Differentiation — Drill Pack (Pure Mathematics

Pure Mathematics — Paper 1 · AQA AS Mathematics (sample) · Drill 2 of 3

Why this topic matters. Differentiation is the single highest-frequency topic on AQA AS Mathematics (sample) Paper 1 — it appears every year without exception and typically accounts for 12–20 marks. Stationary points with the second derivative test, tangents, normals, and first principles all appear regularly. This is your highest-priority drill.

The scaffold to use (DPST Framework)

Differentiate: apply the power rule d/dx(x²) = nxⁿ⁻¹ term by term.
Point: find where dy/dx = 0 (stationary points) or substitute given x to find gradient.
Second Derivative: find d²y/dx²; if positive at a stationary point ⇒ minimum; if negative ⇒ maximum.
Tangent/Normal: use y − y₁ = m(x − x₁); for normal, gradient = −1/m.

Foundational — recall and single-step application

Q1. Differentiate f(x) = 3x⁴ − 5x² + 2x − 7.

Foundational

[3 marks]

M1 Power rule on each term: d/dx(3x⁴) = 12x³; d/dx(−5x²) = −10x; d/dx(2x) = 2; d/dx(−7) = 0. AO1 — applying power rule term by term Common mistake: Differentiating the constant term to −7 instead of 0.

A1 f′(x) = 12x³ − 10x + 2. AO1 — correct derivative

A1 Check: each exponent drops by 1 and multiplies the coefficient. Constant vanishes. ✓ AO2 — verifying the pattern

Q2. Find the gradient of y = x³ − 4x + 1 at the point where x = 2.

Foundational

[3 marks]

M1 dy/dx = 3x² − 4. AO1 — differentiating correctly

A1 At x = 2: dy/dx = 3(4) − 4 = 12 − 4 = 8. AO1 — correct substitution

A1 The gradient at x = 2 is 8. AO2 — stating the result clearly

Q3. Find the x-coordinates of the stationary points of y = x³ − 3x + 2.

Foundational

[3 marks]

M1 dy/dx = 3x² − 3. Set equal to zero: 3x² − 3 = 0 ⇒ x² = 1. AO1 — setting derivative to zero Common mistake: Forgetting to set dy/dx = 0 and instead working with y = 0.

A1 x = 1 or x = −1. AO1 — correct x-values

A1 The stationary points occur at x = 1 and x = −1. AO2 — clear statement of coordinates

Q4. Find d²y/dx² for y = 4x³ − 3x² + x.

Foundational

[2 marks]

M1 dy/dx = 12x² − 6x + 1. Differentiate again. AO1 — first differentiation

A1 d²y/dx² = 24x − 6. AO1 — correct second derivative

Q5. The curve C has equation y = x² − 4x + 7. Find the equation of the tangent to C at the point (3, 4).

Foundational

[4 marks]

M1 dy/dx = 2x − 4. At x = 3: gradient = 2(3) − 4 = 2. AO1 — finding gradient at the point Common mistake: Using the gradient of the normal instead of the tangent.

A1 Tangent through (3, 4) with gradient 2: y − 4 = 2(x − 3). AO1 — applying point-gradient formula

A1 y = 2x − 6 + 4 = 2x − 2. (Or: y − 4 = 2(x − 3).) AO1 — correct equation of tangent

A1 Verify: at x = 3, y = 6 − 2 = 4. ✓ AO2 — substituting back to verify

Q6. Differentiate from first principles: f(x) = x² + 3x. Show all working.

Foundational

[4 marks]

M1 f(x+h) = (x+h)² + 3(x+h) = x² + 2xh + h² + 3x + 3h. AO1 — computing f(x+h) correctly

M1 f(x+h) − f(x) = 2xh + h² + 3h = h(2x + h + 3). Divide by h: 2x + h + 3. AO2 — forming and simplifying the difference quotient Common mistake: Forgetting to divide by h, or not factoring h before cancelling.

A1 As h → 0: f′(x) = 2x + 3. AO1 — correct limit

A1 This confirms the power rule result d/dx(x² + 3x) = 2x + 3. ✓ AO2 — linking to power rule

Mid-tier — application and analysis (exam-bulk questions)

Q7. Find the stationary points of y = x³ − 6x² + 9x + 2 and determine their nature using the second derivative test.

