketch, not to scale. For small input changes, the tangent's predicted height is close to the curve's actual height.
Suppose \(y=f(x)\), and the input changes from \(x_0\) to \(x_0+\Delta x\).
The exact output change is:
\[ \boxed{\Delta y=f(x_0+\Delta x)-f(x_0)}\qquad\text{[I]} \]
The differential is:
\[ \boxed{df(x_0)=f'(x_0)\,dx}\qquad\text{[F]} \]
Here \(dx\) represents the input increment, so we choose \(dx=\Delta x\).
For a differentiable function and a sufficiently small change:
\[ \boxed{\Delta y\approx df(x_0)=f'(x_0)\Delta x}\qquad\text{[D]} \]
The derivative tells you the slope; multiplying by the input change predicts the output change.
Worked example — Production costs
A factory has the cost function:
\[ C(x)=100+4x+0.02x^2 \]
where \(x\) is production in units and \(C(x)\) is measured in CHF.
Production rises from 50 to 53 units. Estimate the additional cost.
Step 1 — Differentiate.
\[ C'(x)=4+0.04x \]
At \(x_0=50\):
\[ C'(50)=4+0.04(50)=6 \]
The marginal cost is CHF 6 per additional unit at that production level.
Step 2 — Apply the differential.
\[ dx=53-50=3 \]
\[ \boxed{dC(50)=6(3)=18\text{ CHF}} \]
Step 3 — Compare with the exact change.
\[ \begin{aligned} C(50)&=350\\ C(53)&=368.18\\ \Delta C&=368.18-350=18.18 \end{aligned} \]
Differential prediction
CHF 18.00
Exact additional cost
CHF 18.18
The differential underestimates the increase by CHF 0.18 because the cost curve bends upward.
Key insight: A differential is a local approximation, not automatically an exact difference. For a linear function, it is exact; for a curved function, accuracy generally improves as \(\Delta x\) approaches zero.
2. Rate of change — measuring relative growth
Concept 2 · Relative changes
Intuition: CHF 100 can be a lot or very little
Imagine two companies each increase monthly revenue by CHF 100.
- Company A previously earned CHF 1,000: growth is 10%.
- Company B previously earned CHF 100,000: growth is 0.1%.
The same absolute increase has a completely different relative meaning.
That's why we divide the derivative by the current value of the function.
\[ \boxed{\rho_f(x_0)=\frac{f'(x_0)}{f(x_0)}}\qquad\text{[F]} \]
The rate of change tells you the instantaneous relative change in \(f\) per one unit of \(x\).
For a small increment, it follows from the differential that:
\[ \boxed{\frac{\Delta y}{f(x_0)} \approx \rho_f(x_0)\Delta x}\qquad\text{[D]} \]
The left-hand side is a relative change. Multiply it by 100 to express it as a percentage.
Worked example — Revenue growth
Suppose revenue grows according to:
\[ R(t)=500e^{0.03t} \]
where \(t\) is in months and revenue is measured in thousands of CHF.
Differentiate:
\[ R'(t)=500(0.03)e^{0.03t} \]
Now divide by revenue:
\[ \rho_R(t)= \frac{500(0.03)e^{0.03t}}{500e^{0.03t}} =\boxed{0.03} \]
Interpretation: Revenue grows at an instantaneous relative rate of 3% per month, at every point in time.
For a small interval of half a month:
\[ \frac{\Delta R}{R(t)} \approx 0.03(0.5)=0.015=1.5\% \]
The exact relative change over half a month is:
\[ e^{0.015}-1\approx1.5113\% \]
The approximation is close, but not identical.
Why the time unit matters
A rate of \(0.03\) per month is not \(0.03\) per year. The input unit is part of the interpretation.
Notice also that \(R'(t)\) increases as revenue grows, while \(\rho_R(t)\) stays constant: the company gains more money in absolute terms while maintaining the same relative growth rate.
3. Elasticity — how strongly does a function react?
Concept 3 · Percentage sensitivity
Intuition: compare percentages, not units
Suppose a shop raises its prices by 1%.
Would you expect sales to decrease by 0.2%, 1%, or 4%?
Elasticity measures exactly this sensitivity.
Instead of comparing the change in sales to the change in price measured in CHF, we compare their percentage changes.
\[ \boxed{\varepsilon_f(x_0)= \frac{x_0f'(x_0)}{f(x_0)}}\qquad\text{[F]} \]
For small percentage changes:
\[ \boxed{ \frac{\Delta y}{f(x_0)} \approx \varepsilon_f(x_0) \frac{\Delta x}{x_0} }\qquad\text{[D]} \]
In simple language:
A 1% input increase produces approximately an \(\varepsilon_f(x_0)\)% output change.
For example, elasticity \(-2\) means a 1% input increase produces approximately a 2% decrease in the output.
Elasticity is dimensionless. Measuring prices in CHF or cents does not change the economic sensitivity.
