Secondary 3 Additional Mathematics | Singapore G3
Differentiation

How a Curve Can Have a Gradient at One Exact Point

A straight line has one gradient everywhere. A curve changes direction continuously—yet calculus lets us define its gradient at one exact point.

That is the central idea of differentiation.

For a curve, the gradient between two points is easy: draw a secant line and use rise over run.

But what if the two points move closer and closer together?

secant gradient → points approach → limiting gradient → tangent gradient.

The derivative captures that limiting rate of change.

The current Singapore G3 Additional Mathematics syllabus includes the derivative as tangent gradient and rate of change; standard derivative notation; derivatives of rational powers, sin x, cos x, tan x, eˣ and ln x; product, quotient and chain rules; increasing/decreasing functions; stationary points; second derivative classification; tangents and normals; connected rates; and maximum/minimum problems.

SEAB 2027 G3 Additional Mathematics syllabus (K341) →

The official syllabus groups differentiation and integration together under Calculus. This page owns the differentiation half; integration should remain a separate teaching owner.

The Topic Job

Measure how a quantity changes at an instant, then use that local rate to understand shape, motion and optimisation.

Average Gradient Comes First

For y=f(x), between x and x+h:

average gradient = [f(x+h)−f(x)]/h.

This is the gradient of the secant joining two points on the curve.

From Secant to Tangent

As h becomes smaller, the second point moves closer to the first.

If the gradients approach a limiting value, that limit defines the derivative:

f′(x)=lim(h→0)[f(x+h)−f(x)]/h.

You are not dividing by h=0. You study what the quotient approaches as h tends to zero.

First-Principles Example — Differentiate x²

Let f(x)=x².

[f(x+h)−f(x)]/h
=[(x+h)²−x²]/h
=[x²+2xh+h²−x²]/h
=[2xh+h²]/h
=2x+h.

As h→0:

f′(x)=2x.

The derivative is a new function. It tells us the gradient of y=x² at every x.

Derivative Notation

• f′(x)
• dy/dx
• d/dx [f(x)]
• y′
• f″(x) or d²y/dx² for the second derivative

These notations emphasise different aspects, but at this level they describe the same derivative object.

The Power Rule

d/dx(xⁿ)=nxⁿ⁻¹

for the rational powers required by the syllabus, wherever the function is defined.

Examples:

• d/dx(x⁵)=5x⁴
• d/dx(x⁻²)=−2x⁻³
• d/dx(√x)=d/dx(x¹ᐟ²)=1/(2√x)

Constants, Sums and Differences

Differentiation is linear:

d/dx[af(x)+bg(x)] = af′(x)+bg′(x).

A constant differentiates to zero because it does not change as x changes.

Standard Trigonometric and Exponential Derivatives

• d/dx(sin x)=cos x
• d/dx(cos x)=−sin x
• d/dx(tan x)=sec²x
• d/dx(eˣ)=eˣ
• d/dx(ln x)=1/x

For the trig derivatives, x must be interpreted in radians for the standard formulas to hold without extra conversion factors.

Why eˣ Is Special

The function eˣ is its own derivative:

d/dx(eˣ)=eˣ.

Its instantaneous rate of change equals its current value. This is why e appears naturally in continuous growth and decay models.

Product Rule

For y=uv:

dy/dx = u(dv/dx)+v(du/dx).

The derivative of a product is not the product of the derivatives.

Example:

y=x²eˣ.

Then:

y′=x²eˣ+2xeˣ=eˣ(x²+2x).

Quotient Rule

For y=u/v:

dy/dx=[v(du/dx)−u(dv/dx)]/v².

A common memory aid may help, but the minus sign and denominator square must survive exactly.

Example:

y=(x²+1)/x.

You can use quotient rule, but an even better first move is simplify:

y=x+x⁻¹.

Then:

y′=1−x⁻².

A strong student simplifies before choosing a heavier rule.