Mid-tier

[7 marks]

M1 dy/dx = 3x² − 12x + 9. Set = 0: 3(x² − 4x + 3) = 0 ⇒ 3(x−1)(x−3) = 0. AO1 — finding stationary x-values

A1 x = 1 and x = 3. AO1 — correct x-values

M1 y-values: at x=1, y = 1−6+9+2 = 6. At x=3, y = 27−54+27+2 = 2. AO1 — computing y-coordinates

M1 d²y/dx² = 6x − 12. At x=1: 6−12 = −6 < 0. At x=3: 18−12 = 6 > 0. AO1 — evaluating second derivative

A1 x=1 gives local maximum at (1, 6); x=3 gives local minimum at (3, 2). AO2 — correct classification with reasoning Common mistake: Reversing maximum and minimum: a negative d²y/dx² means maximum (curve is concave down at that point).

A1 State: the curve has a local maximum of 6 at x = 1 and a local minimum of 2 at x = 3. AO2 — both stationary points fully described

A1 Sanity check: for a cubic with positive leading coefficient, the first stationary point (smaller x) is the local maximum. ✓ AO2 — verifying with knowledge of cubic shape

Q8. Find the equation of the normal to the curve y = 2x³ − x at the point where x = 1.

Mid-tier

[5 marks]

M1 dy/dx = 6x² − 1. At x=1: dy/dx = 6 − 1 = 5. AO1 — finding tangent gradient

M1 Normal gradient: m_n = −1/5. AO2 — applying perpendicularity condition Common mistake: Using the tangent gradient for the normal equation.

A1 Point on curve: y = 2(1)³ − 1 = 1. Point is (1, 1). AO1 — finding the y-coordinate

A1 Normal: y − 1 = −(1/5)(x − 1) ⇒ 5y − 5 = −x + 1 ⇒ x + 5y = 6. AO1 — correct equation of normal

A1 Check: at (1,1): 1 + 5 = 6. ✓ AO2 — verifying the point lies on the normal

Q9. A curve C has equation y = x³ − 6x² + 12x − 4. Show that C has no stationary points.

Mid-tier

[3 marks]

M1 dy/dx = 3x² − 12x + 12. For stationary points: 3x² − 12x + 12 = 0 ⇒ x² − 4x + 4 = 0. AO1 — setting derivative to zero

M1 Discriminant: (−4)² − 4(1)(4) = 16 − 16 = 0. One repeated root at x = 2. AO2 — evaluating discriminant Common mistake: Stopping after finding discriminant = 0 without interpreting it.

A1 Since the discriminant is 0 (not positive), there is only one value of x where dy/dx = 0. However, d²y/dx² = 6x − 12; at x=2: d²y/dx² = 0. A point of inflection, not a stationary max/min. C has no turning points (local max or min). AO2 — correct interpretation of the degenerate case

Q10. The function f(x) = x³ + 3x² − 9x + 5. (a) Find f′(x). (b) Find the range of values of x for which f is a decreasing function.

Mid-tier

[6 marks]

M1 (a) f′(x) = 3x² + 6x − 9 = 3(x² + 2x − 3) = 3(x+3)(x−1). AO1 — differentiating and factorising

A1 f′(x) = 3(x+3)(x−1). AO1 — correct factorised derivative

M1 (b) f decreasing when f′(x) < 0: 3(x+3)(x−1) < 0 ⇒ (x+3)(x−1) < 0. AO2 — setting up inequality

M1 Sign table: negative between the roots. Critical values x = −3 and x = 1. AO2 — finding the sign between roots Common mistake: Writing x < −3 or x > 1 instead of −3 < x < 1.

A1 f is decreasing for −3 < x < 1. AO1 — correct interval

A1 At x = −3: f′ = 0 (boundary); at x = 0: f′ = 3(3)(−1) = −9 < 0 (decreasing). ✓ AO2 — verifying with a test value

Q11. Differentiate from first principles f(x) = 3x² − 2x, and verify using the power rule.

Mid-tier

[5 marks]

M1 f(x+h) = 3(x+h)² − 2(x+h) = 3x² + 6xh + 3h² − 2x − 2h. AO1 — expanding f(x+h)

M1 Numerator: f(x+h) − f(x) = 6xh + 3h² − 2h = h(6x + 3h − 2). Divide by h: 6x + 3h − 2. AO2 — forming difference quotient and cancelling h Common mistake: Not factoring h before dividing.

A1 lim_h→0 (6x + 3h − 2) = 6x − 2. AO1 — correct limit

A1 Power rule check: d/dx(3x² − 2x) = 6x − 2. ✓ AO2 — verification

A1 The first principles result matches the power rule, confirming both methods give the same derivative. AO2 — linking methods

Q12. A particle moves along a straight line. Its displacement s metres from origin at time t seconds is s = t³ − 6t² + 9t. Find: (a) the velocity when t = 2; (b) the times when the particle is at rest.

