Differentiate y = 4x^3 - 2x. [3 marks]

ScotlandMathsSyllabus dot point

How do you find the gradient of a curve at any point, and how does the derivative reveal stationary points and the shape of a graph?

Differentiation of polynomial, root and reciprocal functions and of sine and cosine, the gradient of a curve and the equation of a tangent, increasing and decreasing functions, and stationary points and their nature.

A focused answer to the SQA Higher Mathematics differentiation content, covering differentiating polynomial, root, reciprocal and trigonometric functions, the gradient of a curve and the tangent equation, increasing and decreasing functions, and stationary points and their nature.

Generated by Claude Opus 4.810 min answerUpdated 2026-06-02

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

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What this dot point is asking
The rules
Tangents and increasing or decreasing
Stationary points
Examples in context
Try this

What this dot point is asking

The SQA wants you to differentiate polynomial, root, reciprocal and trigonometric functions, find the gradient of a curve and the equation of a tangent, decide where a function is increasing or decreasing, and find and classify stationary points.

The rules

Differentiation produces the gradient function $f'(x)$ , which gives the slope of the curve at every point. The power rule handles all polynomial, root and reciprocal terms once they are written as powers, and the two trigonometric derivatives complete the Higher toolkit.

Tangents and increasing or decreasing

The gradient at $x = a$ is $f'(a)$ , and the tangent there is $y - f(a) = f'(a)(x - a)$ . To find a tangent you need both the point (substitute into the original equation) and the gradient (substitute into the derivative). A function is increasing where $f'(x) > 0$ and decreasing where $f'(x) < 0$ ; to prove a function is always increasing you show $f'(x) > 0$ for all $x$ , often by completing the square on the derivative.

Show that

f(x) = x^3 + 4x - 1

is always increasing

To show a function is always increasing, we need to prove that its derivative is positive for every real $x$ .

Step 1: Differentiate to get the gradient function

Applying the power rule term by term:

f'(x) = 3x^2 + 4

Step 2: Argue that the derivative is always positive

Any real number squared is non-negative, so $x^2 \ge 0$ for all $x \in \mathbb{R}$ . Therefore $3x^2 \ge 0$ , and adding $4$ gives:

f'(x) = 3x^2 + 4 \ge 0 + 4 = 4 > 0 \text{ for all } x

Final answer: Since $f'(x) > 0$ for all real $x$ , the function $f(x) = x^3 + 4x - 1$ is always increasing, with no stationary points.

Stationary points

A stationary point occurs where $f'(x) = 0$ , meaning the tangent is horizontal. To classify it, either build a nature table of the sign of $f'(x)$ just before and just after the point, or use the second derivative: $f''(x) > 0$ gives a minimum and $f''(x) < 0$ gives a maximum.

Find and classify the stationary points of

y = x^3 - 3x

Find the coordinates of all stationary points and determine the nature of each.

Step 1: Differentiate and set the derivative to zero

Stationary points occur where the gradient is zero. Differentiating and factorising:

\dfrac{dy}{dx} = 3x^2 - 3 = 3(x^2 - 1) = 3(x - 1)(x + 1)

Setting $\dfrac{dy}{dx} = 0$ gives $x = 1$ or $x = -1$ . These are the x-coordinates of the stationary points.

Step 2: Classify each point using the second derivative

Differentiate again to get $\dfrac{d^2y}{dx^2} = 6x$ . A positive second derivative means a minimum; a negative second derivative means a maximum. At $x = 1$ : $\dfrac{d^2y}{dx^2} = 6 > 0$ , so this is a minimum. At $x = -1$ : $\dfrac{d^2y}{dx^2} = -6 < 0$ , so this is a maximum.

Step 3: Find the $y$ -coordinates and state the full points

Substitute each $x$ -value into the original equation:

y(1) = 1 - 3 = -2 \qquad y(-1) = -1 + 3 = 2

Final answer: The stationary points are $(1, -2)$ , a minimum, and $(-1, 2)$ , a maximum.

A stationary point can be a maximum, a minimum or a point of inflection; always state which.

Examples in context

The derivative of $\sin$ and $\cos$ describes oscillating systems. For a pendulum whose displacement is $x = \cos t$ , the velocity is $\dfrac{dx}{dt} = -\sin t$ , which is zero at $t = 0$ (the extreme of the swing, where the bob is momentarily still) and largest in magnitude when $t = \dfrac{\pi}{2}$ (passing through the centre). Reading the stationary points of the displacement against the values of its derivative connects the calculus directly to the physical motion.

Try this

Q1. Differentiate $y = 4x^3 - \dfrac{2}{x}$ . [3 marks]

Cue. Write $-2x^{-1}$ , so $\dfrac{dy}{dx} = 12x^2 + 2x^{-2}$ .

Q2. Find the gradient of $y = x^2 - 5x$ at the point where $x = 3$ . [2 marks]

Cue. $\dfrac{dy}{dx} = 2x - 5$ , so at $x = 3$ the gradient is $1$ .

Exam-style practice questions

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

SQA Higher 20185 marksA curve has equation

y = x^3 - 3x^2 - 9x + 5

. Find the coordinates of the stationary points and determine the nature of each.

Show worked answer →

Differentiate: $\dfrac{dy}{dx} = 3x^2 - 6x - 9$ (1 mark).

Set to zero: $3x^2 - 6x - 9 = 0$ , divide by $3$ to get $x^2 - 2x - 3 = 0$ , so $(x - 3)(x + 1) = 0$ and $x = 3$ or $x = -1$ (1 mark).

y-coordinates: $y(3) = 27 - 27 - 27 + 5 = -22$ and $y(-1) = -1 - 3 + 9 + 5 = 10$ (1 mark).

Nature: $\dfrac{d^2y}{dx^2} = 6x - 6$ . At $x = 3$ , $6(3) - 6 = 12 > 0$ , a minimum at $(3, -22)$ ; at $x = -1$ , $6(-1) - 6 = -12 < 0$ , a maximum at $(-1, 10)$ (2 marks). Markers reward the derivative, solving for both x-values, both y-coordinates, and a valid nature check.

SQA Higher 20214 marksFind the equation of the tangent to the curve

y = 2x^2 - 5x + 1

at the point where

x = 2

Show worked answer →

Find the point: $y(2) = 2(4) - 5(2) + 1 = 8 - 10 + 1 = -1$ , so the point is $(2, -1)$ (1 mark).

Differentiate: $\dfrac{dy}{dx} = 4x - 5$ (1 mark). At $x = 2$ the gradient is $4(2) - 5 = 3$ (1 mark).

Tangent through $(2, -1)$ with gradient $3$ : $y - (-1) = 3(x - 2)$ , so $y = 3x - 7$ (1 mark). Markers reward the point on the curve, the derivative, the gradient at $x = 2$ , and the tangent equation.

Related dot points

Sources & how we know this

SQA Higher Mathematics Course Specification — SQA (2018)