📖 Theory notes

Every note is built the same way: I can…, a quick recall, 🔑 key ideas, ✨ tips, ⚠️ common mistakes, ✏️ worked examples and 🧠 Remember in 3 lines.

🎯 Learn, then practise from two books

Each LO starts with 📘 a learning sheet (worked examples one step at a time), then 📗 Collins practice (Book 1 exercises) and, where Book 2 covers the topic, 📙 Pearson practice (Book 2 exercises, including starred questions).

🧪 LO tests · 30 minutes each

When the practice feels easy, take the timed test for that LO. Tests run in Quiz mode, submit themselves at 00:00 and can be taken once. Your report shows each LO as a pie chart.

About this chapter

On a straight line the gradient never changes, but on a curve it changes all the time. Calculus gives the exact gradient at any point, so you can find the steepest part of a hill, the top of a ball's flight or the moment a car stops. This chapter mixes two books. 📗 Collins Book 1 (Chapter 22) gives the core methods and practice. 📙 Pearson Book 2 (Graphs 9) adds the δx proof, fractional powers, equations of tangents, real-life rates of change and many more turning-point and motion problems.

LO	📗 Collins Book 1	📙 Pearson Book 2 (Graphs 9)
LO 1	22.1 The gradient of a curve	The gradient of a function; Exercise 1, Exercise 2 Q1–4, Q13–18
LO 2	22.2 More complex curves (powers, negative powers)	Differentiation; Exercise 1★, Exercise 2 Q5–12, Exercise 2★
LO 3	22.2 Rate of change; 22A Q8, 22B Q6–7	Equation of the tangent; applied rates of change; Exercise 3, 3★
LO 4	22.3 Turning points	Stationary points; Exercise 4, 4★; Activity 1 (tank)
LO 5	22.4 Motion of a particle	Motion in a straight line; Exercise 5, 5★

Learning objectives

LO 1 · 22.1 Gradient of a curveI can explain the gradient of a curve as the gradient of its tangent, and find dy/dx for a quadratic y = ax² + bx + c to work out gradients and where the gradient has a given value.

LO 2 · 22.2 Differentiating powers of xI can differentiate kxⁿ for positive, negative and fractional n, term by term, after multiplying out brackets or splitting fractions.

LO 3 · 22.2 Tangents and rates of changeI can find the equation of a tangent and use dy/dx as a rate of change in real situations.

LO 4 · 22.3 Turning pointsI can find turning points by solving dy/dx = 0, decide whether each is a maximum or a minimum, and solve problems about largest and smallest values.

LO 5 · 22.4 Motion of a particleI can find velocity v = ds/dt and acceleration a = dv/dt for a particle moving in a straight line, and use them to find when it stops or is furthest away.

Check what you already know

You need	Quick check	Answer
Gradient of a line	Gradient of the line through (1, 2) and (3, 8)	3
Index laws	Write [[1\|\|x²]] as a power of x	x⁻²
Expanding brackets	(x + 4)(x − 2)	x² + 2x − 8
Solving quadratics	3x² − 12 = 0	x = ±2
Substitution	x³ − 3x² when x = 2	−4

22.1 The gradient of a curve

LO 1I can… explain the gradient of a curve as the gradient of its tangent, and find dy/dx for a quadratic y = ax² + bx + c to work out gradients and where the gradient has a given value.

🔁 Quick recall before you start
What is the gradient of y = 3x − 1? (3.) Expand (x + 3)(x − 1). (x² + 2x − 3.)

📗 Collins: 22.1 The gradient of a curve, Exercise 22A
📙 Pearson Book 2: The gradient of a function (the δx method), Examples 1–3 and 5, Exercise 1 and Exercise 2 Q1–4, Q13–18

The gradient of a curve at a point P is the gradient of the tangent at P: the straight line that just touches the curve there. Drawing tangents by hand is slow and not accurate, so we calculate the gradient instead.

🔑 Key idea · dy/dx
The gradient of a curve is written [[dy||dx]] (say "dee y by dee x"). For a quadratic:
y = ax² + bx + c ⇒ [[dy||dx]] = 2ax + b
e.g. y = x² − 3x + 4 ⇒ [[dy||dx]] = 2x − 3, so the gradient at (3, 4) is 3.

✏️ Worked example · Why the gradient of x² is 2x (Pearson)

Take A at x and B at x + δx on y = x².
δy = (x + δx)² − x² = 2x·δx + (δx)², so [[δy||δx]] = 2x + δx.
As B slides down to A, δx → 0, so [[dy||dx]] = 2x.