Mid-tier

[7 marks]

M1 v = ds/dt = 3t² − 12t + 9. AO1 — differentiating displacement

A1 (a) At t=2: v = 3(4) − 12(2) + 9 = 12 − 24 + 9 = −3 m/s. AO1 — correct velocity at t=2

M1 (b) Particle at rest: v = 0 ⇒ 3t² − 12t + 9 = 0 ⇒ t² − 4t + 3 = 0. AO1 — setting v=0

M1 (t−1)(t−3) = 0 ⇒ t = 1 or t = 3. AO1 — solving the quadratic Common mistake: Forgetting to check that t ≥ 0 for physical validity.

A1 Particle at rest at t = 1 s and t = 3 s. AO2 — clear statement with units

A1 At t=1: s = 1−6+9 = 4m. At t=3: s = 27−54+27 = 0m. ✓ AO2 — sanity-checking displacement values

A1 The negative velocity at t=2 (between t=1 and t=3) confirms the particle reverses direction at t=1. AO2 — physical interpretation

Q13. Find the x-coordinates of the points on the curve y = x⁴ − 8x² + 3 where the gradient is zero. Classify each using the second derivative.

Mid-tier

[7 marks]

M1 dy/dx = 4x³ − 16x = 4x(x² − 4) = 4x(x−2)(x+2). Set = 0. AO1 — differentiating and factorising

A1 x = 0, x = 2, x = −2. AO1 — three stationary x-values

M1 d²y/dx² = 12x² − 16. At x=0: −16 < 0 (max). At x=2: 48−16=32 > 0 (min). At x=−2: same as x=2 by symmetry (min). AO1 — computing second derivative at each point Common mistake: Forgetting to check x = −2; symmetry saves time but must be stated.

A1 x = 0: local maximum. x = 2: local minimum. x = −2: local minimum. AO2 — correct classification with evidence

A1 y-values: y(0) = 3; y(2) = 16−32+3 = −13; y(−2) = −13. AO1 — y-coordinates

A1 Local max (0, 3); local minima at (2, −13) and (−2, −13) by symmetry about the y-axis. AO2 — full description including coordinates

A1 The quartic has a positive leading coefficient, so the two outer arms go to +∞. Consistent with two local minima and one local maximum in between. ✓ AO2 — shape argument for verification

Q14. The curve y = ax³ + bx has a stationary point at (1, −2). Find a and b.

Mid-tier

[5 marks]

M1 Point (1,−2) on curve: a + b = −2. [Eq 1] AO1 — using the given point

M1 dy/dx = 3ax² + b. At x=1: 3a + b = 0. [Eq 2] AO1 — stationary point condition

M1 Eq2 − Eq1: 2a = 2 ⇒ a = 1. AO1 — solving the system

A1 b = −2 − a = −3. So a = 1, b = −3. AO1 — correct values

A1 Verify: y = x³−3x. y(1) = 1−3 = −2 ✓. dy/dx = 3x²−3; at x=1: 3−3=0 ✓. AO2 — verification

Q15. Find the equations of the tangent and normal to y = √x (i.e. x^1/2) at the point x = 4. Write your answers in the form ax + by + c = 0.

Mid-tier

[5 marks]

M1 dy/dx = (1/2)x^−1/2. At x=4: dy/dx = 1/(2√4) = 1/4. AO1 — differentiating x^1/2 Common mistake: Using d/dx(√x) = 1/√x instead of 1/(2√x).

A1 Point: (4, 2). Tangent gradient = 1/4. AO1 — coordinates and gradient

M1 Tangent: y − 2 = (1/4)(x − 4) ⇒ 4y − 8 = x − 4 ⇒ x − 4y + 4 = 0. AO1 — tangent equation in ax+by+c=0 form

A1 Normal gradient = −4. Normal: y−2 = −4(x−4) ⇒ y−2 = −4x+16 ⇒ 4x + y − 18 = 0. AO1 — normal equation in ax+by+c=0 form

A1 Check tangent at (4,2): 4−8+4 = 0 ✓. Check normal at (4,2): 16+2−18 = 0 ✓. AO2 — verifying both lines pass through the point

Q16. Find the coordinates of the point on the curve y = x² − 6x + 5 where the tangent is parallel to the line y = 4x − 1.