Chain Rule

If y=f(g(x)), then:

dy/dx = (dy/du)(du/dx).

Think of one function inside another.

Example:

y=(3x+1)⁵.

Outer derivative:

5(3x+1)⁴.

Multiply by inner derivative 3:

y′=15(3x+1)⁴.

The Earliest Chain-Rule Weak Link

Students often differentiate the outside correctly and forget the derivative of the inside.

A useful spoken routine is:

differentiate the outside, keep the inside, multiply by the derivative of the inside.

Tangent Gradient

If y=f(x), the gradient of the tangent at x=a is:

f′(a).

To find the tangent equation:

1. find the point on the curve;
2. differentiate;
3. evaluate the derivative at the point;
4. use point-gradient form.

Worked Example 1 — Tangent

Find the tangent to y=x³−2x at x=2.

Point:

y=8−4=4 → (2,4).

Derivative:

dy/dx=3x²−2.

At x=2:

m=12−2=10.

Tangent:

y−4=10(x−2).

Normals

A normal is perpendicular to the tangent.

If tangent gradient m≠0, normal gradient is:

−1/m.

If the tangent is horizontal, the normal is vertical, so treat that case geometrically rather than forcing a reciprocal of zero.

Increasing and Decreasing

• f′(x)>0 → function increasing locally;
• f′(x)<0 → function decreasing locally;
• f′(x)=0 → stationary candidate.

The derivative acts like a sign map for the curve’s direction.

Stationary Points

A stationary point occurs where:

dy/dx=0.

Possible types include:

• local maximum;
• local minimum;
• stationary point of inflexion.

Zero gradient alone does not tell you which type.

First-Derivative Sign Test

Check the sign of f′ around the stationary point.

Sign change	Interpretation
+ to −	local maximum
− to +	local minimum
same sign across	not a max/min; may be stationary inflexion

Second-Derivative Test

At a stationary point x=a:

• f″(a)>0 → local minimum;
• f″(a)<0 → local maximum;
• f″(a)=0 → test is inconclusive.

Do not write “f″=0 means point of inflexion.” It does not. More evidence is needed.

Worked Example 2 — Stationary Points

Find and classify the stationary points of:

y=x³−3x²−9x+5.

Differentiate:

y′=3x²−6x−9=3(x−3)(x+1).

Stationary when x=−1 or x=3.

Second derivative:

y″=6x−6.

• at x=−1, y″=−12<0 → local maximum;
• at x=3, y″=12>0 → local minimum.

Then calculate the corresponding y-values from the original function.

Differentiation as Rate of Change

If s is displacement and t is time:

v=ds/dt.

If v is velocity:

a=dv/dt=d²s/dt².

The derivative has not changed meaning. It is still instantaneous rate of change; only the quantities have changed.

Units Are Part of the Derivative

If distance is measured in metres and time in seconds:

• ds/dt has units m/s;
• d²s/dt² has units m/s².

A rate is unfinished until we know what is changing with respect to what.

Connected Rates of Change

Suppose the area of a circle is A=πr² and r changes with time.

Differentiate with respect to t using the chain rule:

dA/dt = 2πr · dr/dt.

This connects the rate of change of area to the rate of change of radius.

The crucial question is always:

Which quantities are changing, and with respect to which variable?

Optimisation

Many maximum/minimum problems follow the same architecture:

1. define the quantity to optimise;
2. use constraints to write it in one variable;
3. differentiate;
4. set derivative to zero;
5. classify the stationary point;
6. interpret the answer in context.

Worked Example 3 — Maximum Area

A rectangle has perimeter 40. Let one side be x.

Other side = 20−x.

Area:

A=x(20−x)=20x−x².

Differentiate:

dA/dx=20−2x.

Set zero:

20−2x=0 → x=10.

Second derivative:

d²A/dx²=−2<0.

So area is maximised when both sides are 10: the rectangle is a square.

Domain Still Matters in Optimisation

For the rectangle, 0<x<20.

