AQA Level 2 Further Maths · Level 2 Certificate · Chapter 8: Calculus · Lesson 1

The gradient of a curve and differentiation

About 85 minutesNo calculator75 marks

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Date

Do Now

The first two questions are what today's lesson needs. The others bring back earlier topics.

A1
Needed today
Write $\frac{3}{5 x ^{2}}$ in the form $a x^{n}$ .
[2 marks]
Answer $\dots\dots$
A2
Needed today
Work out the gradient of the straight line through $(- 2, 7)$ and $(3, - 8)$ .
[2 marks]
Answer $\dots\dots$
A3
Last lesson
Work out the length of the longest diagonal of a cuboid $2$ cm by $6$ cm by $9$ cm.
[2 marks]
Answer $\dots\dots$ cm
A4
From chapter 1
Expand and simplify $(2 x - 3) (x^{2} + 4 x - 1)$ .
[2 marks]
Answer $\dots\dots$

Example, then your turn

Level 2

Your teacher works through each example with you. Then try the one beside it.

Example 1

$P (2, 10)$ and $Q$ are points on $y = x^{2} + 3 x$ . Work out the gradient of $P Q$ when $Q$ has $x$ -coordinate $3$ , $2.5$ , $2.1$ and $2.01$ . Hence write down the gradient of the curve at $P$ .

Not drawn accurately

Hint Gradient of

P Q

= change in

y

\div

change in

x

. Watch what the gradients get close to.

B1
$P (1, 1)$ and $Q$ are points on the curve $y = 2 x^{2} - x$ . Work out the gradient of $P Q$ when $Q$ has $x$ -coordinate $1.1$ and then $1.01$ . Hence write down the gradient of the curve at $P$ .
[3 marks]
Not drawn accurately
Answer $\dots\dots$ and $\dots\dots$ ; at $P$ : $\dots\dots$

Example 2

Differentiate $y = 3 x^{4} - \frac{2}{x} + \frac{5}{2 x ^{3}} + 7$ .

Hint Write every term as a number times a power of

x

first. The

2

\frac{5}{2 x ^{3}}

is not cubed, so that term is

\frac{5}{2} x^{- 3}

B2
$f (x) = \frac{x ^{5}}{10} + \frac{4}{x ^{2}} - \frac{1}{6 x ^{3}} - 9 x + 11$ . Work out $f^{'} (x)$ .
[4 marks]
Answer $f^{'} (x) = \dots\dots$

Example 3

Work out $\frac{d y}{d x}$ when (a) $y = (2 x + 1) (x - 3)^{2}$ (b) $y = \frac{4 x ^{3} - 6 x + 1}{2 x ^{2}}$ .

Hint You cannot differentiate a product or a fraction term by term. Expand the brackets, or divide each term on the top by

2 x^{2}

, first.

B3
Work out $\frac{d y}{d x}$ when (a) $y = x (3 x - 2) (x + 4)$ (b) $y = \frac{5 x ^{4} + 3 x ^{2} - 8}{x ^{3}}$ .
[5 marks]
Answer(a) $\dots\dots$ (b) $\dots\dots$

Example 4

The curve $y = x^{3} - \frac{16}{x}$ crosses the positive $x$ -axis at the point $(2, 0)$ . Work out the gradient of the curve at this point.

Not drawn accurately

Hint Work out

\frac{d y}{d x}

, then put

x = 2

into it. Putting

x = 2

into

y

only gives the

0

you were told.

B4
$y = \frac{x ^{5} + 24}{4 x ^{2}}$ . Work out the rate of change of $y$ with respect to $x$ when $x = - 2$ .
[4 marks]
Answer $\dots\dots$

Example 5

Work out the coordinates of the points on the curve $y = x^{3} - 3 x^{2} - 7 x + 4$ where the gradient is $2$ .

Hint Put

\frac{d y}{d x} = 2

and solve. Then use the equation of the curve to find each

y

B5
Work out the exact coordinates of the points on the curve $y = 3 x + \frac{12}{x}$ where the gradient is $- 1$ .
[4 marks]
Answer $(\dots\dots, \dots\dots)$ and $(\dots\dots, \dots\dots)$

Practice

Level 2

C1
$P (3, 3)$ is a point on the curve $y = x^{2} - 2 x$ . $Q$ is the point on the curve with $x$ -coordinate $3 + h$ . Show that the gradient of the chord $P Q$ is $4 + h$ . Hence write down the gradient of the curve at $P$ .
[3 marks]
Not drawn accurately
Answergradient at $P$ : $\dots\dots$
C2
The curve $y = (x + 3) (x^{2} - 2 x + 5)$ crosses the $x$ -axis at one point, $A$ . Work out the gradient of the curve at $A$ .
[4 marks]
Not drawn accurately
Answer $\dots\dots$

