Differentiation and Integration (Module 1)
FoundationThe M1 calculus storyline: understand the derivative through limits, apply the power, exponential and logarithmic formulas with the four rules and the chain rule, classify extrema and concavity with the second derivative, and use integration to recover functions; M1 excludes derivatives of trigonometric functions but prizes complete arguments in applied problems.
- M1 calculus has four stations: the limit-defined derivative, formulas and rules, applications, and indefinite integration to recover functions.
- M1's table has no trigonometric derivatives but includes log_a x and aˣ; keep the M1 and M2 syllabuses separate when revising.
- Second derivative test: f'' > 0 gives a local minimum, f'' < 0 a local maximum; state concavity alongside the classification.
文體小知識：數學概念說明文——整合單元一的微積分主線：導數概念與記法、求導公式與四則鏈式法則、二階導數判別與凹性、變率極值應用題與不定積分，並點出 M1 與 M2 的範圍分別。
§Text Passage
1M1 calculus has four stations: the concept of the derivative (understood through limits), differentiation (formulas and rules), applications (tangents, rates, maxima and minima), and indefinite integration (the reverse of differentiation). The main difference from Module 2 is scope: M1's differentiation formulas cover only algebraic, exponential and logarithmic functions — no derivatives of trigonometric functions and no implicit differentiation — while M1 puts heavy weight on applied problems, whose long questions often chain differentiation and integration together.
2The derivative is defined by a limit: the limit of the secant slope as the horizontal distance tends to zero. M1 requires understanding the concept and the notations — y prime, f prime, dy by dx, and the notation for evaluation at a point — but not lengthy computations from first principles. Geometrically the derivative is the tangent slope; physically it is the instantaneous rate of change; applied questions alternate between these two readings.
3M1's formula table: the derivative of a constant is zero; x to the n differentiates to n times x to the (n−1); e to the x differentiates to itself; the natural logarithm differentiates to one over x; the logarithm to base a differentiates to one over (x ln a); a to the x differentiates to itself times ln a. Note that e to the x and the natural logarithm are the special cases with a equal to e — the formulas echo one another.
4The rules match M2: the sum rule works term by term; the product and quotient rules have fixed structures; the chain rule multiplies outer by inner. The second derivative — the derivative of the derivative — powers the second derivative test: positive means a local minimum, negative a local maximum; its sign also gives concavity. Applied questions must address global as well as local extrema, and the endpoints of the domain must not be forgotten.
5For integration, M1 requires only the power, reciprocal and exponential formulas, with linearity and substitution — no integration by parts. The most typical application recovers f from f prime and an initial condition: integrate to get the family of curves, then substitute the condition to fix C. Whether differentiating or integrating, the answer format of applied questions — define variables, write the function, state the domain, solve, classify, conclude — is where the marks live.
1Summary 重點摘要
The M1 calculus storyline: understand the derivative through limits, apply the power, exponential and logarithmic formulas with the four rules and the chain rule, classify extrema and concavity with the second derivative, and use integration to recover functions; M1 excludes derivatives of trigonometric functions but prizes complete arguments in applied problems.
2Vocabulary cards Vocabulary
The three equivalent notations y', f'(x) and dy/dx; evaluation at a point is written as dy/dx at x = x₀.
Understanding the derivative through lim h→0 [f(x+h)−f(x)]/h; M1 does not require lengthy computations with it.
f'(x₀) is the slope of the tangent to the curve at x = x₀ — the geometric meaning of the derivative.
dy/dx = (dy/du)(du/dx); for composite functions, outer derivative times inner derivative.
At a stationary point, f'' > 0 indicates a local minimum and f'' < 0 a local maximum.
The curve is concave upward when f'' > 0 and concave downward when f'' < 0.
The absolute maximum and minimum over the whole domain, versus relative extrema in a neighbourhood; applied questions must address both.
The derivative as instantaneous rate of change; speed, cost changes and inflation rates in applied questions.
The reverse of differentiation, ∫f(x)dx = F(x) + C; M1 requires only the basic formulas and substitution.
Simplify the integrand with u, integrate, then substitute back; integration by parts is outside M1.
3Step by step Steps
- Step 1: The map of M1 calculusM1 calculus has four stations: the derivative concept (through limits), differentiation (formulas and rules), applications (tangents, rates, maxima and minima), and indefinite integration (reversing differentiation). The main difference from M2: M1's formulas cover only algebraic, exponential and logarithmic functions — no trigonometric derivatives and no implicit differentiation — while M1 prizes applied problems. For example, the derivative of sine belongs to M2 and will not be required in an M1 paper; keeping the two syllabus lists separate saves revision time.
- Step 2: The concept and notation of the derivativeThe derivative is defined by the limit f'(x) = lim h→0 [f(x+h)−f(x)]/h. M1 requires the concept and the notations y', f'(x), dy/dx, plus evaluation at a point written as dy/dx at x = x₀, but not lengthy first-principles computations. For example, for f(x) = x² at x = 3, f'(3) = 6 is the slope of the tangent at x = 3; writing dy/dx at x = 3 equals 6 is equally correct. Hold both readings in mind — geometric slope and rate of change — since applied questions borrow each in turn.
