Use standard differentiation notation including first and second derivatives

Objective

Students will be able to:

• Write the derivative of a function using all standard notations required by the Cambridge IGCSE Additional Mathematics (0606) syllabus.
• Explain the meaning of the first, second and higher‑order derivatives as derived functions.
• Apply the full range of differentiation rules to algebraic, trigonometric, exponential and logarithmic functions.
• Use first‑ and second‑derivative tests (including points of inflection) to locate and classify stationary points.
• Solve basic rate‑of‑change and kinematics problems, and find equations of tangent and normal lines.

1. The Derived Function

If y = f(x) then the derivative of f is itself a function, called the derived function. It is written as

• Leibniz notation: dy/dx (or df/dx)
• Prime notation: f′(x) or y′
• Higher‑order notation: f⁽ⁿ⁾(x) for the n‑th derivative.

The second derivative f″(x) (or d²y/dx²) is the derivative of the derived function and tells how the rate of change itself is changing. Higher‑order derivatives (f⁽³⁾(x), f⁽⁴⁾(x), …) are part of the syllabus even though only the first two are examined.

1.1 Standard Notation (Cambridge required)

Concept	Leibniz notation	Prime notation	Other common forms
First derivative	dy/dx or df/dx	f′(x) or y′	f′(x)
Second derivative	d²y/dx² or d²f/dx²	f″(x) or y″	f″(x)
Higher‑order derivative	dⁿy/dxⁿ	f⁽ⁿ⁾(x)	–

1.2 Meaning of the Derivatives

• First derivative f′(x): gradient of the tangent to the curve at (x, f(x)); instantaneous rate of change of f with respect to x.
• Second derivative f″(x): rate of change of the gradient; tells whether the curve is concave upwards (f″(x) > 0) or concave downwards (f″(x) < 0).
• Higher‑order derivatives give successive rates of change and are useful for series expansions and for checking the behaviour of very smooth curves.

2. Derivatives of Standard Functions

Function	Derivative	Example
Power: f(x)=xⁿ (n any rational number)	f′(x)=n·xⁿ⁻¹	d/dx[x³]=3x², d/dx[x^{½}]=½x^{-½}
Constant: f(x)=c	f′(x)=0	d/dx[5]=0
Exponential (base a): f(x)=aˣ (a > 0, a ≠ 1)	f′(x)=aˣ·ln a	d/dx[2ˣ]=2ˣ·ln 2
Natural exponential: f(x)=eˣ	f′(x)=eˣ	d/dx[eˣ]=eˣ
Natural logarithm: f(x)=ln x (x > 0)	f′(x)=1/x	d/dx[ln x]=1/x
Logarithm base a: f(x)=logₐx	f′(x)=1/(x·ln a)	d/dx[log₅x]=1/(x·ln 5)
Trigonometric: f(x)=sin x	f′(x)=cos x	d/dx[sin x]=cos x
Trigonometric: f(x)=cos x	f′(x)=-sin x	d/dx[cos x]=-sin x
Trigonometric: f(x)=tan x	f′(x)=sec² x	d/dx[tan x]=sec² x
Trigonometric: f(x)=sec x	f′(x)=sec x·tan x	d/dx[sec x]=sec x·tan x
Trigonometric: f(x)=csc x	f′(x)=-csc x·cot x	d/dx[csc x]=-csc x·cot x
Inverse sine: f(x)=sin⁻¹x (|x| ≤ 1)	f′(x)=1/√(1‑x²)	d/dx[sin⁻¹x]=1/√(1‑x²)
Inverse cosine: f(x)=cos⁻¹x	f′(x)=-1/√(1‑x²)	d/dx[cos⁻¹x]=-1/√(1‑x²)
Inverse tangent: f(x)=tan⁻¹x	f′(x)=1/(1+x²)	d/dx[tan⁻¹x]=1/(1+x²)