Mid-tier

[4 marks]

M1 Gradient of line y = 4x − 1 is 4. Set dy/dx = 4: 2x − 6 = 4 ⇒ x = 5. AO1 — equating derivative to line gradient

A1 y-coordinate: y(5) = 25 − 30 + 5 = 0. AO1 — computing y

A1 The point is (5, 0). AO2 — clearly stated coordinate

A1 Verify: dy/dx at x=5 is 2(5)−6 = 4. ✓ AO2 — verification

Stretch — discriminators, twist-traps, top-band signals

Q17. The curve C has equation y = x³ − 3x + k, where k is a constant. (a) Show that C has two distinct stationary points for all real values of k. (b) The point of inflection of C has y-coordinate 5. Find the value of k.

Stretch

[7 marks]

M1 (a) dy/dx = 3x² − 3 = 3(x²−1). Set = 0: x = ±1, which gives two distinct stationary points independent of k. AO2 — argument independent of k Common mistake: Claiming stationary points depend on k by confusing y-values with x-values.

A1 (a) ✓ Two distinct stationary points at x = −1 and x = 1 for any value of k. AO2 — clear statement of the argument

M1 (b) Point of inflection of C where d²y/dx² = 0: for y = x³ − 3x + k, d²y/dx² = 6x = 0 ⇒ x = 0 (and d²y/dx² changes sign through x = 0, so it is a genuine inflection). AO1 — finding inflection by second derivative

A1 y-coordinate of C at x = 0: y(0) = 0³ − 3(0) + k = k. So the inflection of C is at (0, k). AO1 — computing y-coordinate of inflection

M1 The inflection y-coordinate is given as 5, so k = 5. AO2 — applying the passage-through condition

A1 k = 5. AO1 — correct value of k

A1 Verify: C has equation y = x³−3x+5. At x=0: y=5 ✓. AO2 — verification

A1 The stationary points of C (k=5) are at (1, 3) and (−1, 7). AO2 — computing full stationary coordinates

Q18. Differentiate from first principles f(x) = 1/x. Hence state the derivative of x⁻¹ and verify it matches the power rule.

Stretch

[6 marks]

M1 [f(x+h) − f(x)]/h = [1/(x+h) − 1/x]/h. Common denominator: = [x − (x+h)] / [h⋅x(x+h)] = [−h] / [hx(x+h)]. AO1 — forming and simplifying the difference quotient

M1 Cancel h (h ≠ 0): = −1 / [x(x+h)]. AO2 — algebraic simplification Common mistake: Not cancelling h first, leading to 0/0 confusion.

A1 As h → 0: −1/[x(x+0)] = −1/x². AO1 — correct limit

A1 So d/dx(1/x) = −1/x², equivalently d/dx(x⁻¹) = −1⋅x⁻² = −x⁻². AO2 — writing in two equivalent forms

A1 Power rule: d/dx(x⁻¹) = (−1)x⁻² = −1/x². ✓ Matches. AO2 — verification via power rule

A1 The first principles derivation confirms that the power rule holds for negative integer powers. AO2 — connecting specific result to general principle

Q19. The tangent to y = x³ + ax + b at the point P(2, 5) has gradient 15. Find a and b. Find also the x-coordinate of the second point Q where this tangent meets the curve.

Stretch

[8 marks]

M1 dy/dx = 3x² + a. At x=2: 12 + a = 15 ⇒ a = 3. AO1 — using gradient condition

M1 Point (2,5) on curve: 8 + 2a + b = 5 ⇒ 8 + 6 + b = 5 ⇒ b = −9. AO1 — using point condition

A1 a = 3, b = −9. Curve: y = x³ + 3x − 9. AO1 — correct a and b

M1 Tangent: y − 5 = 15(x − 2) ⇒ y = 15x − 25. Set equal to curve: x³+3x−9 = 15x−25 ⇒ x³−12x+16 = 0. AO2 — setting up the intersection equation

M1 We know x=2 is a double root (tangent touches curve at P). Factor: x³−12x+16 = (x−2)²(x+4). AO2 — recognising the double root and factorising

A1 (x−2)²(x+4) = 0 ⇒ x = 2 (double) or x = −4. AO1 — finding the third root

A1 Q is at x = −4. y = 15(−4)−25 = −85. Q = (−4, −85). AO2 — full coordinates of Q

A1 Verify: y(−4) = −64−12−9 = −85. Tangent: 15(−4)−25 = −85. ✓ AO2 — verification

Q20. The curve C: y = x⁴ − 4x³ + 4x². Show that C has a point of inflection at x = 1 but NOT at x = 0. Find the third stationary point of C (other than x = 0 and x = 2) and classify it.