Three types of elasticity
Inelastic
\(|\varepsilon|<1\)Smaller % response
Unit elastic
\(|\varepsilon|=1\)Equal % response
Elastic
\(|\varepsilon|>1\)Larger % response
Conceptual comparison of response magnitudes, not graphs of three specific demand functions. Classification uses the absolute value of elasticity.
The sign tells you the direction of the response. The absolute value tells you its strength.
Worked example — Price elasticity of demand
A company faces the demand function:
\[ q(p)=120-3p,\qquad 0\<p<40 \]
where \(p\) is the price in CHF and \(q\) is the quantity demanded.
Step 1 — Differentiate demand.
\[ q'(p)=-3 \]
Every CHF 1 price increase reduces demand by 3 units.
Step 2 — Calculate elasticity.
\[ \varepsilon_q(p) =\frac{p\,q'(p)}{q(p)} =\boxed{-\frac{3p}{120-3p}} \]
Step 3 — Compare different prices.
Notice something important: the derivative is always \(-3\), but elasticity changes with price.
At higher prices, the same decrease of 3 units represents a much larger percentage of remaining demand.
At \(p=30\), suppose the company raises its price by 5%.
\[ \frac{\Delta q}{q}\approx -3(5\%)=-15\% \]
The model predicts approximately 15% fewer units sold.
Because this particular demand function is linear, the prediction is actually exact for this price change. For nonlinear demand, finite percentage changes generally produce approximation errors.
The hidden connection
Look at the three formulas:
\[ df=f'(x)\,dx \]
\[ \rho_f=\frac{f'}{f} \]
\[ \varepsilon_f=\frac{xf'}{f} \]
The relationship is:
\[ \boxed{\varepsilon_f(x)=x\,\rho_f(x)}\qquad\text{[D]} \]
Elasticity is simply the relative rate of change multiplied by the current input.
Quick mental check
If elasticity equals \(-0.5\), a small 2% input increase produces approximately a 1% output decrease.
If elasticity equals \(-3\), the same increase produces approximately a 6% output decrease.
You don't need a new formula for either calculation.
4. Choose the right mathematical tool
Concept 4 · Exam decision-making
Imagine an exam question describing a production change. Before calculating anything, identify what the question is actually measuring.
Five exam mistakes to avoid
1. Treating the differential as an exact difference
\(df\approx\Delta y\) for small changes, but usually \(df\ne\Delta y\). State clearly when you're approximating.
2. Mixing percentages with decimals
A 5% price change is \(0.05\), not \(5\). With elasticity \(-2\), the predicted relative output change is \(-2(0.05)=-0.10=-10\%\).
3. Ignoring the absolute value in classification
Elasticity \(-3\) is elastic, not inelastic. Compare \(|-3|=3\) with \(1\).
4. Dividing by zero
Both \(\rho_f(x_0)\) and \(\varepsilon_f(x_0)\) require \(f(x_0)\ne0\). Percentage-input interpretations additionally require \(x_0\ne0\).
5. Confusing elasticity with the derivative
A derivative can depend on the measurement units. Elasticity compares proportions and is independent of units.
What the exam formulary gives you
The formulary supplies some special derivatives, but not every differentiation rule you need.
5. Optional deeper understanding
Extra depth · Skip on your first pass if the core concepts are clear
Why logarithms simplify growth rates
For \(f(x)>0\), the chain rule gives:
\[ \frac{d}{dx}\ln(f(x)) =\frac{f'(x)}{f(x)} =\rho_f(x) \]
So the rate of change is the derivative of the logarithm of the function.
More generally, for a differentiable nonzero function, we may use \(\ln|f(x)|\).
[D] Derived identity, not printed as a standalone formulary formula.
This is especially useful for products and quotients, since logarithms turn multiplication into addition and division into subtraction.
A function with constant elasticity
Consider:
\[ f(x)=Ax^k,\qquad A>0,\ x>0 \]
Then:
\[ f'(x)=Akx^{k-1} \]
and therefore:
\[ \varepsilon_f(x) =\frac{xAkx^{k-1}}{Ax^k} =\boxed{k} \]
The elasticity is constant, regardless of \(x\).
For example, demand \(q(p)=100p^{-2}\) always has price elasticity \(-2\). A small 1% increase in price therefore produces approximately a 2% reduction in demand at any positive price.
[I] Derive independently using the power rule and the printed elasticity formula.
Why local approximations become less accurate
The definition of differentiability implies:
\[ f(x_0+h)=f(x_0)+f'(x_0)h+o(h) \]
where the error becomes negligible relative to \(h\) as \(h\to0\).
This explains why differentials work locally, without requiring Taylor polynomials, which are introduced in Week 6.
For twice-differentiable functions, curvature helps explain the approximation error: positive curvature generally makes the function rise above its tangent line.
Check your understanding
Five questions · Conceptual and applied · Try before opening the explanations
You are not being tested on memorising the three printed formulas. Focus on choosing methods, interpreting results and recognising limitations.
Q1 — Production cost
A company has C(q) = 80 + 6q + 0.1q². Production increases from 10 to 11 units. Estimate the additional cost with a differential, compute the exact change, and explain the difference.
