Differentiation is the language Additional Mathematics uses to describe how a quantity is changing at an instant. A graph can tell us where a function is. Differentiation adds another layer: how steeply it is moving, whether that motion is increasing or decreasing, where it momentarily becomes flat, and how local behaviour reveals the shape of a curve.
This Secondary 4 Additional Mathematics Learning Guide develops differentiation as a connected reasoning system rather than a list of derivative formulas. It belongs to the Additional Mathematics Learning Hub and connects functions, algebra, graph interpretation, calculus, optimisation and examination control.
Function → gradient → derivative → stationary behaviour → optimisation → interpretation.
A derivative measures local change
If y=f(x), then dy/dx or f′(x) describes the gradient of the curve at a particular x-value. For a straight line, the gradient is constant. For a curve, the gradient changes as x changes. The derivative is therefore itself a function.
This is the first conceptual shift students need to hold. Differentiation does not merely produce a new expression. It transforms one description of a relationship into another description of the same relationship.
| Original object | Derivative tells us |
|---|---|
| Position s(t) | Velocity ds/dt |
| Velocity v(t) | Acceleration dv/dt |
| Curve y=f(x) | Gradient dy/dx |
| Area or volume formula | How rapidly the quantity changes |
From average gradient to instantaneous gradient
The gradient of a chord joining two nearby points gives an average rate of change over an interval. As the second point moves toward the first, the chord approaches the tangent. Differentiation captures the limiting gradient of that tangent.
Average rate asks what changed over a span. Differentiation asks what the change is doing now.
The power rule
For y=xn, dy/dx=nxn−1. This becomes the basic engine for polynomials and for expressions that can first be rewritten as powers.
If y=5x4−3x2+7x−9, then dy/dx=20x3−6x+7. The constant disappears because a constant does not change as x changes.
Rewrite before differentiating
If y=3/x2+2√x, rewrite it as 3x−2+2x1/2. Then dy/dx=−6x−3+x−1/2. A-Math rewards representation choice: the expression you are given is not always the expression you should operate on.
Worked Example 1 | Differentiate and evaluate a gradient
Given y=2x3−5x2+4, find the gradient at x=2.
dy/dx=6x2−10x. At x=2, dy/dx=24−20=4.
The answer 4 is a gradient, not the y-coordinate. Keeping those roles separate prevents tangent and normal errors.
The chain rule
If y=(3x+1)5, there is an inner function 3x+1 inside an outer fifth power. Differentiate the outer structure and multiply by the derivative of the inner structure:
dy/dx=15(3x+1)4.
The chain factor captures the fact that the inside is changing too.
Worked Example 2 | Chain rule with negative power
For y=(2x−5)−2, dy/dx=−2(2x−5)−3×2=−4(2x−5)−3.
Product structures
When two x-dependent factors are multiplied, differentiating each and multiplying the answers is generally wrong. For y=uv:
dy/dx=u(dv/dx)+v(du/dx).
Worked Example 3 | Product rule
Differentiate y=x2(3x−4)3. Let u=x² and v=(3x−4)³. Then du/dx=2x and dv/dx=9(3x−4)².
dy/dx=9x²(3x−4)²+2x(3x−4)³. Factorising gives x(3x−4)²(15x−8).
The factorised form is especially useful when the next instruction is to set dy/dx=0.
Tangents: turn derivative information into a line
- Find the point on the curve.
- Differentiate and evaluate the gradient at that point.
- Use a straight-line equation with the correct point and gradient.
Worked Example 4 | Equation of a tangent
For y=x³−2x+1, find the tangent at x=2. The point is (2,5). dy/dx=3x²−2, so the gradient is 10. Using y−5=10(x−2), the tangent is y=10x−15.
Normals: perpendicularity introduces a second gradient
If the tangent gradient is m, the normal gradient is −1/m when both gradients are finite. In the previous example, the normal gradient is −1/10, not −10.
Stationary points
A stationary point occurs where the tangent is horizontal, so dy/dx=0. But “stationary” does not automatically mean maximum or minimum. The derivative condition finds candidates; classification requires more evidence.
- positive to negative gradient: local maximum;
- negative to positive gradient: local minimum;
- same sign on both sides: possible stationary point of inflection.
Worked Example 5 | Find and classify stationary points
For y=x³−3x²−9x+5, dy/dx=3x²−6x−9=3(x−3)(x+1). Hence x=−1 or 3. The points are (−1,10) and (3,−22).
d²y/dx²=6x−6. At x=−1, the second derivative is −12, so (−1,10) is a local maximum. At x=3, it is 12, so (3,−22) is a local minimum.
The second derivative
| At a stationary point | Interpretation |
|---|---|
| d²y/dx²>0 | Local minimum |
| d²y/dx²<0 | Local maximum |
| d²y/dx²=0 | Test is inconclusive |
A zero second derivative does not prove a stationary point of inflection. It only means another classification method is needed.
