What is quotient-rule order?

The numerator is v\,u' - u\,v', not the other way round; the wrong order flips the sign.

What is not rewriting roots first?

1x^2 should be written x^-2 before differentiating to -2x^-3.

Differentiate y = 1x by writing it as a power. [2 marks]

SingaporeAdditional MathematicsSyllabus dot point

What is differentiation, and how do the power, chain, product and quotient rules let us find a derivative?

Differentiate powers of x and use the power, chain, product and quotient rules to find derivatives

A focused answer to the N(A)-Level Additional Mathematics outcome on differentiation. The meaning of the derivative as a gradient, the power rule, and the chain, product and quotient rules for finding derivatives.

Generated by Claude Opus 4.89 min answerUpdated 2026-06-06

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking
The answer
Examples in context
Try this

What this dot point is asking

SEAB wants you to differentiate: to find the derivative $\dfrac{dy}{dx}$ , which gives the gradient of a curve at any point. You need the power rule for terms like $x^n$ , and the chain, product and quotient rules for more involved expressions. Differentiation is the engine behind tangents, normals and stationary points, so fluency here pays off across the whole calculus strand.

The answer

What the derivative means

The derivative $\dfrac{dy}{dx}$ is the gradient of the curve at a point, that is, how fast $y$ changes as $x$ changes. Where the curve is steep the derivative is large; where it is flat the derivative is zero.

The power rule

For a power of $x$ , multiply by the power and reduce the power by one:

y = x^n \quad\Rightarrow\quad \frac{dy}{dx} = nx^{n-1}

A constant multiple stays in front, and a constant on its own differentiates to $0$ . Differentiate a sum term by term. Rewrite roots and reciprocals as indices first (for example $\sqrt{x} = x^{1/2}$ ) so the rule applies.

The chain rule

To differentiate a "function of a function", such as $(2x + 1)^5$ , set the inside as $u$ , differentiate the outside and the inside separately, and multiply:

\frac{dy}{dx} = \frac{dy}{du}\times\frac{du}{dx}

In practice: differentiate the outer function, keep the inside unchanged, then multiply by the derivative of the inside.

The product rule

For a product $y = uv$ of two functions:

\frac{dy}{dx} = u\frac{dv}{dx} + v\frac{du}{dx}

Differentiate one factor at a time, keeping the other, and add.

The quotient rule

For a quotient $y = \dfrac{u}{v}$ :

\frac{dy}{dx} = \frac{v\dfrac{du}{dx} - u\dfrac{dv}{dx}}{v^2}

Mind the order in the numerator: it is "bottom times derivative of top, minus top times derivative of bottom", all over the bottom squared.

Examples in context

Example 1. Speed from a distance graph. If distance is given as a function of time, its derivative is the speed at any instant. This is why differentiation is the tool behind the kinematics topic, where velocity is the derivative of displacement.

Example 2. Marginal change in economics. A cost function's derivative gives the marginal cost, the extra cost of producing one more unit. Differentiation turns a total-quantity function into a rate-of-change function, useful far beyond pure mathematics.

Try this

Q1. Differentiate $y = 5x^3$ . [1 mark]

Cue. $\dfrac{dy}{dx} = 15x^2$ .

Q2. Differentiate $y = (4x - 3)^2$ using the chain rule. [2 marks]

Cue. $2(4x - 3) \times 4 = 8(4x - 3)$ .

Q3. Differentiate $y = \dfrac{1}{x}$ by writing it as a power. [2 marks]

Cue. $y = x^{-1}$ , so $\dfrac{dy}{dx} = -x^{-2} = -\dfrac{1}{x^2}$ .

Exam-style practice questions

Practice questions written in the style of SEAB exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

Original3 marksDifferentiate

y = 3x^4 - 2x + 5

with respect to

x

Show worked answer →

Differentiate term by term using the power rule (multiply by the power, reduce the power by one):

$\dfrac{dy}{dx} = 3(4)x^3 - 2 = 12x^3 - 2$ .

The constant $5$ differentiates to $0$ .

What markers reward: applying the power rule to each term, differentiating $-2x$ to $-2$ , and the constant to $0$ .

Original4 marksDifferentiate

y = (2x + 1)^5

using the chain rule.

Show worked answer →

Let the inside be $u = 2x + 1$ , so $y = u^5$ .

Then $\dfrac{dy}{du} = 5u^4$ and $\dfrac{du}{dx} = 2$ .

Chain rule: $\dfrac{dy}{dx} = 5u^4 \times 2 = 10(2x + 1)^4$ .

What markers reward: identifying the inside function, differentiating outer and inner separately, and multiplying to get $10(2x + 1)^4$ .