Need a hint?Show explanation
Q2 — Relative growth
Revenue is R(t) = 600e^(0.04t), where t is measured in years. At t = 5, estimate the percentage increase over the next three months. Explain why treating 0.04 as a monthly rate would be wrong.
Need a hint?Show explanation
Q3 — Same slope, different sensitivity
Two companies have demand functions qA(p) = 100 − 2p and qB(p) = 30 − 2p. At p = 10, which company has more price-elastic demand? Explain why equal derivatives do not imply equal elasticities.
Need a hint?Show explanation
Q4 — Domain trap
For f(x) = x − 4 at x₀ = 4, can you calculate the differential, rate of change and elasticity? Explain which quantities exist and which are undefined.
Need a hint?Show explanation
Q5 — Approximation versus reality
For f(x) = x² at x₀ = 3, the input rises by 20%. Use elasticity to predict the output percentage change. Then calculate the exact percentage change. Why do the results differ?
Need a hint?Show explanation
Explained answer key — independent review
Use this section to check your reasoning or as a separate correction reference.
Q1. A differential can underestimate additional cost
\[ C'(q)=6+0.2q,\qquad C'(10)=8 \]
Therefore:
\[ dC(10)=8(1)=\boxed{8} \]
The exact increase is:
\[ \begin{aligned} C(10)&=150\\ C(11)&=158.1\\ \Delta C&=\boxed{8.1} \end{aligned} \]
The tangent underestimates cost because \(C''(q)=0.2>0\). The function bends upward. The key skill is recognising when a marginal quantity is only an approximation of a finite change.
Q2. Relative rates depend on the unit of time
\[ \rho_R(t)=\frac{R'(t)}{R(t)}=0.04 \]
Since three months equals \(0.25\) years:
\[ \frac{\Delta R}{R}\approx0.04(0.25) =\boxed{1\%} \]
The exact change is \(e^{0.01}-1\approx1.005\%\). A quantity expressed per year cannot be interpreted as per month without converting units.
Q3. Elasticity accounts for the size of the initial output
At \(p=10\):
\[ \varepsilon_A =\frac{10(-2)}{80} =\boxed{-0.25} \]
\[ \varepsilon_B =\frac{10(-2)}{10} =\boxed{-2} \]
Company A is inelastic; Company B is elastic. Both lose the same number of units for a small equal price increase, but Company B loses a much larger proportion of its sales.
Q4. Differentiability does not guarantee defined elasticity
For \(f(x)=x-4\), we have \(f'(4)=1\), so:
\[ \boxed{df(4)=dx} \]
However:
\[ f(4)=0 \]
Consequently, both \(\rho_f(4)\) and \(\varepsilon_f(4)\) are undefined due to division by zero. In exams, check the domain and denominator before simplifying an expression.
Q5. Elasticity is local
\[ \varepsilon_f(x) =\frac{x(2x)}{x^2}=2,\qquad x>0 \]
The elasticity estimate for a 20% input increase is:
\[ \boxed{40\%} \]
The exact output increase is:
\[ \frac{f(3.6)-f(3)}{f(3)} =\frac{12.96-9}{9} =\boxed{44\%} \]
The difference is caused by curvature and the finite size of the change. This is why an elasticity calculation should not be presented as an exact forecast without checking the model.
Before you unlock Week 5 exercises
You should now be able to explain the difference between an absolute change, a relative growth rate and an elasticity; use the three printed formulas correctly; interpret the sign and magnitude of elasticity; and identify when a differential estimate is unreliable or a ratio is undefined.
Your next step is Exercise Series 3, especially the officially listed applications on elasticity (A6) and rates of change (A9). These follow the Week 4 differentiation foundations.
Sources and verification
HS2026 Mathematics with Statistics A — English Information Document: lecture schedule, PDF p. 9; Exercise Series 3 mapping, PDF p. 10.
01_COURSE_INFORMATION.md
Mathematics A — Formulary: differential, rate of change and elasticity on verified PDF p. 2.
06_FORMULARY.md
MONASTERY_FRIENDS_PRELAUNCH_AUDIT_2026-10-11.md
Mathematics A — Midterm Review: differential (p. 33), rate of change (p. 34), elasticity and classification (pp. 35–36).
07_MIDTERM_REVIEW.md
Historical Mathematics A exercise bank: supplementary examples used for topic cross-checking, not as HS2026 syllabus authority.
14_LEGACY_EXERCISES_02_FINANCE_FUNCTIONS_CALCULUS.md
Coverage limitation: The archived Lecture 5 entry states that its English attachment was locked until 9 October 2026 at 23:59. A subsequently released lecture PDF was not available to verify in the retrieved Project materials. This lesson therefore covers the confirmed Week 5 syllabus and supplied formulary/review concepts, but cannot certify that every example, emphasis or possible extension in the newly released slides is represented. The supplied formulary's page layout is verified; its status as the final binding HS2026 examination aid remains to be confirmed against the official examination rules.
03_LECTURES_04_06.md
MONASTERY_FRIENDS_PRELAUNCH_AUDIT_2026-10-11.md