## Question 8:

### Part (a) — Way 1:

| Answer/Working | Marks | Guidance |
|---|---|---|
| $y = \ln\left(\frac{e^x+1}{e^x-1}\right) = \ln(e^x+1) - \ln(e^x-1) \Rightarrow \frac{dy}{dx} = \frac{e^x}{e^x+1} - \frac{e^x}{e^x-1}$ | M1A1 | M1: Uses correct log rule and attempts derivative using chain rule. A1: Correct derivative |
| $= \frac{e^{2x}-e^x-e^{2x}-e^x}{e^{2x}-1} = \frac{-2e^x}{e^{2x}-1}$ * | dM1A1* | dM1: Attempt single fraction and uses $(e^x-1)(e^x+1) = e^{2x}-1$. Dependent on first method mark. A1: Completes correctly with no errors |

### Part (a) — Way 2:

| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{dy}{dx} = \frac{e^x-1}{e^x+1}\left(\frac{e^x(e^x-1)-e^x(e^x+1)}{(e^x-1)^2}\right)$ or $\frac{dy}{dx} = \frac{e^x-1}{e^x+1}\left(e^x(e^x-1)^{-1} - e^x(e^x+1)(e^x-1)^{-2}\right)$ | M1A1 | M1: Uses chain and quotient or product rules. A1: Correct derivative |
| $= \frac{1}{e^x+1}\left(\frac{-2e^x}{e^x-1}\right) = -\frac{2e^x}{e^{2x}-1}$ * | dM1A1* | dM1: Cancels $e^x-1$ and uses $(e^x-1)(e^x+1)=e^{2x}-1$. Dependent on first method mark. A1: Completes correctly with no errors |

### Part (a) — Way 3:

| Answer/Working | Marks | Guidance |
|---|---|---|
| $y = \ln\left(\frac{e^x+1}{e^x-1}\right) \Rightarrow e^y = \frac{e^x+1}{e^x-1} \Rightarrow e^y\frac{dy}{dx} = \frac{e^x(e^x-1)-e^x(e^x+1)}{(e^x-1)^2}$ | M1A1 | M1: Removes logs correctly and differentiates implicitly using chain and quotient rules. A1: Correct differentiation |
| $\frac{dy}{dx} = -\frac{2e^x}{(e^x-1)^2} \times \frac{e^x-1}{e^x+1} = -\frac{2e^x}{e^{2x}-1}$ * | dM1A1 | dM1: Divides by $e^y$ in terms of $x$. Dependent on first method mark. A1: Completes correctly with no errors |

### Part (a) — Way 4:

| Answer/Working | Marks | Guidance |
|---|---|---|
| $y = \ln\left(\frac{e^x+1}{e^x-1}\right) = \ln\left(\coth\frac{1}{2}x\right) \Rightarrow \frac{dy}{dx} = \frac{1}{\coth\frac{1}{2}x} \times -\frac{1}{2}\text{cosech}^2\frac{1}{2}x$ | M1A1 | M1: Writes as $\ln\left(\coth\frac{1}{2}x\right)$ and differentiates using chain rule. A1: Correct differentiation |
| $= \left(\frac{e^x-1}{e^x+1}\right) \times \frac{-2e^x}{(e^x-1)^2} = -\frac{2e^x}{e^{2x}-1}$ | dM1A1 | dM1: Substitutes correct exponential forms. Dependent on first method mark. A1: Completes correctly with no errors |

### Part (a) — Way 5:

| Answer/Working | Marks | Guidance |
|---|---|---|
| $\text{artanh}\,x = \frac{1}{2}\ln\left(\frac{1+x}{1-x}\right) \Rightarrow y = 2\,\text{artanh}(e^{-x})$ $\frac{dy}{dx} = \frac{2}{1-(e^{-x})^2} \times -e^{-x}$ | M1A1 | M1: Writes $y$ correctly in terms of artanh and attempts to differentiate using chain rule. A1: Correct differentiation |
| $\frac{dy}{dx} = \frac{-2e^{-x}}{1-e^{-2x}} = \frac{-2e^x}{e^{2x}-1}$ * | dM1A1 | dM1: Multiplies numerator and denominator by $e^{2x}$. Dependent on first method mark. A1: Completes correctly with no errors |

### Part (a) — Way 6:

| Answer/Working | Marks | Guidance |
|---|---|---|
| $y = \ln\left(1+\frac{2}{e^x-1}\right) \Rightarrow \frac{dy}{dx} = \frac{1}{1+2(e^x-1)^{-1}} \times -2e^x(e^x-1)^{-2}$ | M1A1 | M1: Writes $\frac{e^x+1}{e^x-1}$ as $1+\frac{2}{e^x-1}$ and differentiates using chain rule. A1: Correct differentiation |
| $= \frac{-2e^x}{(e^x-1)^2+2(e^x-1)} = -\frac{2e^x}{e^{2x}-1}$ | dM1A1 | dM1: Multiplies denominator by $(e^x-1)^2$. Dependent on first method mark. A1: Completes correctly with no errors |

---

### Part (b):

| Answer/Working | Marks | Guidance |
|---|---|---|
| $L = \int\sqrt{1+\left(\pm\frac{2e^x}{e^{2x}-1}\right)^2}\,dx$ | M1 | Uses correct arc length formula with $\pm$ the result from part (a). Note omission of minus sign on fraction is condoned |
| $= \int\sqrt{\frac{e^{4x}-2e^{2x}+1+4e^{2x}}{(e^{2x}-1)^2}}\,dx$ | dM1 | Attempt single fraction. Dependent on first method mark |
| $L = \int\sqrt{\frac{(e^{2x}+1)^2}{(e^{2x}-1)^2}}\,dx = \int\frac{e^{2x}+1}{e^{2x}-1}\,dx$ | A1 | Correct integral with square root removed. No limits required |
| $= \int\coth x\,dx$, $\int\frac{e^x+e^{-x}}{e^x-e^{-x}}\,dx$, $\int 1+\frac{2e^{-2x}}{1-e^{-2x}}\,dx$, $\frac{1}{2}\int\frac{2}{u}-\frac{1}{u+1}\,du\,(u=e^{2x}-1)$, $\frac{1}{2}\int\frac{2}{u-1}-\frac{1}{u}\,du\,(u=e^{2x})$, $\int\frac{1}{u+1}+\frac{1}{u-1}-\frac{1}{u}\,du\,(u=e^x)$, $\frac{1}{2}\int\frac{2}{u-2}-\frac{1}{u-1}\,du\,(u=e^{2x}+1)$ | | Various equivalent forms shown |
| $= [\ln\sinh x]$, $[\ln(e^x-e^{-x})]$, $[x+\ln(1-e^{-2x})]$, $[\ln u - \ln\sqrt{1+u}]$, $[\ln(u-1)-\ln\sqrt{u}]$, $\left[\ln\frac{u^2-1}{u}\right]$, $[\ln(u-2)-\ln\sqrt{u-1}]$ | A1 | Correct integration |
| $= \ln\sinh(\ln 3) - \ln\sinh(\ln 2)$ $\left(= \ln\frac{4}{3} - \ln\frac{3}{4}\right)$ | ddM1 | Correct use of limits e.g. $\ln 3$ and $\ln 2$ for $x$, and e.g. 3 and 8 if $u=e^{2x}-1$. Must be correct limits for their method if substitution used. Dependent on both previous method marks |
| $= \ln\frac{16}{9}$ | A1 | cao |

**Total: 10 marks** (4 for part (a), 6 for part (b))