P4, from zero

Toolkit: the P3 results P4 leans on

Block 0used in almost every question

Almost every P4 mark needs one of these. They’re P3 content, so if P3 is new to you too, get this box solid first.

Differentiate

d/dx(xⁿ) = nxⁿ⁻¹d/dx(e^kx) = ke^kxd/dx(ln x) = 1/xsin kx → k cos kxcos kx → −k sin kxtan kx → k sec²kxsec x → sec x tan xcosec x → −cosec x cot xcot x → −cosec²x

• Chain rule: differentiate the outside, keep the inside, multiply by the derivative of the inside. d/dx(3 + 4 sin x)⁻² = −2(3 + 4 sin x)⁻³ × 4 cos x.
• Product rule: (uv)′ = u′v + uv′. The quotient rule, (u/v)′ = (u′v − uv′)/v², is in the booklet.
• d/dx ln(f(x)) = f′(x)/f(x), so d/dx ln(2x + 1) = 2/(2x + 1).

Integrate (and add +c)

∫xⁿ dx = xⁿ⁺¹/(n + 1)∫(ax + b)ⁿ dx = (ax + b)ⁿ⁺¹/(a(n + 1))∫1/(ax + b) dx = (1/a) ln|ax + b|∫e^ax+b dx = (1/a)e^ax+b∫cos ax dx = (1/a) sin ax∫sin ax dx = −(1/a) cos ax∫sec²ax dx = (1/a) tan ax

• Spot f′/f. If the top is a multiple of the derivative of the bottom, the answer is a log: ∫6x/(3x² + 5) dx = ln(3x² + 5) + c.
• Spot the reverse chain rule. ∫f′(x)[f(x)]ⁿ dx = [f(x)]ⁿ⁺¹/(n + 1) + c, so ∫sin²t cos t dt = ⅓ sin³t + c.

Identities and exact values

sin²x + cos²x = 11 + tan²x = sec²xsin 2x = 2 sin x cos xcos 2x = 1 − 2sin²x = 2cos²x − 1sin²x = ½(1 − cos 2x)cos²x = ½(1 + cos 2x)

• Logs: ln a + ln b = ln ab, ln a − ln b = ln(a/b), k ln a = ln a^k, e^{ln x} = x. Every “exact answer in the form p ln q” uses these.
• Exact values you’ll substitute as limits: sin(π/6) = ½, cos(π/6) = √3/2, sin(π/4) = cos(π/4) = √2/2, tan(π/4) = 1, tan(π/3) = √3, sec(π/3) = 2.

Binomial expansion

Block 1≈7 marks a paperevery paperQuestion 1 in 11 of 18

Learn

• For any rational power n, negative or a fraction, valid for |x| < 1 (this is in the booklet):
  (1 + x)ⁿ = 1 + nx + n(n − 1)2!x² + n(n − 1)(n − 2)3!x³ + …
• Whatever replaces x goes in brackets. For (1 − 5x/4)^−1/2, the x² term uses (−5x/4)² = 25x²/16. Dropping these brackets is the most common lost mark.
• If the bracket doesn’t start with 1, take the first number out to the power: (a + bx)ⁿ = aⁿ(1 + bx/a)ⁿ. So (4 − 5x)^−1/2 = ½(1 − 5x/4)^−1/2.
• Validity: the expansion of (a + bx)ⁿ is valid for |bx/a| < 1, which means |x| < |a/b|. When two expansions are combined, the narrower range wins.
• Roots and reciprocals are powers: √(…) = (…)^1/2, 1/(…)² = (…)⁻², 1/∛(…) = (…)^−1/3. These series never stop, so you’re asked for the first few terms.