Stretch

[8 marks]

M1 dy/dx = 4x³−12x²+8x = 4x(x²−3x+2) = 4x(x−1)(x−2). Stationary points: x=0, 1, 2. AO1 — finding all stationary x-values

M1 d²y/dx² = 12x²−24x+8. At x=0: 8 > 0. At x=1: 12−24+8=−4 < 0. At x=2: 48−48+8=8 > 0. AO1 — evaluating second derivative

A1 x=0: d²y/dx² = 8 > 0 ⇒ local minimum. x=2: 8 > 0 ⇒ local minimum. x=1: −4 < 0 ⇒ local maximum. AO2 — classifying all three stationary points

M1 Point of inflection where d²y/dx² = 0 AND sign changes: 12x²−24x+8 = 0 ⇒ 3x²−6x+2 = 0 ⇒ x = (6±√12)/6 = 1 ± 1/√3. AO2 — finding inflection by solving d²y/dx²=0

A1 Check x=1: d²y/dx²=−4 ≠ 0, so x=1 is NOT an inflection point — it is a local maximum. AO2 — correcting the premise of the question Common mistake: Confusing a local maximum (where d²y/dx²<0) with a point of inflection.

A1 Check x=0: d²y/dx²=8 ≠ 0, so x=0 is also NOT an inflection point — it is a local minimum. AO2 — completing the disproof for x=0

A1 Inflection points are at x = 1 − 1/√3 ≈ 0.423 and x = 1 + 1/√3 ≈ 1.577. AO1 — correct inflection x-values

A1 The question asks for the third stationary point: x=1 is the local maximum at (1, y(1)). y(1) = 1−4+4 = 1. Local maximum at (1, 1). AO2 — answering the final part clearly

Stuck on a question?

Differentiation — Primer

Pure Mathematics — Paper 1 · AQA AS Mathematics (sample)

What this primer is for. A revision-grade refresher. Focus: what scores marks in AQA Paper 1, not derivation theory.

What this topic is

Differentiation gives the instantaneous rate of change of a function. The three key applications on AQA Paper 1 are: Power Rule (differentiating polynomials); Stationary Points (finding and classifying maxima, minima, inflections); and Tangents & Normals (equations of lines touching a curve). First Principles is examined periodically (roughly 1 in 3 years).

Why it matters in this paper

Differentiation is a very high-frequency topic — it appears on most AQA AS Paper 1. Stationary point classification with the second derivative and tangent/normal equations together account for roughly 12–15 marks. First principles is examined for 4–6 marks when it appears. This is the single most reliable topic to prepare thoroughly.

How to tackle this topic

Choose your method from the question stem:

Power Rule Differentiation — d/dx(ax^n) = anxⁿ⁻¹. Apply term by term; constants differentiate to zero; bring down the power, reduce by 1. Always simplify first (e.g. rewrite √x as x^1/2, 1/x² as x⁻²). Use when: any polynomial or simple algebraic function, tangent gradient, increasing/decreasing intervals.
First Principles Proof — form f′(x) = lim_h→0 [f(x+h)−f(x)]/h. Expand f(x+h) fully, subtract f(x), factor out h from the numerator, cancel h (valid since h≠0 in the limit), then take h→0. Use when: the question explicitly asks for “from first principles” or “using the definition of the derivative”.
Stationary Point Classify — set dy/dx = 0 and solve; find y-coordinates; compute d²y/dx² at each stationary x: if positive ⇒ minimum (curve concave up), if negative ⇒ maximum (curve concave down), if zero ⇒ use sign change or higher derivatives to decide. Use when: find stationary/turning points; maximum/minimum of a function; range of increasing or decreasing.

The key facts you must know

Power rule: d/dx(ax^n) = anxⁿ⁻¹. Constants differentiate to 0.
d/dx(√x) = 1/(2√x) = (1/2)x^−1/2. d/dx(1/x) = −1/x².
Stationary point: dy/dx = 0. Not every solution of dy/dx=0 is a turning point (could be inflection).
Second derivative test: d²y/dx² > 0 at SP ⇒ minimum; d²y/dx² < 0 ⇒ maximum; = 0 is inconclusive.
Tangent at (x₁, y₁): y − y₁ = m(x − x₁) where m = dy/dx at x₁.
Normal at (x₁, y₁): gradient = −1/m (perpendicular to tangent).
First principles: f′(x) = lim_h→0 [f(x+h)−f(x)]/h. Must show full working.
Velocity = ds/dt; acceleration = dv/dt = d²s/dt².

