NCEA Level 2Differentiating polynomialsAS 91262

Differentiating: 7 exam questions solved step by step

What this lesson covers

  • A polynomial is differentiated term by term, and a constant term differentiates to zero.
  • The derivative is a new function; substituting into it gives a gradient, not a y value.
Key idea

7 questions, one rule

  • 1. Expand any brackets first.
  • 2. Each term: power down, power − 1.
  • 3. A constant goes to 0.
  • 4. Substitute into dy/dx for a gradient.
Your turn

Question 1 · Achieved

  • y = 5x⁴ − 3x² + 7x − 2
  • Find dy/dx.
  • Find the gradient at x = 1/2.
Worked example

Q1: y = 5x⁴ − 3x² + 7x − 2

  • One term at a time.
  • Bracket the 1/2.
  1. 5x⁴ → 4 × 5x³ = 20x³
  2. −3x² → 2 × (−3x) = −6xkeep the sign
  3. 7x → 7
  4. −2 → 0a constant
  5. dy/dx = 20x³ − 6x + 7
  6. 20(1/2)³ − 6(1/2) + 7x = 1/2
  7. = 20(1/8) − 3 + 7
  8. = 2.5 − 3 + 7 = 6.5
Your turn

Question 2 · Achieved

  • f(x) = x⁴/2 − 2x³/3 + 5x
  • Find f′(−1).
  • Careful with the minus signs.
Worked example

Q2: f(x) = x⁴/2 − 2x³/3 + 5x

  • Multiply, then simplify.
  • Square first, then minus.
  1. x⁴/2 → 4x³/2 = 2x³cancel the 2
  2. −2x³/3 → −6x²/3 = −2x²cancel the 3
  3. 5x → 5
  4. f′(x) = 2x³ − 2x² + 5
  5. f′(−1) = 2(−1)³ − 2(−1)² + 5bracket the −1
  6. = 2(−1) − 2(1) + 5
  7. = −2 − 2 + 5 = 1
Your turn

Question 3 · Merit

  • y = (3x − 2)(x + 5)
  • Find the gradient at x = −1.
Worked example

Q3: y = (3x − 2)(x + 5)

  • Not bracket by bracket.
  • Expand, then differentiate.
  1. dy/dx = 3 × 1 = 3brackets not expanded
  2. y = 3x² + 15x − 2x − 10expand
  3. = 3x² + 13x − 10collect x terms
  4. dy/dx = 6x + 13term by term
  5. x = −1: 6(−1) + 13
  6. = −6 + 13 = 7
Your turn

Question 4 · Merit

  • y = x³ − 3x² + 2
  • Find the points where the gradient is 9.
Worked example

Q4: where is the gradient 9?

  • Set dy/dx equal to 9.
  • y from the curve.
  1. dy/dx = 3x² − 6x
  2. 3x² − 6x = 9gradient 9
  3. 3x² − 6x − 9 = 0−9 both sides
  4. x² − 2x − 3 = 0÷ 3
  5. (x − 3)(x + 1) = 0factorise
  6. x = 3 or x = −1
  7. x = 3: y = 27 − 27 + 2 = 2
  8. x = −1: y = −1 − 3 + 2 = −2
  9. (3, 2) and (−1, −2)
Your turn

Question 5 · Merit, in context

  • A farm tank holds V = 40t − t² litres
  • after t minutes (0 ≤ t ≤ 20).
  • a) How fast is it filling at t = 5?
  • b) When is it filling at 10 L/min?
Worked example

Q5: the filling tank

  • A rate is dV/dt, not V.
  • Answer in words, with units.
  1. V(5) = 200 − 25 = 175litres, not a rate
  2. dV/dt = 40 − 2trate of change
  3. t = 5: 40 − 2(5) = 30
  4. 30 litres per minute
  5. 40 − 2t = 10rate = 10
  6. 2t = 30, so t = 15
  7. 0 ≤ 15 ≤ 20
  8. after 15 minutes
Your turn

Question 6 · Excellence

  • y = 2x³ + 3x² − 12x + 1
  • For which values of x is
  • the curve decreasing?
Worked example

Q6: where is it decreasing?

  • Decreasing: dy/dx < 0.
  • Test a value either side.
  1. dy/dx = 6x² + 6x − 12
  2. 6x² + 6x − 12 < 0decreasing
  3. 6(x² + x − 2) < 0take out 6
  4. 6(x + 2)(x − 1) < 0factorise
  5. zero at x = −2 and x = 1
  6. x = 0: 0 + 0 − 12 = −12
  7. x = 2: 24 + 12 − 12 = 24
  8. −2 < x < 1
Your turn

Question 7 · Excellence

  • y = ax² + bx + 3 has gradient 7 at x = 1
  • and gradient −1 at x = −1.
  • Find a and b.
Worked example

Q7: form two equations

  • a and b are numbers.
  • One equation per gradient.
  1. dy/dx = 2ax + ba, b are constants
  2. x = 1: 2a + b = 7gradient 7
  3. x = −1: −2a + b = −1gradient −1
Worked example

Q7: solve, then check

  • Add to remove a.
  • Check both gradients.
  1. 2a + b = 7
  2. −2a + b = −1
  3. 2b = 6, so b = 3add the equations
  4. 2a + 3 = 7, so a = 2substitute b
  5. y = 2x² + 3x + 3
  6. dy/dx = 4x + 3
  7. 4(1) + 3 = 7
  8. 4(−1) + 3 = −1
Summary

The method that solved all seven

  • Expand, differentiate term by term, then substitute with care.
StepHow
Each termpower down, power minus one
Constantsgo to zero
Fractionsmultiply, then simplify
Bracketsexpand before differentiating
Substitutebrackets around negatives
Gradient = kset dy/dx = k and solve

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