3. Differentiation Rules (Algebraic)

Rule	Formula	When to use
Constant rule	d/dx[c]=0	Any constant term
Constant multiple rule	d/dx[c·u]=c·du/dx	Factor a constant out of a term
Sum/Difference rule	d/dx[u±v]=du/dx ± dv/dx	Any sum or difference of functions
Product rule	d/dx[u·v]=u·dv/dx + v·du/dx	Product of two non‑constant functions
Quotient rule	d/dx[u/v]=(v·du/dx – u·dv/dx)/v²	Division of two functions (v ≠ 0)
Chain rule	d/dx[f(g(x))]=f′(g(x))·g′(x)	Composite functions, e.g. (3x+2)⁴, sin(2x²), e^{3x²}, log₂(5x+1)

3.1 Worked Example – Product Rule

Find dy/dx for y = (4x² + 3)·sin x.

Let u = 4x² + 3, v = sin x
u′ = 8x, v′ = cos x
y′ = u·v′ + v·u′ = (4x²+3)·cos x + 8x·sin x

3.2 Worked Example – Quotient Rule

Differentiate y = (x² + 1)/sin x.

u = x² + 1  u′ = 2x
v = sin x  v′ = cos x
y′ = (v·u′ – u·v′)/v²
    = (sin x·2x – (x²+1)·cos x) / sin² x

3.3 Worked Example – Chain & Power Rules

Differentiate y = √(5x² + 3).

y = (5x²+3)^{1/2}
g(x)=5x²+3  g′(x)=10x
f(u)=u^{1/2}  f′(u)=½u^{-1/2}
y′ = f′(g(x))·g′(x) = (½)(5x²+3)^{-1/2}·10x
    = 5x / √(5x²+3)

3.4 Worked Example – Composite Exponential & Logarithmic

Find dy/dx for y = e^{3x²} and for y = log₂(5x+1).

y = e^{3x²}
Let u = 3x²  du/dx = 6x
dy/dx = e^{u}·du/dx = e^{3x²}·6x

y = log₂(5x+1) = ln(5x+1)/ln 2
Let u = 5x+1  du/dx = 5
dy/dx = (1/(u·ln 2))·du/dx = 5 / [(5x+1)·ln 2]

4. Stationary Points and the First‑/Second‑Derivative Tests

4.1 Definitions

• A stationary point occurs where f′(x)=0 (provided the derivative exists).
• Types of stationary points:
  • Local maximum – function rises before the point and falls after it.
  • Local minimum – function falls before the point and rises after it.
  • Point of inflection – gradient changes sign but the function does not change from increasing to decreasing (or vice‑versa).

4.2 First‑Derivative Test (full justification)

1. Find all solutions of f′(x)=0.
2. Choose a test value in each interval defined by the solutions.
3. Determine the sign of f′(x) in each interval.

Interpretation:

• Sign change “+ → –” ⇒ local maximum.
• Sign change “– → +” ⇒ local minimum.
• No sign change ⇒ point of inflection (stationary inflection).

4.3 Second‑Derivative Test (Cambridge shortcut)

For a stationary point x = a (so f′(a)=0):

• If f″(a) > 0 ⇒ local minimum.
• If f″(a) < 0 ⇒ local maximum.
• If f″(a) = 0 ⇒ test is inconclusive; revert to the first‑derivative test.

4.4 Example – Classification of Stationary Points

Find and classify the stationary points of f(x)=x³‑3x²+2.

f′(x)=3x²‑6x = 3x(x‑2) → stationary at x=0 and x=2
Second derivative: f″(x)=6x‑6
f″(0)=‑6 < 0 → maximum at (0, 2)
f″(2)=6  > 0 → minimum at (2, ‑2)

4.5 Example – Stationary Point of Inflection

Consider g(x)=x³.

g′(x)=3x² → g′(0)=0  (stationary)
g″(x)=6x → g″(0)=0  (second‑derivative test inconclusive)
First‑derivative test: g′(x)=3x² is ≥0 on both sides, so the gradient does not change sign.
Conclusion: x=0 is a stationary point of inflection.