✏️ Worked example · Gradient at a point and where the gradient is 0

y = 0.5x² + 4x − 3 ⇒ [[dy||dx]] = x + 4.
At (2, 7) the gradient is 6.
Gradient 0: x + 4 = 0 ⇒ x = −4, y = −11: the point (−4, −11).

✨ Tip · Special cases
y = k (a constant) is a horizontal line: [[dy||dx]] = 0. y = kx is a straight line: [[dy||dx]] = k.

⚠️ Common mistake · Brackets
Never differentiate the brackets in (3x + 1)(2x − 3) separately. Multiply out first: 6x² − 7x − 3, then [[dy||dx]] = 12x − 7.

🧠 Remember in 3 lines

Gradient of a curve = gradient of its tangent.
y = ax² + bx + c ⇒ dy/dx = 2ax + b.
Put x in for the gradient; solve dy/dx = k for the point.

Memory hook: Tangent tells the truth. The tangent's slope is the curve's slope.

22.2 Differentiating powers of x

LO 2I can… differentiate kxⁿ for positive, negative and fractional n, term by term, after multiplying out brackets or splitting fractions.

🔁 Quick recall before you start
Write √x and [[1||x³]] as powers of x. (x^½ and x⁻³.)

📗 Collins: 22.2 More complex curves, Examples 2–3, Exercise 22B
📙 Pearson Book 2: Differentiation, Examples 4, 6, 7, 8, Exercise 1★, Exercise 2 Q5–12, Exercise 2★

🔑 Key idea · The power rule
y = xⁿ ⇒ [[dy||dx]] = nxⁿ⁻¹ and y = kxⁿ ⇒ [[dy||dx]] = nkxⁿ⁻¹. Bring the power down, multiply, then take 1 off the power. It works for any power: positive, negative or a fraction.

y	write as	[[dy\|\|dx]]
x³	x³	3x²
5x²	5x²	10x
[[1\|\|x]]	x⁻¹	−x⁻² = −[[1\|\|x²]]
[[2\|\|x²]]	2x⁻²	−4x⁻³
√x	x^½	[[1\|\|2√x]]
7	7x⁰	0

✏️ Worked example · Term by term

y = x³ − 3x² + 4x + 7 ⇒ [[dy||dx]] = 3x² − 6x + 4.
At (2, 11): 12 − 12 + 4 = 4.

✏️ Worked example · Negative powers

y = x + [[2||x]] = x + 2x⁻¹ ⇒ [[dy||dx]] = 1 − 2x⁻² = 1 − [[2||x²]].
At (2, 3): 1 − [[2||4]] = [[1||2]].

✏️ Worked example · Split a fraction first

y = [[(x + 1)(2x + 1)||x]] = [[2x² + 3x + 1||x]] = 2x + 3 + x⁻¹.
[[dy||dx]] = 2 − x⁻²; at x = 1 the gradient is 1.

⚠️ Common mistake · Negative powers go further down
x⁻² differentiates to −2x⁻³ (−2 − 1 = −3), not −2x⁻¹.

✨ Tip · Typing answers
Type powers with ^, e.g. 3x^2-6x+4 or -2x^-3. For roots use √, e.g. 1/(2√x).

🧠 Remember in 3 lines

Rewrite: roots and fractions become powers of x.
Multiply by the power, then subtract 1 from it.
Constants vanish; do each term in turn.

Memory hook: Down and drop. Bring the power down, then drop it by one.

22.2 Tangents and rates of change

LO 3I can… find the equation of a tangent and use dy/dx as a rate of change in real situations.

🔁 Quick recall before you start
Find c if y = 4x + c passes through (2, 5). (c = −3.)

📗 Collins: 22.2 rate of change; Exercise 22A Q8, 22B Q2, Q6, Q7
📙 Pearson Book 2: Equation of the tangent (Example 9), applied rates of change (Example 10), Exercise 3 and 3★

🔑 Key idea · Equation of a tangent
1. Find the point: put x into y.
2. Find the gradient m: put x into [[dy||dx]].
3. Use y = mx + c with the point to find c.

✏️ Worked example · Tangent to y = x³ − 3x² + 2 at x = 3

Point P: y = 27 − 27 + 2 = 2, so P(3, 2).
[[dy||dx]] = 3x² − 6x = 9 at x = 3.
2 = 9 × 3 + c ⇒ c = −25. Tangent: y = 9x − 25.

🔑 Key idea · Rate of change
[[dy||dx]] is the rate of change of y with respect to x. With time t on the horizontal axis, [[dy||dt]] tells you how fast y is changing at that moment.

