**(a)** $(1 - 2x)^2 = 1 - 4x + 4x^2$ | M1 |

$\frac{d}{dx}(1 - 2x)^2 = \frac{d}{dt}(-4x + 4x^2) = -4 + 8x$ o.e. | M1A1 |

| | **(3 marks)** |

| **Alternative method using chain rule:** | Answer of $-4(1 - 2x)$ | M1M1A1 |

| | **(3 marks)** |

**(b)** $\frac{x^2 + 6\sqrt{x}}{2x^2} = \frac{x^2}{2x^2} + \frac{6\sqrt{x}}{2x^2} = \frac{1}{2}x^3 + 3x^{-\frac{1}{2}}$ | M1, A1 |

Attempts to differentiate $x^{-\frac{3}{2}}$ to give $k x^{-\frac{5}{2}}$ | M1 |

$= \frac{3}{2}x^2 - \frac{9}{2}x^{-\frac{5}{2}}$ o.e. | A1 |

| **Quotient Rule (May rarely appear) – See note below.** | | |

| | **(7 marks)** |

| **Notes:** | |
|---|---|
| **(a)** | M1: Attempt to multiply out bracket. Must be $3$ or $4$ term quadratic and **must have constant term $1$**. |
| | M1: $x^n \to x^{n-1}$. Follow through on any term in an incorrect expression. Accept a constant $\to 0$. |
| | A1: $-4 + 8x$ Accept $-4(1 - 2x)$ or equivalent. This is not cso and may follow error in the constant term. Following correct answer by $-2 + 4x$ – apply isw. |
| | **Correct answer with no working – assume chain rule and give M1M1A1 i.e. 3/3.** |
| | **Common errors:** $(1 - 2x)^2 = 2 - 4x + 4x^2$ is M0, then allow M1A1 for $-4 + 8x$. $(1 - 2x)^2 = 1 - 4x^2$ is M0 then $-8x$ earns M1A0 or $(1 - 2x)^2 = 1 - 2x^2$ is M0 then $-4x$ earns M1A0. |
| | **Use of Chain Rule:** M1M1: first M1 for complete method so $2 \times (±2)(1 - 2x)$ second M1 for $(1 - 2x)$ (as power reduced). So (i) $2(1 - 2x)$ gets M0 M1A0 or (ii) $2 \times 2(1 - 2x)$ gets M1 M1A0 for reducing power and (ii) $2 \times 2(1 - 2x)$ gets M1 M1A0. |
| **(b)** | M1: An attempt to divide by $2x^2$ first. This can be implied by the sight of the following $\frac{x^3}{...}$ or $(x^2 + 6\sqrt{x})(2x^2)^{-1}$ **leading to** $ax^b + bx^d$ in either case. Or can be implied by $\frac{1}{2}x^3 + 3x^{-\frac{1}{2}}$ (after no working) i.e. both coefficients correct and power $3$ correct. |
| | **Common error:** $(x^3 + 6\sqrt{x})(2x^2)$ is M0 (may earn next M mark for differentiation $x^{-\frac{3}{2}} \to x^{-\frac{5}{2}}$). |
| | A1: Writing the given expression as $\frac{1}{2}x^3 + 3x^{-\frac{1}{2}}$ or $0.5x + \frac{6}{x}$ or $0.5x + \frac{6}{x}$ or $0.5x + \frac{6}{x}$ or etc…. |
| | M1: $x^{-\frac{3}{2}} \to x^{-\frac{5}{2}}$ | A1: Cao $\frac{3}{2}x^2 - \frac{9}{2}x^{-\frac{5}{2}}$ o.e. e.g. $\frac{3}{2}x^2 - \frac{9}{2x^{\frac{5}{2}}}$ then isw. Allow factorised form. Do not penalise $+- \frac{9}{2}x^{-\frac{5}{2}}$. |
| | **Use of Quotient Rule:** | |
| | M1,A1: Reaching $\frac{2x^3(5x^4 + 3x^{-\frac{1}{2}}) - 4x(x^2 + 6x^{\frac{1}{2}})}{4x^4}$, = $\frac{6x^3 - 18x^2}{4x^3}$. |
| | **Send to review if doubtful.** | M1A1: Simplifying (e.g. dividing numerator and denominator by $2$) to reach $\frac{3x^3 - 9x^2}{2x^3}$ o.e. |

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