(1 + x)ⁿ series (booklet)(a + bx)ⁿ = aⁿ(1 + bx/a)ⁿvalid for |x| < |a/b|

What they ask

• Expand to 3 or 4 terms, coefficients as simplified fractions (4–5 marks). Write the full unsimplified line first, brackets and all, then simplify. That line earns the method mark even if you slip later.
• Find unknowns from given coefficients (October 2020, June 2021, June 2022, January 2025, January 2026). Expand with the letter left in, then set coefficients equal. With two unknowns, such as (1 + ax)ⁿ, the x term gives na = … and the x² term gives n(n − 1)a²/2 = …. Square the first and divide to get rid of a.
• Multiply by a bracket, such as (2 + kx)(½ + 5x/16 + 75x²/256 + …). Build only the terms you need: the x² coefficient is 2 × 75/256 + k × 5/16.
• Approximate a root (7 of 18). Choose the x that turns the bracket into the number you want, and check it’s inside the valid range. October 2021: √(1 − 4x²) ≈ 1 − 2x² − 2x⁴ − 4x⁶, and x = ¼ gives √3/2 ≈ 0.86621, so √3 ≈ 1.7324.
• State the range of validity, 1 mark, asked in 4 of the last 5 papers.
• Partial fractions first, then expand each fraction and add (January 2023, June 2026). See the next section.

Traps

• Taking out a instead of aⁿ. 4^−1/2 = ½, not 4.
• Signs: (−5x/4)² is positive, (−5x/4)³ is negative.
• Writing |x| < 1 for (1 − 5x/4). It’s |x| < 4/5.
• “First three non-zero terms” when the bracket holds x²: the terms go 1, x², x⁴.
• Decimals where the question says “simplified fractions”.

Drill: Oct 2022 Q4, Jun 2022 Q1, Oct 2024 Q1, Jan 2022 Q2

Partial fractions

Block 1≈5 marks a paper17 of 18 papersthen integrated in 11

Learn

• Three forms, where A, B, C, D are constants to find:
  distinct:  px + q(ax + b)(cx + d) ≡ Aax + b + Bcx + d
  repeated:  …(ax + b)(cx + d)² ≡ Aax + b + Bcx + d + C(cx + d)²
  improper (top’s power ≥ bottom’s):  cubicx(x + 3) ≡ Ax + B + Cx + Dx + 3
• Method: multiply through by the whole denominator to get an identity with no fractions. Substitute the x that makes each bracket zero; each one hands you a constant. For whatever is left (the repeated-factor B, or a polynomial part), compare coefficients of the highest power of x.
• Check by putting x = 0 into both sides. Ten seconds, and it catches most slips.

What they ask

• Write in partial fractions (3–5 marks), usually part (a), so the rest of the question depends on it.
• Then integrate (11 of 18): ∫A/(ax + b) dx = (A/a) ln|ax + b| and ∫C/(ax + b)² dx = −C/(a(ax + b)). For an exact answer, merge the logs: 2 ln 3 − ln 5 = ln(9/5).
• Then a differential equation (January 2021, June 2022, October 2023, October 2024) or a binomial expansion (January 2023, June 2026), where you expand each fraction separately and add.
• Improper fractions (October 2020, October 2021, October 2024): the polynomial part comes first.

Traps

• Leaving out the C/(cx + d)² term for a repeated factor, or the polynomial part of an improper fraction.
• ∫1/(2x + 1) dx = ½ ln|2x + 1|, not ln|2x + 1|.
• Arithmetic with negative fractions such as x = −½. That’s what the x = 0 check is for.

Drill: Oct 2022 Q2, Jun 2023 Q3, Jun 2024 Q5(c), Jan 2023 Q1

Implicit differentiation

Block 2≈8 marks a paperevery paper“find the point where…” in 7 of the last 13

Learn

• Differentiate every term with respect to x. Any term with y in it picks up a dy/dx: d/dx(y³) = 3y² dy/dx, d/dx(e^2y) = 2e^2y dy/dx, d/dx(tan y) = sec²y dy/dx.
• A product of x and y needs the product rule: d/dx(x²y) = 2xy + x² dy/dx, d/dx(xy²) = y² + 2xy dy/dx, d/dx(ye^−2x) = −2ye^−2x + e^−2x dy/dx.
• Constants go to 0. Then collect the dy/dx terms on one side, factorise, divide.
• A power with x in it, like y = x^{sin x} (October 2020): take ln first, ln y = sin x ln x, then differentiate both sides: (1/y) dy/dx = cos x ln x + (sin x)/x.
• Product and chain rule together (January 2026): d/dx(x² tan y) = 2x tan y + x² sec²y dy/dx.

d/dx(yⁿ) = nyⁿ⁻¹ dy/dxd/dx(xy) = y + x dy/dxnormal gradient = −1 ÷ tangent gradient