5. Applications of Differentiation

5.1 Rate of Change

If a quantity y depends on time t, then dy/dt is the instantaneous rate of change (e.g., speed, growth rate).

Example – Exponential growth

Population P(t)=1200e^{0.03t}. The rate of increase is

dP/dt = 1200·e^{0.03t}·0.03 = 0.03·P(t)

5.2 Uniformly Accelerated Motion

For motion along a straight line:

• Displacement s(t) – given function.
• Velocity v(t)=ds/dt.
• Acceleration a(t)=dv/dt = d²s/dt².

Worked Problem

Given s(t)=5t³‑2t²+4t (metres, t in seconds), find s(t), v(t) and a(t), then sketch the four standard graphs (s‑t, v‑t, a‑t, distance‑t).

v(t)=ds/dt = 15t²‑4t+4
a(t)=dv/dt = 30t‑4
Distance travelled from t=0 to t=2:
   Since v(t)≥0 on [0,2], distance = ∫₀² v(t) dt = s(2)‑s(0) = 5·8‑2·4+4·2 = 40‑8+8 = 40 m
Graphs:
 • s‑t : cubic curve, increasing faster as t grows.
 • v‑t : parabola opening upwards, crossing the t‑axis at t≈0.13 s (momentary rest).
 • a‑t : straight line with slope 30, intercept –4.
 • distance‑t : same as s‑t because v(t) never becomes negative on the interval.

5.3 Tangent and Normal Lines

For y = f(x) at the point (a, f(a)):

• Tangent line: y – f(a) = f′(a)(x – a)
• Normal line (perpendicular to the tangent): y – f(a) = –1/f′(a)·(x – a) (provided f′(a) ≠ 0)

Worked Example – Tangent & Normal

Find the equations of the tangent and normal to y = x³‑3x at x = 1.

f(1)=1‑3 = –2
f′(x)=3x²‑3 → f′(1)=0
Since f′(1)=0, the tangent is horizontal:
   Tangent: y = –2
The normal is vertical (slope undefined):
   Normal: x = 1

6. Common Pitfalls & How to Avoid Them

• Missing the chain rule: always check whether the function is a composition before applying the power rule.
• Incorrect notation: f′(dx) is meaningless; keep the variable separate from the prime.
• Power‑rule slip: remember the exponent drops by one (n xⁿ⁻¹), and the rule works for any rational exponent.
• Quotient‑rule denominator: the entire denominator is squared, not just the numerator.
• Sign errors in product/quotient rules: write the formula first, then substitute.
• Second‑derivative test misuse: it only works at stationary points where f′(a)=0. If f″(a)=0, revert to the first‑derivative test.
• For kinematics: differentiate the displacement function twice – never try to “integrate” the acceleration directly unless the initial conditions are given.

7. Exam Tips (Cambridge IGCSE 0606)

1. State the rule you are using (e.g., “Using the product rule…”) – marks are awarded for method.
2. Work consistently in one notation (Leibniz or prime) and keep it throughout the answer.
3. After differentiating, simplify algebraic expressions where possible; a tidy answer gains marks.
4. When asked for “gradient at x = a”, first find the first derivative, then substitute a – do not substitute before differentiating.
5. For stationary‑point questions, show both the first‑derivative test (sign chart) and, where appropriate, the second‑derivative test. Explain why a point is a maximum, minimum or inflection.
6. In kinematics problems, label the three graphs (s‑t, v‑t, a‑t) clearly and indicate any points of rest or change of direction.
7. If a question involves a vertical tangent, remember that the derivative is undefined (division by zero) – state this explicitly.
8. Check the domain of inverse‑trigonometric and logarithmic functions before differentiating; an undefined point invalidates a stationary‑point claim.