Step-by-step worked example

Question: Find the stationary points of y = 2x³ − 9x² + 12x − 5 and classify each.

Step 1. dy/dx = 6x² − 18x + 12 = 6(x²−3x+2) = 6(x−1)(x−2).
Step 2. Set dy/dx = 0: x = 1 or x = 2.
Step 3. d²y/dx² = 12x − 18. At x=1: −6 < 0 ⇒ local maximum. At x=2: 6 > 0 ⇒ local minimum.
Step 4. y(1) = 2−9+12−5 = 0. y(2) = 16−36+24−5 = −1.
Step 5. Local maximum at (1, 0); local minimum at (2, −1).

Why this lands top-band: Computing d²y/dx² at each point (not just stating the formula) and giving the full coordinates earns the classification marks.

Common mistakes (from examiner reports)

Not finding y-coordinates of stationary points (the question usually wants full coordinates).
Using d²y/dx² = 0 as inconclusive and not applying a sign-change test as a follow-up. — “Candidates who found d²y/dx² = 0 often incorrectly stated there was no stationary point.”
In first principles, not expanding f(x+h) correctly, especially for quadratics with middle terms.
Confusing the gradient of the tangent and normal (normal gradient is −1/m_tangent).
Forgetting to simplify the function (e.g. multiply out brackets) before differentiating.

Quick self-check

Can you differentiate y = 3x⁴ − 7x² + 5/x in under 60 seconds?
Can you carry out first principles on f(x) = x² + 2x from memory?
At a stationary point, what does a positive d²y/dx² tell you?

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Differentiation High-frequency topic

The scaffold to use (DPST Framework)

Q1. Differentiate f(x) = 3x⁴ − 5x² + 2x − 7.

Q2. Find the gradient of y = x³ − 4x + 1 at the point where x = 2.

Q3. Find the x-coordinates of the stationary points of y = x³ − 3x + 2.

Q4. Find d²y/dx² for y = 4x³ − 3x² + x.

Q5. The curve C has equation y = x² − 4x + 7. Find the equation of the tangent to C at the point (3, 4).

Q6. Differentiate from first principles: f(x) = x² + 3x. Show all working.

Q7. Find the stationary points of y = x³ − 6x² + 9x + 2 and determine their nature using the second derivative test.

Q8. Find the equation of the normal to the curve y = 2x³ − x at the point where x = 1.

Q9. A curve C has equation y = x³ − 6x² + 12x − 4. Show that C has no stationary points.

Q10. The function f(x) = x³ + 3x² − 9x + 5. (a) Find f′(x). (b) Find the range of values of x for which f is a decreasing function.

Q11. Differentiate from first principles f(x) = 3x² − 2x, and verify using the power rule.

Q12. A particle moves along a straight line. Its displacement s metres from origin at time t seconds is s = t³ − 6t² + 9t. Find: (a) the velocity when t = 2; (b) the times when the particle is at rest.

Q13. Find the x-coordinates of the points on the curve y = x⁴ − 8x² + 3 where the gradient is zero. Classify each using the second derivative.

Q14. The curve y = ax³ + bx has a stationary point at (1, −2). Find a and b.

Q15. Find the equations of the tangent and normal to y = √x (i.e. x1/2) at the point x = 4. Write your answers in the form ax + by + c = 0.

Q16. Find the coordinates of the point on the curve y = x² − 6x + 5 where the tangent is parallel to the line y = 4x − 1.

Q17. The curve C has equation y = x³ − 3x + k, where k is a constant. (a) Show that C has two distinct stationary points for all real values of k. (b) The point of inflection of C has y-coordinate 5. Find the value of k.

Q18. Differentiate from first principles f(x) = 1/x. Hence state the derivative of x−1 and verify it matches the power rule.

Q19. The tangent to y = x³ + ax + b at the point P(2, 5) has gradient 15. Find a and b. Find also the x-coordinate of the second point Q where this tangent meets the curve.

Q20. The curve C: y = x⁴ − 4x³ + 4x². Show that C has a point of inflection at x = 1 but NOT at x = 0. Find the third stationary point of C (other than x = 0 and x = 2) and classify it.

Q15. Find the equations of the tangent and normal to y = √x (i.e. x^1/2) at the point x = 4. Write your answers in the form ax + by + c = 0.

Q18. Differentiate from first principles f(x) = 1/x. Hence state the derivative of x⁻¹ and verify it matches the power rule.