- Step 3: The M1 formula tableM1 formulas: (C)' = 0; (xⁿ)' = nxⁿ⁻¹; (eˣ)' = eˣ; (ln x)' = 1/x; (log_a x)' = 1/(x ln a); (aˣ)' = aˣ ln a. Note that e to the x and the natural logarithm are the a = e special cases. For example, the derivative of y = 2ˣ is 2ˣ ln 2; the derivative of y = log₁₀ x is 1/(x ln 10). Factor out constants first, convert surds and fractions to powers first — these two habits cut most of the workload in M1 computations.
- Step 4: The four rules and the chain ruleThe sum rule works term by term; the product rule d(uv)/dx = v(du/dx) + u(dv/dx); the quotient rule d(u/v)/dx = [v(du/dx) − u(dv/dx)]/v²; the chain rule dy/dx = (dy/du)(du/dx). For example, y = x² eˣ uses the product rule to give eˣ(x² + 2x); y = (ln x)² uses the chain rule to give 2 ln x ÷ x. M1 questions often stack two rules — say x times e to the 3x — so identify the structure first (product plus chain) before computing; clear ordering prevents dropped terms.
- Step 5: Second derivatives, concavity and extremaThe second derivative f''(x) is the derivative of the derivative, written d²y/dx². The second derivative test: at a stationary point, f'' > 0 gives a local minimum and f'' < 0 a local maximum; the sign of f'' also gives concavity. Applied questions must state global as well as local extrema and check endpoints and the domain. For example, A(x) = 20x − x² has A'' = −2 < 0, so the stationary point x = 10 is a local maximum; since the parabola opens downward with a single stationary point and zero area at the endpoints, it is also the global maximum. Classification and checking are where the marks sit.
- Step 6: The routine for rate and optimisation problemsApplied questions follow six fixed steps: define variables, write the objective function, state the domain, differentiate and solve for stationary points, classify, and conclude. For example, fencing a rectangle with 40 metres: let the length be x metres and the width 20 − x metres (0 < x < 20), with area A(x) = x(20 − x); A'(x) = 20 − 2x = 0 gives x = 10, A'' < 0 confirms a maximum, and the greatest area is 100 square metres, when the rectangle is a square. In word problems the marks fall on stating the domain and classifying — the computation is rarely the hardest part.
- Step 7: Indefinite integration in M1M1 integration formulas: ∫xⁿ dx = xⁿ⁺¹/(n+1) + C for n ≠ −1, ∫(1/x)dx = ln x + C, ∫eˣ dx = eˣ + C; with linearity and substitution — no integration by parts. The typical application recovers f(x) from f'(x) and an initial condition. For example, f'(x) = 6x² + eˣ with f(0) = 4: integrating gives f(x) = 2x³ + eˣ + C; substituting gives 1 + C = 4, so C = 3 and f(x) = 2x³ + eˣ + 3. Differentiating out and integrating back within one question is the classic shape of an M1 long question.
5Worked example Example
Example: a rectangular garden bed is fenced with 40 metres of fencing. Find the greatest possible area and prove that it is indeed the maximum.
- Let the length be x metres; then the width is (40 − 2x)÷2 = (20 − x) metres, with domain 0 < x < 20
- Write the area function: A(x) = x(20 − x) = 20x − x²
- Differentiate: A'(x) = 20 − 2x; solving A'(x) = 0 gives x = 10
- Classify: A''(x) = −2 < 0, concave downward, so x = 10 is a local maximum; A has a single stationary point in (0, 20) and vanishes at the endpoints, so it is the global maximum
- Compute: A(10) = 10 × 10 = 100
- Answer: The greatest area is 100 square metres, when the bed is a 10-metre by 10-metre square. The solution must contain five parts — variables, domain, differentiation and stationary points, classification, and conclusion; omitting the classification or the domain commonly loses marks.
6Common mistakes Mistakes
7Quick check Quick check
Q1Which of the following derivatives lies outside the M1 syllabus: (sin x)', (ln x)', (2ˣ)'?
Q2Let f(x) = 2eˣ − 3x². Find f'(x).
Q3Let y = ln(4x + 1). Find dy/dx.
Q4At a stationary point, what can be concluded if f''(a) < 0?
Q5Given f'(x) = 6x² + eˣ and f(0) = 4, find f(x).
8Recap Recap
- M1 calculus has four stations: the limit-defined derivative, formulas and rules, applications, and indefinite integration to recover functions.
- M1's table has no trigonometric derivatives but includes log_a x and aˣ; keep the M1 and M2 syllabuses separate when revising.
- Second derivative test: f'' > 0 gives a local minimum, f'' < 0 a local maximum; state concavity alongside the classification.
- Applied questions in six steps: define variables, write the function, state the domain, differentiate and solve, classify, conclude.
- Always write C when integrating; use any initial condition to find C, and verify by differentiating.