A stationary point outside that interval would not describe a possible rectangle.

Calculus finds mathematical candidates. The context decides whether they are admissible.

The Earliest Weak Link

What you see	Likely weak link
Power rule applied correctly but chain factor missing	composite-function structure not seen
Uses u′v′ for a product	product rule not understood
Finds x-coordinate of stationary point but not y	forgets stationary point is a coordinate
Writes f″=0 therefore inflexion	second derivative test misinterpreted
Connected-rate equation mixes dr/dt and dA/dr	“with respect to” variable not tracked
Optimisation gives impossible length	domain/context check missing

Common Mistakes to Repair

• Derivative of a product = product of derivatives. False.
• Forgetting inner derivative in chain rule.
• Forgetting minus sign in d(cos x)/dx.
• Using degree mode conceptually for standard trig derivatives. Standard formulas assume radians.
• Setting y=0 instead of dy/dx=0 for stationary points.
• Classifying from f″=0. Inconclusive.
• Giving only a stationary x-value when a point is requested.
• Optimising before reducing to one independent variable.

Retrieval Check

1. What does f′(a) represent geometrically?
2. Differentiate x⁷.
3. Differentiate x⁻³.
4. Differentiate sin x, cos x and eˣ.
5. State the product rule.
6. What condition defines a stationary point?
7. What does f″(a)<0 tell you if f′(a)=0?
8. What is the relation between tangent and normal gradients?

Transfer Set

1. Differentiate y=(2x−1)⁶.
2. Differentiate y=x³ln x.
3. Differentiate y=(x²+1)/(x−1).
4. Find the tangent and normal to y=x²+3x at x=1.
5. Find and classify the stationary points of y=x³−6x²+9x.
6. A circle’s radius increases at 2 cm/s. Find dA/dt when r=5 cm.
7. A rectangle has area A=x(30−2x). Find its maximum area over the physically meaningful domain.

Answer outline — open only after attempting

1. 12(2x−1)⁵.
2. 3x²ln x+x².
3. Use quotient rule or first simplify if possible: y′=[2x(x−1)−(x²+1)]/(x−1)².
4. Point (1,4); derivative 2x+3 gives tangent gradient 5, normal gradient −1/5.
5. y′=3(x−1)(x−3), so x=1,3; y″=6x−12, classify x=1 max, x=3 min; calculate y-values.
6. dA/dt=2πr dr/dt=20π cm²/s.
7. A=30x−2x²; A′=30−4x=0 → x=7.5; A″=−4, maximum A=112.5.

Independent Checks

• Estimate a tangent gradient from nearby graph points and compare with the derivative.
• Substitute a simple x-value into a derivative to check sign and rough size.
• Expand a simple chain-rule example and differentiate the expanded form as a second route.
• For stationary points, verify f′=0 at the reported x-values.
• For optimisation, inspect endpoints/domain as well as stationary candidates.
• For connected rates, verify units.

For Parents and Tutors — What This Topic Is Really Testing

Differentiation should not begin as a table of derivative rules.

The child first needs one stable meaning:

the derivative tells how fast the output is changing with respect to the input, at that point.

Then ask:

• What are the input and output quantities?
• What are the units of the derivative?
• Is this one function, a product, quotient or composition?
• Does the derivative sign match the graph’s direction?
• Is a stationary point actually a max/min?
• Does the optimisation answer fit the physical domain?

The existing BTT reflection A Gradient Is Not Finished Until We Know What Is Changing explores the conceptual habit. This page owns the formal teaching machinery.

The teaching target is meaning → rule selection → derivative → interpretation.

Where This Connects Next

• Coordinate Geometry
• Trigonometric Functions
• Exponential & Logarithmic Functions
• Integration
• Secondary 3 A-Math Complete Map
• Additional Mathematics Directory

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Bukit Timah Tutor Mathematics
The derivative is not a symbol-manipulation trick. It is a precise way to ask what a changing quantity is doing right now.