E2.12Extended tier0580 · Section 2: Algebra and graphs

Differentiation — derivatives of ax^n, stationary/turning points

Key learnings

Stationary points: set dy/dx = 0
At a maximum or minimum the gradient is 0. Solve $\dfrac{dy}{dx} = 0$ for $x$ , then substitute back to get $y$ .
Differentiate term by term (power rule)
For $y = ax^n$ , $\dfrac{dy}{dx} = anx^{\,n-1}$ : multiply by the power, then drop the power by one. A constant differentiates to 0.
dy/dx is the gradient
The derivative gives the gradient of the curve at any $x$ . For a tangent or velocity, evaluate $\dfrac{dy}{dx}$ at the required point.

Common mistakes

Drawn from real examiner reports.

Dropping the ± when square-rooting

From $x^2 = 4$ students gave only $x = 2$ , ignoring $x = -2$ ; conversely $\sqrt[3]{8} = 2$ only (not $\pm 2$ ). Know which sign(s) are valid.

Seen on recent higher-tier exams.

Setting dy/dx = 0 when a gradient is given

When asked for points with a given gradient, students equated $\dfrac{dy}{dx}$ to 0 instead of to that gradient and lost the method mark.

Seen on a recent higher-tier exam.

A point, not the tangent equation

Asked for the equation of the tangent, students stopped at the coordinates $(2, -96)$ ; subsequent incorrect working also lost the final mark.

Seen on a recent higher-tier exam.

Second-derivative sign rule reversed

To classify a stationary point, $\dfrac{d^2y}{dx^2} > 0$ is a MINIMUM and $\dfrac{d^2y}{dx^2} < 0$ is a MAXIMUM. Swapping the two, or stopping after finding the $x$ -values without classifying, loses marks.

Substituting before differentiating

To find a gradient at $x=4$ , differentiate FIRST, then substitute: for $y=x^2-6x+4$ , $\dfrac{dy}{dx}=2x-6$ , giving $2$ at $x=4$ . Putting $x=4$ into $y$ before differentiating gives a value, not a gradient.

Exam tips

Recognise when a problem needs calculus
Questions about gradients, tangents, turning points, maximum/minimum values or velocity are calculus — differentiate first to gain at least one mark.
Tangent at a stationary point is horizontal
At a stationary point the gradient is $0$ , so the tangent is horizontal: its equation is $y=$ (the $y$ -value there), e.g. $y=-48$ — the equation, not the point $(2,-48)$ .
Use a stationary point to find a constant
Given a stationary point at $x=1$ with an unknown $k$ , differentiate, put $x=1$ into $\dfrac{dy}{dx}$ , set it equal to $0$ , and solve for $k$ . Differentiate before substituting, never the other way round.

Revision notes

For $y = ax^n$ : $\dfrac{dy}{dx} = a\,n\,x^{\,n-1}$ — multiply by the power, then subtract 1 from the power. Differentiate each term; a constant goes to 0. $y = x^2 - 6x + 4 \;\Rightarrow\; \frac{dy}{dx} = 2x - 6.$

Flashcards

Foundation

What does dy/dx represent?

1 of 1

Test yourself — 13 questionsQuick multiple-choice check with instant feedback and a full review.

← All revision topics

Differentiation — derivatives of ax^n, stationary/turning points

Key learnings

Common mistakes

Exam tips

Revision notes

The power rule

dy/dx is the gradient

Stationary points — set the gradient to 0

Equation of a tangent

Flashcards

Practice questions