C3
Work out the values of $x$ at which the curves $y = x^{3} + x$ and $y = 2 x^{2} + 5 x - 1$ have the same gradient.
[3 marks]
Answer $x = \dots\dots$ or $x = \dots\dots$
C4
A rectangle has width $x$ cm and area $36$ cm². Its perimeter is $P$ cm. Show that $P = 2 x + \frac{72}{x}$ . Work out the rate of change of $P$ with respect to $x$ when $x = 4$ .
[4 marks]
Not drawn accurately
Answer $\dots\dots$

Exam-style questions

AQA exam style

D1
$y = \frac{2}{x ^{3}}$ . Which of these is $\frac{d y}{d x}$ ? Circle your answer.
$- \frac{6}{x ^{4}} \frac{6}{x ^{4}} - \frac{6}{x ^{2}} \frac{2}{3 x ^{2}}$
[1 mark]

D2
Mia is working out the gradient of the curve $y = (x^{2} + 1) (3 x - 5)$ . She writes: " $\frac{d y}{d x} = 2 x \times 3 = 6 x$ ."
Explain Mia's mistake. Work out the gradient of the curve at the point where $x = 2$ .
[4 marks]
Answergradient $= \dots\dots$
D3
A curve has equation $y = a x^{3} + b x^{2}$ , where $a$ and $b$ are constants. The gradient of the curve is $4$ at the point where $x = 1$ , and $20$ at the point where $x = 2$ .
(a)
Show that $3 a + 2 b = 4$ .
[2 marks]
(b)
Work out the values of $a$ and $b$ .
[3 marks]
Answer $a = \dots\dots b = \dots\dots$
(c)
Hence work out the coordinates of the other point on the curve where the gradient is $4$ .
[3 marks]
Answer $(\dots\dots, \dots\dots)$
Total for D3: 8 marks
D4
Water drains out of a tank. The volume of water in the tank, $V$ litres, $t$ minutes after it starts to drain is $V = 2 t^{3} - 30 t^{2} + 1200$ for $0 \leq t \leq 10$ .
(a)
Work out $\frac{d V}{d t}$ .
[2 marks]
Answer $\frac{d V}{d t} = \dots\dots$
(b)
Work out the rate of change of the volume when $t = 4$ . Is the volume increasing or decreasing at this time?
[2 marks]
Answer $\dots\dots$ litres per minute, $\dots\dots$
(c)
Show that there is exactly one time at which the volume is decreasing at $150$ litres per minute. Work out this time.
[3 marks]
Answer $t = \dots\dots$
Total for D4: 7 marks
D5
A curve has equation $y = \frac{x ^{3} + 8}{2 x}$ .
(a)
Work out $\frac{d y}{d x}$ .
[3 marks]
Answer $\frac{d y}{d x} = \dots\dots$
(b)
Show that there is only one point on the curve where the gradient is $1$ , and work out its coordinates. You must show your working.
[6 marks]
Answer $(\dots\dots, \dots\dots)$
Total for D5: 9 marks

Extension

Stretch

No route is given. The hints are at the end of the sheet. Use one at a time.

E1
Prove that, for every quadratic curve $y = a x^{2} + b x + c$ , the chord joining the points where $x = p$ and $x = q$ is parallel to the tangent at the point where $x = \frac{p + q}{2}$ .
[4 marks]
Hint 1 · What to try
Two lines are parallel when their gradients are equal. Work out the gradient of the chord and the gradient of the curve at $x = \frac{p + q}{2}$ , and compare them.
Hint 2 · The first line
The gradient of the chord is $\frac{( a q ^{2} + b q + c ) - ( a p ^{2} + b p + c )}{q - p} = \frac{a ( q ^{2} - p ^{2} ) + b ( q - p )}{q - p}$ .
Hint 3 · The full method
$q^{2} - p^{2} = (q - p) (q + p)$ , so the chord's gradient is $a (p + q) + b$ . $\frac{d y}{d x} = 2 a x + b$ , which at $x = \frac{p + q}{2}$ is $a (p + q) + b$ . The gradients are equal, so the lines are parallel.

How did it go?

Colour one circle on each line.

	Not yet	Nearly	Yes
I can find the gradient of a curve at a point from the gradients of chords.
I can differentiate powers of $x$ , including negative powers.
I can expand or divide before differentiating.
I can use $\frac{d y}{d x}$ to find gradients, rates of change and points with a given gradient.

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