What they ask

• Find or show dy/dx in terms of x and y (4–6 marks).
• Tangent or normal at a point. Put the point into the derivative to get the gradient m (the normal uses −1/m), then y − y₁ = m(x − x₁), rearranged to the form asked, such as ax + by + c = 0 with integers. If you’re only given x, find y from the curve first. A point “on the y-axis” means x = 0.
• Where that tangent or normal meets an axis (June 2023, October 2024): put y = 0 into the line.
• A condition on the gradient, the current favourite:
  • stationary point, turning point, “furthest north or south”: dy/dx = 0, so the numerator is 0;
  • vertical tangent, “the maximum value of x”, “where the curve isn’t defined”: the denominator is 0;
  • the gradient equals a number (October 2025), or the normal is a given line (October 2023): set them equal.
  Each one gives a link between x and y. Substitute that link back into the curve’s equation and solve. Skipping that step is where most people stop scoring.

Traps

• Forgetting dy/dx on a y-term, or on the y part of a product.
• Differentiating the constant on the right to anything but 0.
• Using the tangent gradient for a normal.
• Stopping at “numerator = 0”. You still have to go back to the curve.
• Not giving integers a, b, c when the question asks for them.

Drill: Jan 2022 Q1, Oct 2022 Q11, Jan 2024 Q3, Oct 2024 Q4

Parametric equations

Block 2≈9 marks a paperevery papertangent or normal in 12 of 18Cartesian equation in 11 of 18

Learn

• x and y are both given in terms of a parameter t (or θ). Differentiate each with respect to t, then divide: dy/dx = (dy/dt) ÷ (dx/dt).
• A point: put t into x and y. A gradient: put t into dy/dx. Given x or y instead of t? Solve for t first, and check it’s in the allowed interval.
• Cartesian equation (get rid of t):
  • Algebraic: make t the subject of the simpler equation and substitute it into the other. Fractions like x = (t + 2)/(2t − 1) lead to answers like y = (ax + b)/(cx + d).
  • Trig: write both in matching functions, then use an identity. cos 2t = 1 − 2sin²t (June 2021), sec²t = 1 + tan²t (October 2021, January 2024), sin²t + cos²t = 1.
• Domain and range of the Cartesian function: the domain is the set of x-values the curve covers, the range is its y-values. Work out x and y at both ends of the t-interval and at any turning point between them.
• Areas and volumes under parametric curves are in Volumes and areas.

dy/dx = dy/dtdx/dtcos 2t = 1 − 2sin²tsec²t = 1 + tan²t

What they ask

• Gradient, tangent or normal at a given t, often as “show that the tangent is …”, so you can check yourself.
• Where the tangent meets the curve again (October 2020, January 2025): substitute x(t) and y(t) into the line’s equation to get a polynomial in t. The point you started at is a repeated root, so (t − t₀)² is a factor.
• Cartesian equation, then the domain or range (11 of 18).
• Where the curve meets the axes: set y = 0 or x = 0 and solve for t.
• A point with a given gradient (October 2025): set dy/dx equal to it and solve for t.

Traps

• Dividing the wrong way up. dy/dt goes on top.
• Chain rule slips inside: d/dt(sin 2t) = 2 cos 2t, d/dt(sec²t) = 2 sec²t tan t.
• Taking a range from the end values only and missing a turning point in the middle, like t = 0 above.
• Giving the range in x. The range of f is y-values, so write it with f(x) or y.

Drill: Jun 2022 Q7, Oct 2022 Q6, Jan 2024 Q9, Oct 2024 Q3

Connected rates of change

Block 2≈4 marks a paper13 of 18 papers

Learn

• Chain the rates: dV/dt = (dV/dh) × (dh/dt). Write the chain linking the rate you want to the rate you’re given; the middle derivative comes from differentiating the formula.
• “Water flows in at 160 cm³ s⁻¹” means dV/dt = 160. Decreasing means negative.
• Get the formula into one variable before differentiating. In a cone of fixed shape, use similar triangles to replace r with a multiple of h (October 2024: r = 2h/5).
• Time to fill at a constant rate = full volume ÷ rate.

dV/dt = (dV/dh)(dh/dt)sphere V = 4πr³/3, S = 4πr²cone V = πr²h/3

The sphere and cone formulas are usually printed in the question, but not always.

