Derivative toolkit: 7 exam questions solved step by step

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What this lesson covers

  • Every Level 3 derivative is built from a few standard results and three rules for combining them.
  • A dropped factor or a missing bracket loses the whole question, even if later working is right.
Key idea

7 questions, one method

  • 1. Roots and fractions as powers.
  • 2. Choose: chain, product or quotient.
  • 3. One line per rule, with brackets.
  • 4. Substitute or simplify if asked.
Your turn

Question 1 · Achieved

  • y = 2x√x − 6/√x
  • Find dy/dx, and the gradient
  • of the curve at x = 4.
Worked example

Q1: y = 2x√x − 6/√x

  • Powers first.
  • Then the power rule.
  1. y = 2x√x − 6/√x
  2. = 2x^³ᐟ² − 6x^−½x × x^½ = x^³ᐟ²
  3. dy/dx = 2(3/2)x^½ − 6(−½)x^−³ᐟ²
  4. = 3x^½ + 3x^−³ᐟ²power rule
  5. x = 4: 3(4)^½ + 3(4)^−³ᐟ²
  6. = 3(2) + 3(1/8)4^½ = 2, 4^³ᐟ² = 8
  7. = 6 + 3/8 = 51/8
Your turn

Question 2 · Achieved

  • Differentiate y = cot(x²) + sec 4x
  • You do not need to simplify.
Worked example

Q2: y = cot(x²) + sec 4x

  • A chain rule for each term.
  • Not simplified: stop.
  1. d/du cot u = −cosec² ubooklet
  2. cot(x²): u = x², u′ = 2x
  3. → −cosec²(x²) × 2xchain rule
  4. d/du sec u = sec u tan ubooklet
  5. sec 4x: u = 4x, u′ = 4
  6. → sec 4x tan 4x × 4chain rule
  7. dy/dx = −2x cosec²(x²)
  8. + 4sec 4x tan 4x
Your turn

Question 3 · Merit

  • y = x³ ln(2x)
  • Find the gradient of the curve
  • at x = e/2.
Worked example

Q3: y = x³ ln(2x)

  • Product rule.
  • ln(2x) needs the chain.
  1. u = x³, u′ = 3x²product rule
  2. v = ln(2x)
  3. v′ = 1/(2x)chain factor dropped
  4. v′ = (1/(2x)) × 2 = 1/xchain: × 2
  5. dy/dx = 3x² ln(2x) + x³(1/x)
  6. = 3x² ln(2x) + x²x³/x = x²
  7. x = e/2: ln(2x) = ln e = 1
  8. 3(e/2)² + (e/2)²
  9. = 3e²/4 + e²/4 = e²
Your turn

Question 4 · Merit

  • y = e³ˣ/(x + 2)
  • Find dy/dx, and the gradient
  • at x = 0.
Worked example

Q4: y = e³ˣ/(x + 2)

  • Quotient rule.
  • Brackets on every part.
  1. u = e³ˣ, u′ = 3e³ˣchain: × 3
  2. v = x + 2, v′ = 1
  3. dy/dx = (u′v − uv′)/v²quotient rule
  4. = (3e³ˣ(x + 2) − e³ˣ(1))/(x + 2)²
  5. = e³ˣ(3x + 6 − 1)/(x + 2)²take out e³ˣ
  6. = e³ˣ(3x + 5)/(x + 2)²
  7. x = 0: e⁰(5)/2² = 5/4e⁰ = 1
Your turn

Question 5 · Merit, in context

  • Tea cools in a Wellington café.
  • After t minutes its temperature is
  • T = 20 + 70e^−0.05t (°C).
  • How fast is it cooling at t = 10?
Worked example

Q5: dT/dt at t = 10

  • Chain rule.
  • Negative rate: falling.
  1. T = 20 + 70e^−0.05t
  2. dT/dt = 70(−0.05)e^−0.05tchain; 20 → 0
  3. = −3.5e^−0.05t
  4. t = 10: −3.5e^−0.5
  5. = −2.1229calculator
  6. ≈ −2.123 s.f.
  7. cooling at about 2.12 °C per min
Your turn

Question 6 · Excellence

  • y = e^−x sin 2x
  • Show that y″ + 2y′ + 5y = 0
Worked example

Q6: first derivative

  • Product rule.
  • Keep the −1 and the 2.
  1. u = e^−x, u′ = −e^−xchain: × −1
  2. v = sin 2x, v′ = 2cos 2xchain: × 2
  3. y′ = u′v + uv′product rule
  4. = −e^−x sin 2x + 2e^−x cos 2x
  5. = e^−x(2cos 2x − sin 2x)take out e^−x
Worked example

Q6: second derivative

  • Product rule again.
  • Brackets round each part.
  1. y′ = e^−x(2cos 2x − sin 2x)
  2. u = e^−x, u′ = −e^−x
  3. v = 2cos 2x − sin 2x
  4. v′ = −4sin 2x − 2cos 2xchain: × 2 each
  5. y″ = −e^−x(2cos 2x − sin 2x)
  6. + e^−x(−4sin 2x − 2cos 2x)
  7. = e^−x(−4cos 2x − 3sin 2x)collect
Worked example

Q6: show it is zero

  • Like terms together.
  • Each set cancels.
  1. y″ = e^−x(−4cos 2x − 3sin 2x)
  2. 2y′ = e^−x(4cos 2x − 2sin 2x)
  3. 5y = 5e^−x sin 2x
  4. cos 2x: −4 + 4 = 0the cos terms
  5. sin 2x: −3 − 2 + 5 = 0the sin terms
  6. y″ + 2y′ + 5y = 0as required
Your turn

Question 7 · Excellence

  • k is a constant. y = eᵏˣ cos 2x
  • The gradient is 0 at x = π/8.
  • Find k.
Worked example

Q7: y = eᵏˣ cos 2x

  • Product rule with k.
  • eᵏˣ is never zero.
  1. u = eᵏˣ, u′ = keᵏˣchain: × k
  2. v = cos 2x, v′ = −2sin 2xchain: × 2
  3. dy/dx = keᵏˣ cos 2x − 2eᵏˣ sin 2x
  4. x = π/8: 2x = π/4cos = sin = √2/2
  5. e^(kπ/8)(√2/2)(k − 2) = 0factorise
  6. e^(kπ/8) > 0, √2/2 ≠ 0neither is 0
  7. k − 2 = 0, so k = 2
  8. e^(π/4)(√2 − √2) = 0k = 2
Summary

The method that solved all seven

  • Powers first, name the rule, one line each, brackets always.
StepHow
Surdspowers first, then bring down
Chaintimes the inside’s derivative
Productu′v + uv′
Quotient(u′v − uv′)/v², in brackets
In contexta rate, with its units
Show thatcollect like terms to 0

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