Vertical axis	Horizontal axis	Gradient means
distance (m)	time (s)	speed (m/s)
speed (m/s)	time (s)	acceleration (m/s²)
temperature (°C)	time (min)	°C per minute
profit ($)	time (h)	$ per hour

✏️ Worked example · A harbour (Pearson)

Depth y = 4 + 3t − t² ⇒ [[dy||dt]] = 3 − 2t.
At 13:00 (t = 1): +1 m/hr (rising). At 15:00 (t = 3): −3 m/hr (falling).

⚠️ Common mistake · Signs matter
A negative rate of change means the quantity is going down. Keep the minus sign and say what it means.

🧠 Remember in 3 lines

Point, gradient, then c.
dy/dx is a rate of change: units are y-units per x-unit.
Negative rate = decreasing.

Memory hook: P, m, c. Find the Point, find m, then find c.

22.3 Turning points

LO 4I can… find turning points by solving dy/dx = 0, decide whether each is a maximum or a minimum, and solve problems about largest and smallest values.

🔁 Quick recall before you start
Solve 3x² − 12 = 0. (x = ±2.) Solve 6(x − 2)(x + 1) = 0. (x = 2 or −1.)

📗 Collins: 22.3 Turning points, Example 4, Exercise 22C
📙 Pearson Book 2: Stationary points, Example 11, Exercise 4 and 4★, Activity 1

🔑 Key idea · Turning points
At a turning (stationary) point the tangent is horizontal, so [[dy||dx]] = 0. Solve this for x, then find y.

Gradient just before	at	just after	Type
+	0	−	maximum ∩
−	0	+	minimum ∪

✏️ Worked example · Find and classify

y = x³ − 12x + 4 ⇒ [[dy||dx]] = 3x² − 12 = 0 ⇒ x = ±2.
(2, −12): gradient −1.17 at x = 1.9 and +1.23 at 2.1 ⇒ minimum.
(−2, 20): gradient + then − ⇒ maximum.

✨ Tip · Use the shape
A quadratic with a > 0 is ∪ (one minimum); a < 0 is ∩ (one maximum). A cubic with a > 0 rises, falls, rises: the left turning point is the maximum.

✏️ Worked example · Largest area

A rectangle has perimeter 30 cm: A = x(15 − x) = 15x − x².
dA/dx = 15 − 2x = 0 ⇒ x = 7.5, A = 56.25 cm² (a maximum): the best rectangle is a square.

⚠️ Common mistake · Answer the question asked
Often you need the maximum value (the y-value), not just x. Put x back into the original formula, not into dy/dx.

🧠 Remember in 3 lines

Turning point ⇔ dy/dx = 0.
Solve for x, then find y from the original equation.
+0− is a maximum; −0+ is a minimum.

Memory hook: Flat at the top, flat at the bottom. Every hilltop and valley floor has gradient 0.

22.4 Motion of a particle

LO 5I can… find velocity v = ds/dt and acceleration a = dv/dt for a particle moving in a straight line, and use them to find when it stops or is furthest away.

🔁 Quick recall before you start
If s = 3 + 4t − t², what is s when t = 2? (7.)

📗 Collins: 22.4 Motion of a particle, Example 5, Exercise 22D
📙 Pearson Book 2: Motion in a straight line, Examples 12–13, Exercise 5 and 5★

🔑 Key idea · s, v and a
Displacement s is the position from a fixed point O (it can be negative).
v = [[ds||dt]] (gradient of the displacement–time graph)
a = [[dv||dt]] (gradient of the velocity–time graph)

✏️ Worked example · s = 3 + 4t − t²

v = 4 − 2t and a = −2 m/s².
v = 0 when t = 2: the particle stops and turns round, 7 m from O.
At t = 5: s = −2 m, on the other side of O.

✏️ Worked example · Highest point (Pearson)

A pebble: s = 10t − 5t² ⇒ v = 10 − 10t.
At the top v = 0 ⇒ t = 1 s, s = 5 m.

✨ Tip · Maximum velocity
Velocity is greatest when its gradient is 0, i.e. when a = 0.

⚠️ Common mistake · Velocity or speed?
A negative velocity means moving back towards O. Speed is the size of the velocity, so v = −6 m/s is a speed of 6 m/s.

🧠 Remember in 3 lines

Differentiate s to get v.
Differentiate v to get a.
v = 0: at rest; a = 0: velocity is greatest or smallest.

Memory hook: s → v → a: each step is one differentiation.