What they ask

• A container with V in terms of h (June 2021, June 2026, almost the same question): the time to fill (2 marks), then dh/dt at a given depth (5 marks).
• Spheres and balloons (January 2023, January 2025, June 2025, January 2026), a cube (October 2023), cones (January 2024, October 2024), a circle and cylinder (June 2022), a segment of a circle (June 2024), an icosahedron (January 2022). The shape changes; the chain doesn’t.
• Show a proportionality (June 2025): with dV/dt = −k and dV/dr = 4πr², you get dr/dt = −k/(4πr²), which is inversely proportional to r².
• Build a differential equation (October 2021, January 2026): rate in minus rate out gives dV/dt; convert it to dh/dt or dr/dt, then solve as in Differential equations.

Traps

• Differentiating a product badly. Expand it first.
• Dividing by the wrong derivative. Check the units: cm³ s⁻¹ ÷ cm³ per cm = cm s⁻¹.
• Putting the value of h into the formula before differentiating.
• Missing units, or the minus sign when they ask for the rate of change of something shrinking.

Drill: Jan 2022 Q4, Oct 2023 Q2, Jan 2024 Q4, Oct 2024 Q5

Integration: substitution, parts and the rest

Block 3≈10 marks a paper on its ownsubstitution in every paperparts in 16 of 18

Choose the method, checking in this order

1. A standard result, an f′/f or a reverse chain rule? Do it directly (see Toolkit).
2. Does the question give a substitution? Use it. P4 almost always tells you which one.
3. A power of x times e^x, sin or cos, or anything with ln x? By parts.
4. A fraction with a factorised denominator? Partial fractions, then logs.
5. sin²x, cos²x, sin x cos x or tan²x? Use an identity first.

Substitution, step by step

1. Write u = …, differentiate, and rearrange to get dx = … du.
2. Replace every x in the integrand, writing x in terms of u where needed, and replace dx.
3. Change the limits by putting each x-limit into u = ….
4. Simplify to something you can integrate (powers of u, or partial fractions in u), integrate, and use the u-limits. Never go back to x in a definite integral.
5. Indefinite integral? Substitute back to x at the end and add +c.

The substitutions that have come up:

By parts

• ∫u (dv/dx) dx = uv − ∫v (du/dx) dx is in the booklet.
• Let u be the part that gets simpler when you differentiate it: ln x every time, otherwise the power of x. So u = ln x in ∫x² ln x dx, and u = x in ∫x cos 3x dx.
• x²e^x or x² cos 2x: parts twice.
• e^x sin x or e^x cos x, the “loop”: parts twice, the original integral reappears, call it I and solve for it.
• ∫ln x dx: write it as ∫1 × ln x dx with u = ln x, giving x ln x − x + c.

Trig integrals

∫sin²x dx = x/2 − (sin 2x)/4∫cos²x dx = x/2 + (sin 2x)/4∫sin x cos x dx = −(cos 2x)/4∫tan²x dx = tan x − x

Each comes from an identity: sin²x = ½(1 − cos 2x), cos²x = ½(1 + cos 2x), sin x cos x = ½ sin 2x, tan²x = sec²x − 1. Add +c to each.

Traps

• Not replacing dx, or swapping it for a bare du. It’s the most common zero on substitution questions.
• Using the x-limits after changing to u.
• Parts the wrong way round with ln x. Setting dv/dx = ln x leads nowhere.
• Sign slips: ∫sin x dx = −cos x, and the minus in front of the second integral in parts.
• “Exact” means keep e, ln, π and surds. “Show all stages” means every substitution and both limits written down.

Drill: Jan 2021 Q5, Jun 2022 Q5, Jan 2023 Q4, Oct 2023 Q3, Jun 2024 Q1, Jan 2024 Q5(a)

Volumes and areas

Block 4≈7 marks a papera volume in 17 of 18none in June 2026

Learn

• Area between a curve and the x-axis: A = ∫y dx.
• Volume when a region turns 360° (2π radians, the same thing) about the x-axis: V = π∫y² dx. Square y first, then integrate. The limits are x-values, usually where the curve meets an axis or a given line.
• Parametric: A = ∫y (dx/dt) dt and V = π∫y² (dx/dt) dt. Turn the x-limits into t-limits by solving x(t) = each limit. If x grows as t shrinks, the integral comes out negative: swap the limits and drop the minus, exactly as the mark schemes do.
• Regions that aren’t simply “under the curve”: subtract. For a region between a curve and a line, find the curve’s volume and take away the cylinder (πr²h) or cone (⅓πr²h) that the line sweeps out. For a region bounded by a curve and a tangent or normal (January 2023, June 2023), split it where they meet and add a triangle (for an area) or a cone (for a volume).

