# Question 6 [Proof by Induction - Differentiation]:

| Answer | Mark | Guidance |
|--------|------|----------|
| $\frac{dy}{dx} = 2e^{2x}\sinh x + e^{2x}\cosh x = ae^{2x}\sinh x + be^{2x}\cosh x$ or $e^{2x}(a\sinh x + b\cosh x)$ | M1 | 2.2a |
| $\frac{dy}{dx} = e^{2x}(2\sinh x + \cosh x)$; for $n=1$: $\frac{dy}{dx} = e^{2x}\left(\frac{3+1}{2}\sinh x + \frac{3-1}{2}\cosh x\right)$ {so result true for $n=1$} | A1 | 2.4 |
| Assume true for $n=k$; attempt product rule with $k$'s in all terms: $\frac{d^{k+1}y}{dx^{k+1}} = Ae^{2x}(f(k)\sinh x + g(k)\cosh x) + e^{2x}(f(k)\cosh x + g(k)\sinh x)$ | M1 | 2.1 |
| $\frac{d^{k+1}y}{dx^{k+1}} = 2e^{2x}\left(\frac{3^k+1}{2}\sinh x + \frac{3^k-1}{2}\cosh x\right) + e^{2x}\left(\frac{3^k+1}{2}\cosh x + \frac{3^k-1}{2}\sinh x\right)$ | | |
| $= e^{2x}\left(\left(3^k+1+\frac{3^k-1}{2}\right)\sinh x + \left(3^k-1+\frac{3^k+1}{2}\right)\cosh x\right)$ or $= e^{2x}\left(\frac{3\times3^k+1}{2}\sinh x + \frac{3\times3^k-1}{2}\cosh x\right)$ | dM1 | 1.1b |
| $= e^{2x}\left(\frac{3^{k+1}+1}{2}\sinh x + \frac{3^{k+1}-1}{2}\cosh x\right)$ | A1 | 2.1 |
| If true for $n=k$ then true for $n=k+1$, and as also true for $n=1$, so result is true for all positive integers or true $n\in\mathbb{N}$ | A1 | 2.4 |

**Notes Q6:**
- **M1:** Differentiates to a correct form
- **A1:** Correct derivative; reaches appropriate form to deduce result true for $n=1$, minimum $\frac{dy}{dx}=e^{2x}\left(\frac{4}{2}\sinh x+\frac{2}{2}\cosh x\right)$
- **M1:** Makes inductive assumption and attempts $(k+1)$th derivative from $k$th derivative; allow slips in coefficients but must show use of product rule; must be of the form $Ae^{2x}(f(k)\sinh x+g(k)\cosh x)+e^{2x}(f(k)\cosh x+g(k)\sinh x)$
- **dM1:** Dependent on previous M1; factors out exponential and gathers $\sinh x$ and $\cosh x$ terms; accept either form or equivalent
- **A1:** Reaches correct form from correct work; must have "$k+1$" showing; depends on previous two method marks
- **A1:** Appropriate concluding sentence covering points indicated; depends on all method marks; must have reached at least second line of dM1 and made attempt at $n=1$; requires all necessary brackets throughout

**Alternative (exponential definitions):**

| Answer | Mark | Guidance |
|--------|------|----------|
| $y=e^{2x}\sinh x \Rightarrow y=e^{2x}\left(\frac{e^x-e^{-x}}{2}\right)=\frac{1}{2}(e^{3x}-e^x)$; $\frac{dy}{dx}=\frac{1}{2}(3e^{3x}-e^x)$ | M1 | 2.2a |
| $n=1$ verified using exponential definitions of $\sinh x$ and $\cosh x$ | A1 | 2.4 |
| $\frac{d^k y}{dx^k}=e^{2x}\left(\frac{3^k+1}{2}\left(\frac{e^x-e^{-x}}{2}\right)+\frac{3^k-1}{2}\left(\frac{e^x+e^{-x}}{2}\right)\right) = e^{2x}\left(\frac{3^k}{2}e^x-\frac{1}{2}e^{-x}\right)=\frac{3^k}{2}e^{3x}-\frac{1}{2}e^x$ | M1 | 2.1 |
| Differentiates and simplifies: $\frac{d^{k+1}y}{dx^{k+1}}=e^{2x}\left(\frac{3^{k+1}}{2}e^x-\frac{1}{2}e^{-x}\right)=\frac{3^{k+1}}{2}e^{3x}-\frac{1}{2}e^x$ | dM1 | 1.1b |
| $\frac{d^{k+1}y}{dx^{k+1}}=e^{2x}\left(\frac{3^{k+1}+1}{2}\left(\frac{e^x-e^{-x}}{2}\right)+\frac{3^{k+1}-1}{2}\left(\frac{e^x+e^{-x}}{2}\right)\right) = 3\times\frac{3^k}{2}e^{3x}-\frac{1}{2}e^x=\frac{3^{k+1}}{2}e^{3x}-\frac{1}{2}e^x$ | A1 | 2.1 |
| Appropriate conclusion | A1 | 2.4 |