# Question 2:

## Part (a) — Main Method

| Answer/Working | Mark | Guidance |
|---|---|---|
| $y = \ln(\tanh 2x) \Rightarrow \frac{dy}{dx} = \frac{1}{\tanh 2x}\times 2\text{sech}^2 2x$ **or** $y=\ln(\tanh 2x)\Rightarrow e^y = \tanh 2x \Rightarrow e^y\frac{dy}{dx}=2\text{sech}^2 2x \Rightarrow \frac{dy}{dx}=\frac{2\text{sech}^2 2x}{\tanh 2x}$ | M1A1 | M1: Applies chain rule or eliminates "ln" and differentiates implicitly to obtain $\frac{dy}{dx} = \frac{k\,\text{sech}^2 2x}{\tanh 2x}$. A1: Correct derivative in any form. |
| $= \frac{2\cosh 2x}{\sinh 2x}\times\frac{1}{\cosh^2 2x} = \frac{2}{\sinh 2x \cosh 2x}$ | M1 | Converts to $\sinh 2x$ and $\cosh 2x$ correctly to obtain $\frac{k}{\sinh 2x \cosh 2x}$ |
| $= \frac{2}{\frac{1}{2}\sinh 4x} = 4\,\text{cosech}\,4x$ | A1 | Correct answer. Note this is not a given answer so allow e.g. a sinh becoming a sin if then recovered, but if there are any obvious errors mark should be withheld. |

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## Part (a) — Alternative (Exponentials)

| Answer/Working | Mark | Guidance |
|---|---|---|
| $y = \ln(\tanh 2x) = \ln\!\left(\frac{e^{2x}-e^{-2x}}{e^{2x}+e^{-2x}}\right)$ $\frac{dy}{dx} = \frac{e^{2x}+e^{-2x}}{e^{2x}-e^{-2x}}\left(\frac{(e^{2x}+e^{-2x})(2e^{2x}+2e^{-2x})-(e^{2x}-e^{-2x})(2e^{2x}-2e^{-2x})}{(e^{2x}+e^{-2x})^2}\right)$ **or** $y = \ln(e^{2x}-e^{-2x})-\ln(e^{2x}+e^{-2x})$, $\frac{dy}{dx}=\frac{2e^{2x}+2e^{-2x}}{e^{2x}-e^{-2x}}-\frac{2e^{2x}-2e^{-2x}}{e^{2x}+e^{-2x}}$ | M1A1 | M1: Writes $\tanh 2x$ correctly in terms of exponentials and applies chain rule and quotient rule, or uses subtraction law of logs and applies chain rule. A1: Correct derivative in any form. |
| $= \frac{2(e^{2x}+e^{-2x})^2 - 2(e^{2x}-e^{-2x})^2}{e^{4x}-e^{-4x}} = \frac{8}{e^{4x}-e^{-4x}}$ | M1 | Obtains $\frac{k}{e^{4x}-e^{-4x}}$ |
| $= \frac{4}{\sinh 4x} = 4\,\text{cosech}\,4x$ | A1 | Correct answer. Note this is not a given answer so allow e.g. a sinh becoming a sin if then recovered, but if there are any obvious errors mark should be withheld. |

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## Part (b) — Way 1

| Answer/Working | Mark | Guidance |
|---|---|---|
| $4\,\text{cosech}\,4x = 1 \Rightarrow \sinh 4x = 4 \Rightarrow 4x = \ln\!\left(4+\sqrt{4^2+1}\right)$ | M1 | Changes to $\sinh 4x = \ldots$ and uses the **correct** logarithmic form of arsinh to reach $4x = \ldots$ |
| $x = \frac{1}{4}\ln\!\left(4+\sqrt{17}\right)$ | A1 | This value only. Allow e.g. $x = \ln\!\left(4+\sqrt{17}\right)^{\frac{1}{4}}$ |

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## Part (b) — Way 2

| Answer/Working | Mark | Guidance |
|---|---|---|
| $4\,\text{cosech}\,4x = 1 \Rightarrow 4\times\frac{2}{e^{4x}-e^{-4x}}=1 \Rightarrow e^{8x}-8e^{4x}-1=0$ | M1 | Changes to correct exponential form $\frac{k}{e^{4x}-e^{-4x}}$, obtains a 3TQ in $e^{4x}$, solves and takes ln's to reach $4x = \ldots$ |
| $x = \frac{1}{4}\ln\!\left(4+\sqrt{17}\right)$ | A1 | This value only. Allow e.g. $x = \ln\!\left(4+\sqrt{17}\right)^{\frac{1}{4}}$ |