Solutions — Differentiating Polynomials

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1. Q1 — Power rule

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   (a) f'(x) = 6x⁵

   (b) f'(x) = 4x³

   (c) f'(x) = 10x⁹

   (d) f'(x) = 1 (since x = x¹, d/dx = 1·x⁰ = 1)

   (e) f'(x) = 0 (derivative of a constant is zero)

2. Q2 — Constant multiple rule

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   (a) f'(x) = 5 × 3x² = 15x²

   (b) f'(x) = −2 × 4x³ = −8x³

   (c) f'(x) = ½ × 2x = x

   (d) f'(x) = 8

   (e) f'(x) = −3 × 5x⁴ = −15x⁴

3. Q3 — Differentiating polynomials

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   (a) dy/dx = 6x + 5

   (b) dy/dx = 3x² − 12x

   (c) dy/dx = 16x³ − 9x² + 4x − 1

   (d) dy/dx = −6x² + 9

4. Q4 — Evaluating the derivative

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   (a) f'(x) = 6x − 2. f'(4) = 24 − 2 = 22

   (b) f'(x) = 3x² + 1. f'(2) = 12 + 1 = 13

   (c) f'(x) = −2x + 6. f'(3) = −6 + 6 = 0 (stationary point)

   (d) f'(x) = 8x³ − 10x. f'(1) = 8 − 10 = −2

5. Q5 — Expand then differentiate

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   (a) y = x² − 3x − 10 → dy/dx = 2x − 3

   (b) y = 4x² − 4x + 1 → dy/dx = 8x − 4

   (c) y = x³ − 3x² → dy/dx = 3x² − 6x

   (d) y = (x² − x − 2)(x + 3) = x³ + 2x² − 5x − 6 → dy/dx = 3x² + 4x − 5

6. Q6 — Finding where gradient has a given value

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   (a) dy/dx = 2x − 4 = 6 → 2x = 10 → x = 5

   (b) dy/dx = 3x² − 3 = 0 → x² = 1 → x = ±1

   (c) dy/dx = 6x² − 6x − 12 = −12 → 6x² − 6x = 0 → 6x(x − 1) = 0 → x = 0 or x = 1

7. Q7 — Gradient at a specific point

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   (a) dy/dx = 3x² − 2. At x=2: m = 12 − 2 = 10

   (b) dy/dx = 6x − 1. At x=−1: m = −6 − 1 = −7

   (c) dy/dx = 4x³ − 4. At x=1: m = 4 − 4 = 0. Gradient = 0 → neither positive nor negative (horizontal tangent). Equation: y = −3.

8. Q8 — Second derivative

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   (a) dy/dx = 6x² − 12x;   d²y/dx² = 12x − 12

   (b) f'(x) = 4x³ − 4x;   f''(x) = 12x² − 4

   (c) f'(x) = 3x² − 12x + 9;   f''(x) = 6x − 12 = 0 → x = 2. f''(x) = 0 at x = 2. (This is a possible inflection point.)

9. Q9 — Tangents parallel to a given line

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   (a) dy/dx = 3x² − 6x − 9

   (b) Parallel to y = −12x + 5 means gradient = −12. Set 3x² − 6x − 9 = −12 → 3x² − 6x + 3 = 0 → x² − 2x + 1 = 0 → (x−1)² = 0 → x = 1. y = 1−3−9+2 = −9. Tangent: y − (−9) = −12(x − 1) → y = −12x + 3

   (c) Horizontal tangent: dy/dx = 0 → 3x² − 6x − 9 = 0 → x² − 2x − 3 = 0 → (x−3)(x+1) = 0 → x = 3 or x = −1

10. Q10 — Area and rate problems

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    (a) dA/dx = 2x. This represents the rate of change of area with respect to side length. At x=3: dA/dx=6 cm²/cm (area increases by approximately 6 cm² per 1 cm increase in side length).

    (b) dV/dx = 3x². At x=3: dV/dx = 27 cm³/cm. This represents the rate of change of cube volume per unit increase in side length at x=3 cm.

    (c) V'(x) = 3x² − 4x + 3. At x=2: V'(2) = 12 − 8 + 3 = 7 units³ per unit increase in x.