V = π∫y² dxA = ∫y (dx/dt) dtV = π∫y² (dx/dt) dt

What they ask

• “Show that the volume is π∫… dt” (4–7 marks, with the answer given so you can check it), then “hence find the exact volume” (3–5 marks).
• “Find the exact volume”, where y² needs expanding (October 2020), integrating by parts (October 2021, January 2022, June 2022, June 2025), a substitution (January 2024, June 2024) or partial fractions (October 2024). Volumes are the main way they test the integration methods.
• A real object: a doorknob (January 2022), a paperweight made of two solids (June 2025), a density, which is mass ÷ volume (June 2022). Do exactly what the context says with the volume.
• Working backwards: the volume is given and you find a limit k (October 2022).

Traps

• Leaving π out of a volume, or putting it into an area.
• Squaring y wrongly. (e^0.5x − 2)² has a middle term, −4e^0.5x.
• Using x-limits in a t-integral.
• Forgetting dx/dt in a parametric integral.
• Not checking the sketch for whether the region sits under the curve or between the curve and a line.

Drill: Jun 2021 Q2, Oct 2021 Q8, Jan 2022 Q7, Oct 2024 Q7, Jan 2021 Q9, Oct 2024 Q10

Differential equations

Block 4≈8 marks a paperevery papera long-term value in 8 of 18

Learn

1. Separate: everything with y (and dy) on one side, everything with x (and dx) on the other. Move things across by multiplying or dividing, never adding or subtracting.
2. Integrate both sides, with one +c.
3. Find c straight away from the given values. “Initially” means t = 0.
4. Rearrange to the form asked. With logs, ln y = 2x + c gives y = e^{2x + c} = Ae^2x. Merge logs into one before removing them.

• Proportional wording: “the radius decreases at a rate inversely proportional to r²” is dr/dt = −k/r² with k > 0. A second given value finds k.
• Long-term: let t → ∞, so e^−kt → 0 and 1/t → 0.
• The y-side often integrates to a log: ∫1/(12 − 3h) dh = −⅓ ln|12 − 3h|. The x-side sometimes needs partial fractions or parts.

separate → integrate, one +c → find c → rearrange”∝ 1/r²” means dr/dt = ±k/r²

What they ask

• Solve with a starting value, giving y = … or y² = … in a stated form (6–7 marks).
• Use the solution: the time when something happens, the value at a given time, or the long-term limit (8 of 18). Give units and answer in the context.
• Form the equation from words first (October 2022 melting ice, January 2026 deflating balloon, October 2021 tank).
• With partial fractions (January 2021, June 2022, October 2023, October 2024), with parts (June 2021, January 2024, October 2025), or with a given derivative to point the way (January 2022).
• Interpret: “will the container ever become full?” means compare the long-term limit with the capacity.

Traps

• A constant on both sides, or none at all.
• Dropping the −⅓ in ∫1/(12 − 3h) dh.
• Turning ln y = x + c into y = e^x + c. It’s y = Ae^x.
• Flipping a sum term by term: 1/y = a + b does not give y = 1/a + 1/b. Make it one fraction first.
• Missing units or the sentence in context at the end.

Drill: Jan 2021 Q10, Oct 2022 Q10, Jun 2023 Q6, Oct 2023 Q7, Jun 2024 Q7, Oct 2024 Q9

Vectors

Block 5≈11 marks a paper, the biggest topicevery paper

Learn

• A line: r = a + λd, where a is the position vector of a point on it and d is its direction. Through A and B: r = a + λ(b − a).
• A general point on the line has its coordinates written with λ, such as (4 − 4λ, 2 − 3λ, −3 + 5λ). Most routines start here.
• AB = b − a. The length of (p, q, r) is √(p² + q² + r²). A unit vector is a vector divided by its length.
• Scalar product: a·b = a₁b₁ + a₂b₂ + a₃b₃ = |a||b| cos θ. Perpendicular means a·b = 0.
• Parallel means one direction vector is a multiple of the other.

