**(i)(a)**

$\frac{dy}{dx} = 3x^2 \times \ln 2x + x^3 \times \frac{1}{2x} \times 2$ | M1A1A1 | Applies the product rule $vu' + uv'$ to $x^3 \ln 2x$. If the rule is quoted it must be correct. There must have been some attempt to differentiate both terms. If the rule is not quoted (nor implied by their working, with terms written out $u=...., u' =....., v=...., v' =...., \text{followed by their } vu'+uv'$) then only accept answers of the form $Ax^2 \times \ln 2x + x^3 \times \frac{B}{x}$ where $A, B$ are constants$\neq 0$. | (3 marks)

| | | One term correct, either $3x^2 \times \ln 2x$ or $x^3 \times \frac{1}{2x} \times 2$ |

| | | Cao. $\frac{dy}{dx} = 3x^2 \times \ln 2x + x^3 \times \frac{1}{2x} \times 2$. The answer does not need to be simplified. For reference the simplified answer is $\frac{dy}{dx} = 3x^2 \ln 2x + x^2 = x^2(3\ln 2x + 1)$ |

**(i)(b)**

Sight of $(x + \sin 2x)^2$ | B1M1A1 | | (3 marks)

| | B1 | Sight of $(x + \sin 2x)^2$ |
| | M1 | For applying the chain rule to $(x + \sin 2x)^3$. If the rule is quoted it must be correct. If it is not quoted possible forms of evidence could be sight of $C(x + \sin 2x)^2 \times (1 \pm D \cos 2x)$ where $C$ and $D$ are non-zero constants. Alternatively accept $u = x + \sin 2x$, $u'=$ followed by $Cu^2 \times$ their $u'$. **Do not accept** $C(x + \sin 2x)^2 \times 2\cos 2x$ **unless you have evidence that this is their** $u'$. Allow 'invisible' brackets for this mark, ie. $C(x + \sin 2x)^2 \times 1 \pm D \cos 2x$ |
| | A1 | Cao $\frac{dy}{dx} = 3(x + \sin 2x)^2 \times (1 + 2\cos 2x)$. There is no requirement to simplify this. |

**You may ignore subsequent working (isw) after a correct answer in part (i)(a) and (b)**

**(ii)**

M1 | Writing the derivative of $\cot y$ as $-\cosec^2 y$. It must be in terms of $y$. | 

A1 | $\frac{dx}{dy} = -\cosec^2 y$ **or** $1 = -\cos ec^2 y \frac{dy}{dx}$. Both lhs and rhs must be correct. |

M1 | Using $\frac{dy}{dx} = \frac{dx}{dy}$ |

M1 | Using $\cosec^2 y = 1 + \cot^2 y$ **and** $x = \cot y$ to get $\frac{dy}{dx}$ **or** $\frac{dx}{dy}$ **just in terms of** $x$. |

cso $\frac{dy}{dx} = -\frac{1}{\cosec^2 y} = -\frac{1}{1 + \cot^2 y} = -\frac{1}{1 + x^2}$ | M1, A1* | | (5 marks)

| (11 marks) |

**Alternative to (a)(i) when ln(2x) is written ln x+ln 2**

M1 | Writes $x^3 \ln 2x$ as $x^3 \ln 2 + x^3 \ln x$. Achieves $Ax^2$ for differential of $x^3 \ln 2$ and applies the product rule $vu'+uv'$ to $x^3 \ln x$. |

A1 | Either $3x^2 \times \ln 2 + 3x^2 \ln x$ **or** $x^3 \times \frac{1}{x}$ |

A1 | A correct (un simplified) answer. Eg $3x^2 \times \ln 2 + 3x^2 \ln x + x^3 \times \frac{1}{x}$ |

**Alternative to 5(ii) using quotient rule**

M1 | Writes $\cot y$ as $\frac{\cos y}{\sin y}$ and applies the quotient rule, a form of which appears in the formula book. If the rule is quoted it must be correct. There must have been some attempt to differentiate both terms. If the rule is not quoted (nor implied by their working, meaning terms are written out $u=...., u' =....., v=...., v' =...., \text{followed by their } \frac{vu'-uv'}{v^2}$) only accept answers of the form $\frac{\sin y \times \pm \sin y - \cos y \times \pm \cos y}{(\sin y)^2}$ |

A1 | Correct un simplified answer with both lhs and rhs correct. $\frac{dx}{dy} = \frac{\sin y \times - \sin y - \cos y \times \cos y}{(\sin y)^2} = \{-1 - \cot^2 y\}$ |

M1 | Using $\frac{dy}{dx} = \frac{dx}{dy}$ |

M1 | Using $\sin^2 y + \cos^2 y = 1$, $\frac{1}{\sin^2 y} = \cosec^2 y$ and $\cosec^2 y = 1 + \cot^2 y$ to get $\frac{dy}{dx}$ **or** $\frac{dx}{dy}$ **in** $x$ |

A1 | cso $\frac{dy}{dx} = -\frac{1}{1+x^2}$ |

**Alternative to 5(ii) using a triangle – last M1**

M1 | Uses triangle with $\tan y = \frac{1}{x}$ to find $\sin y$ and get $\frac{dy}{dx}$ **or** $\frac{dx}{dy}$ **just in terms of** $x$. |

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