Mathematics · AQA 7357 · Topic G

Differentiation

Finding gradients and rates of change: the rules for differentiating, tangents, stationary points, the product, quotient and chain rules, and implicit and parametric curves.

6 sub-topics
1 in Year 12, 3 in Year 13
118 practice questions
6 animated lessons

Practice questions (118) Lessons Worksheets All AQA Mathematics topics

What the specification says

Every sub-topic AQA can examine in this topic, with its reference number from the official specification. The Y12 and Y13 tags show when each sub-topic is usually taught.

G1Gradients, first principles and second derivativesY13
G2Differentiating standard functionsY13
G3Tangents, normals and stationary pointsY12
G4The product, quotient and chain rulesY13
G5Implicit and parametric differentiationY13
G6Construct simple differential equationsY13

Try 3 questions

Easy, medium and hard questions from this topic, each with a worked solutions and the mark for every step. All 118 questions in this topic, free, no account needed.

Differentiation

Easy

Gradient of a curve · Year 12 · 3 marks

A curve has equation

y = 2x^3 - 5x^2 + 4

.

Find the gradient of the curve at the point where $x = 2$ .
(3)

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Differentiate each term: multiply by the power, then reduce the power by 1. The constant disappears. $\dfrac{\mathrm{d}y}{\mathrm{d}x}$ ${}= 6x^2 - 10x$
The gradient at a point is the derivative evaluated there.At $x = 2$ : $24 - 20 = 4$

Total3

MMethod mark: for a correct method, even with a slip.
AAccuracy mark: for a correct answer from a correct method.
BIndependent mark: for a correct fact or result on its own, with no method needed.

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Differentiation

Medium

Stationary points · Year 12 · 6 marks

A curve has equation

y = x^3 - 6x^2 + 9x + 1

.

Find the coordinates of the stationary points of the curve and determine their nature.
(6)

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Stationary points are where the gradient is zero, so differentiate first. $\dfrac{\mathrm{d}y}{\mathrm{d}x}$ ${}= 3x^2 - 12x + 9$
Set the derivative to zero and solve. $3(x - 1)(x - 3)$ ${}= 0$ ${}\Rightarrow x$ ${}= 1,\ x$ ${}= 3$
Substitute each $x$ into the original equation to get $y$ . $(1, 5)$ and $(3, 1)$
The second derivative tells you the shape at each point. $\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2}$ ${}= 6x - 12$
Negative second derivative means the curve is bending down.At $x = 1$ : $-6 < 0$ , so $(1, 5)$ is a maximum.
Positive second derivative means the curve is bending up.At $x = 3$ : $6 > 0$ , so $(3, 1)$ is a minimum.

Total6

MMethod mark: for a correct method, even with a slip.
AAccuracy mark: for a correct answer from a correct method.
BIndependent mark: for a correct fact or result on its own, with no method needed.

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Differentiation

Hard

Making the biggest box · Year 12 · 7 marks

An open box is made from a 24 cm square of card by cutting a square of side

x

cm from each corner and folding up the sides.

(a)
Show that the volume is $V = 4x^3 - 96x^2 + 576x$ .
(2)
(b)
Find the maximum volume, showing that it is a maximum.
(5)

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Part (a)2 marks

The base is $24 - 2x$ each way and the height is $x$ . $V = x(24 - 2x)^2$
$= x(576 - 96x + 4x^2)$ ${}= 4x^3 - 96x^2 + 576x$

Part (b)5 marks

$\dfrac{\mathrm{d}V}{\mathrm{d}x}$ ${}= 12x^2 - 192x + 576$ ${}= 12(x - 4)(x - 12)$
When $x = 12$ the base has width 0. $x = 4$ ( $x = 12$ gives no box)
Negative second derivative means a maximum. $\dfrac{\mathrm{d}^2V}{\mathrm{d}x^2}$ ${}= 24x - 192$ ${}= -96 < 0$ , so a maximum
$V = 4(16)^2 = 1024$ cm³

Total7

MMethod mark: for a correct method, even with a slip.
AAccuracy mark: for a correct answer from a correct method.
BIndependent mark: for a correct fact or result on its own, with no method needed.

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