Optimisation
Optimisation problems test more than differentiation. The crucial step is often building the function that should be differentiated. A reliable sequence is:
Model → reduce to one variable → differentiate → solve → classify → return to context.
Worked Example 6 | Maximum area with fixed perimeter
A rectangle has perimeter 40 cm. Let sides be x and y. Then x+y=20, so y=20−x. Area A=x(20−x)=20x−x².
dA/dx=20−2x=0 gives x=10. d²A/dx²=−2<0, so this is a maximum. Hence y=10 and the maximum-area rectangle is a 10 cm by 10 cm square.
Rates of change
If s is displacement in metres and t is time in seconds, ds/dt is velocity in m/s. If v is velocity, dv/dt is acceleration in m/s². Units help detect nonsense even when algebra looks tidy.
Worked Example 7 | Velocity and acceleration
A particle has s=t³−6t²+9t. Then v=3t²−12t+9 and a=6t−12. At t=2, v=−3 m/s and a=0 m/s². Zero acceleration does not mean zero velocity.
Exponential and trigonometric derivatives
Secondary 4 differentiation extends beyond polynomials. For example, d/dx[e3x]=3e3x and d/dx[sin(2x)]=2cos(2x). The recurring principle is the same: identify the outer function, identify the inner function, and preserve the dependency.
Worked Example 8 | Composite exponential
For y=4e2x−1, dy/dx=8e2x−1.
Worked Example 9 | Composite trigonometric function
For y=3sin(4x), dy/dx=12cos(4x).
Curve sketching from derivative information
- Intercepts tell where the graph meets axes.
- Stationary points show horizontal tangents.
- The sign of dy/dx shows increasing or decreasing intervals.
- The sign of d²y/dx² supports local concavity.
- End behaviour comes from the dominant structure.
Common differentiation failure modes
| Error | Likely cause | Repair |
|---|---|---|
| Chain factor missing | Composite structure not recognised | Mark inner and outer functions |
| Tangent uses y-value as gradient | Point and derivative roles confused | Write point and gradient separately |
| Normal gradient written as −m | Perpendicular rule incomplete | Use m₁m₂=−1 |
| Stationary point not classified | dy/dx=0 treated as complete | Use second derivative or sign analysis |
| Optimisation still has two variables | Constraint not used | Reduce to one variable first |
| Correct x, wrong point | y-coordinate not recovered | Substitute into original function |
| Rates lose units | Derivative treated symbolically only | Attach variable meaning and units |
Differentiation decision tree
- What is the function structure?
- Can the expression be simplified first?
- Which derivative rule is load-bearing?
- What does the derivative represent?
- What is the question asking: derivative, tangent, normal, stationary point, optimum or rate?
- How will the result be verified?
Examination control
Many differentiation questions contain a short calculus step followed by substantial algebra. Differentiate cleanly, simplify only as much as the next operation needs, then solve the resulting algebra with the same care used in an algebra chapter.
Derivative technique earns access to the question. Interpretation and algebra convert that access into marks.
Independent practice
- Differentiate y=4x⁵−3x²+7x−8.
- Differentiate y=(5x−2)⁴.
- Differentiate y=x³(2x+1).
- For y=x²+3x−4, find the tangent at x=1.
- For the same curve, find the normal at x=1.
- Find and classify the stationary points of y=x³−6x²+9x+1.
- A rectangle has perimeter 24 cm. Find the maximum area.
- A particle has s=2t³−9t²+12t. Find v and a.
- Differentiate y=5e3x.
- Differentiate y=2cos(3x).
Explained answers
1. 20x⁴−6x+7. 2. 20(5x−2)³. 3. 8x³+3x². 4. At x=1, point (1,0), gradient 5, so y=5x−5. 5. Normal gradient −1/5, so y=−(1/5)(x−1). 6. x=1 and 3; (1,5) is a local maximum and (3,1) a local minimum. 7. 36 cm². 8. v=6t²−18t+12, a=12t−18. 9. 15e3x. 10. −6sin(3x).
Teaching sequence
Begin with gradient meaning and polynomial derivatives, then move to rewritten powers, composite functions, products and quotients. Tangents and normals should follow only after point, derivative and straight-line gradient are clearly separated. Stationary points connect derivative technique to graph behaviour. Optimisation should follow classification. Motion and rates make units explicit. Finally remove the chapter label and mix differentiation with functions, exponentials, trigonometry and algebra.
How this guide connects to Secondary 4 A-Math
- Integration, definite integrals and area under curves
- Trigonometric functions, identities and equations
- Exponential and logarithmic functions and equations
Final thought
Differentiate the structure. Interpret the rate. Verify the behaviour.
Return to the Additional Mathematics Learning Hub.