r = a + λdcos θ = a·b ÷ (|a||b|)triangle = ½|AB||AC| sin Areflection P′ = 2X − P

The five routines

1. Do the lines meet? Set the two general points equal: three equations, two unknowns, λ and μ. Solve two of them, then check the third. If it works, they meet, so substitute back for the point. If it fails and the directions aren’t parallel, the lines are skew. If a line contains an unknown constant, the third equation finds it.
2. Angle. Between two lines, use only their direction vectors. For angle BAC in a triangle, use AB and AC, both pointing away from A. For the acute angle, if the cosine comes out negative, use its size.
3. Foot of the perpendicular from a point P to a line, which is also the closest point and gives the shortest distance. Write the general point X, form PX, set PX·d = 0, solve for λ and substitute. The shortest distance is |PX|.
4. Area. A triangle is ½|AB||AC| sin A: find cos A with the scalar product, then sin A. A parallelogram is |AB||BC| sin B. If you’ve already found a foot of a perpendicular, ½ × base × height is quicker.
5. Reflection of P in a line: find the foot X. It’s the midpoint of PP′, so P′ = 2X − P (October 2021, June 2023).

Other things they ask

• A point at a given distance (June 2024, |OA| = 5√10): the general point’s length squared equals the distance squared. That’s a quadratic in λ, so expect two answers.
• Two possible positions from an area ratio or an isosceles triangle (October 2023, January 2025): the point can sit on either side, so λ takes two values.
• Vectors in a ratio (June 2021, show c = 3b − 2a): write each vector as a difference of position vectors.

Traps

• Using position vectors instead of direction vectors for the angle between two lines.
• Giving both lines the same parameter letter. Each needs its own (λ and μ).
• Claiming the lines meet after checking only two of the three equations.
• Mixing up “position vector” and “coordinates”. Give the one the question asks for.
• Rounding part-way through when the question wants an exact surd.

Drill: Oct 2022 Q9, Jan 2023 Q6, Jun 2023 Q4, Jan 2024 Q6, Oct 2024 Q8, Jun 2024 Q6

Proof by contradiction

Block 6≈5 marks a paperevery paper

Learn the frame

1. “Assume that…” the opposite of what you’re proving, stated precisely. It’s a mark on its own. For “if P then Q”, assume P is true and Q is false: “Assume there is an integer n such that n³ is even and n is odd.” For “there are no…”, assume there is one.
2. Do the algebra.
3. Say what’s contradicted: “this contradicts the assumption that…” or “which is impossible because…”.
4. Conclude: “so the original statement is true.”

The five families

• Parity and divisibility (6 of 18). Odd: n = 2k + 1. Even: n = 2k. Not a multiple of 3: n = 3k + 1 and n = 3k + 2, both cases. Expand and show the form: (2k + 1)³ = 2(4k³ + 6k² + 3k) + 1, which is odd.
• Irrational roots (3 of 18). Assume ∛2 = p/q with no common factors. Cube it: p³ = 2q³, so p³ is even and p is even. Write p = 2m, so q³ = 4m³ and q is even too. A common factor of 2 contradicts the assumption. Square roots work the same way.
• No integer solutions (2 of 18). Factorise, often as a difference of two squares, list the factor pairs of the number, solve each pair, and show none gives (positive) integers.
• Never meets, no stationary points, no real value (5 of 18). Assume it does, set the equations equal (or dy/dx = 0), and reach something impossible: a negative discriminant, a square equal to a negative number, x² = −5. For lines in 3D (June 2026), use routine 1 from Vectors.
• Inequalities (2 of 18). Assume the opposite inequality and rearrange it into something that can’t happen, such as (k − 3)² < 0.

Traps

• Assuming the wrong thing: the statement itself, or “for all n” when it should be “there is an n”.
• Checking only one case, such as 3k + 1 without 3k + 2.
• Writing “contradiction” without saying what’s contradicted.
• In an irrational-root proof, leaving out “with no common factors”. The whole contradiction rests on it.

Drill: Jun 2021 Q9(ii), Oct 2021 Q10, Jan 2023 Q9, Oct 2023 Q4, Jan 2024 Q8, Oct 2024 Q2