| Exam Board | Edexcel |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Year | 2017 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hyperbolic functions |
| Type | Differentiate inverse hyperbolic functions |
| Difficulty | Standard +0.8 This is a Further Maths question requiring chain rule application with inverse hyperbolic functions and manipulation of hyperbolic identities. While the technique is standard (differentiate arsinh, apply chain rule), it requires knowledge of the derivative of arsinh(u) and careful algebraic manipulation of hyperbolic functions—more demanding than typical A-level calculus but routine for FP3 students who have learned these formulas. |
| Spec | 4.07a Hyperbolic definitions: sinh, cosh, tanh as exponentials4.07e Inverse hyperbolic: definitions, domains, ranges |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\sinh y = \tanh x\) | B1 | |
| \(\cosh y \frac{dy}{dx} = \text{sech}^2 x\) or \(\cosh y = \text{sech}^2 x \frac{dx}{dy}\) | M1A1 | M1: \(\pm\cosh y\) or \(\pm\text{sech}^2 x\); A1: All correct |
| \(\frac{dy}{dx} = \frac{\text{sech}^2 x}{\cosh y} = \frac{\text{sech}^2 x}{\sqrt{1+\sinh^2 y}}\) | M1 | Uses a correct identity to express \(\frac{dy}{dx}\) in terms of \(x\) only |
| \(= \frac{\text{sech}^2 x}{\sqrt{1+\tanh^2 x}}\) * | A1* | cso. No errors, no missing/inconsistent variables, no missing h's |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(t = \tanh x \Rightarrow y = \text{arsinh}\, t\) | B1 | Replaces \(\tanh x\) by e.g. \(t\) |
| \(\frac{dt}{dx} = \text{sech}^2 x,\quad \frac{dy}{dt} = \frac{1}{\sqrt{1+t^2}}\) | M1A1 | M1: \(\frac{dt}{dx} = \pm\text{sech}^2 x\), \(\frac{dy}{dt} = \pm\frac{1}{\sqrt{1+t^2}}\); A1: Both correct and correctly labelled |
| \(\frac{dy}{dx} = \frac{dy}{dt}\cdot\frac{dt}{dx} = \frac{\text{sech}^2 x}{\sqrt{1+t^2}}\) | M1 | Uses correct form of chain rule for their variables to express \(\frac{dy}{dx}\) in terms of \(x\) only |
| \(= \frac{\text{sech}^2 x}{\sqrt{1+\tanh^2 x}}\) * | A1* | cso. No errors, no missing/inconsistent variables, no missing h's |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(u = \tanh x \Rightarrow \frac{du}{dx} = \text{sech}^2 x\) | B1 | Correct derivative |
| \(\int \frac{\text{sech}^2 x}{\sqrt{1+\tanh^2 x}}\,dx = \int \frac{\text{sech}^2 x}{\sqrt{1+u^2}}\cdot\frac{1}{\text{sech}^2 x}\,du\) | M1A1 | M1: Complete substitution including the "dx"; A1: Fully correct substitution |
| \(= \int \frac{1}{\sqrt{1+u^2}}\,du = \text{arsinh}\,u (+c)\) | M1 | Reaches \(\text{arsinh}\,u\) |
| \(y = \text{arsinh}(\tanh x)(+c)\) | A1* | Reaches \(y = \text{arsinh}(\tanh x)\) with/without \(+c\), no errors |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(y = \text{arsinh}(\tanh x) \Rightarrow \frac{dy}{dx} = \frac{1}{\sqrt{1+\tanh^2 x}} \times \text{sech}^2 x = \frac{\text{sech}^2 x}{\sqrt{1+\tanh^2 x}}\) | M1A1 | \(\text{sech}^2 x\) must appear separate from the fraction. To score more than 2 marks using a chain rule method, a third variable must be introduced |
## Question 1: $y = \text{arsinh}(\tanh x)$
### Way 1
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\sinh y = \tanh x$ | B1 | |
| $\cosh y \frac{dy}{dx} = \text{sech}^2 x$ or $\cosh y = \text{sech}^2 x \frac{dx}{dy}$ | M1A1 | M1: $\pm\cosh y$ or $\pm\text{sech}^2 x$; A1: All correct |
| $\frac{dy}{dx} = \frac{\text{sech}^2 x}{\cosh y} = \frac{\text{sech}^2 x}{\sqrt{1+\sinh^2 y}}$ | M1 | Uses a correct identity to express $\frac{dy}{dx}$ in terms of $x$ only |
| $= \frac{\text{sech}^2 x}{\sqrt{1+\tanh^2 x}}$ * | A1* | cso. No errors, no missing/inconsistent variables, no missing h's |
### Way 2
| Answer/Working | Mark | Guidance |
|---|---|---|
| $t = \tanh x \Rightarrow y = \text{arsinh}\, t$ | B1 | Replaces $\tanh x$ by e.g. $t$ |
| $\frac{dt}{dx} = \text{sech}^2 x,\quad \frac{dy}{dt} = \frac{1}{\sqrt{1+t^2}}$ | M1A1 | M1: $\frac{dt}{dx} = \pm\text{sech}^2 x$, $\frac{dy}{dt} = \pm\frac{1}{\sqrt{1+t^2}}$; A1: Both correct and correctly labelled |
| $\frac{dy}{dx} = \frac{dy}{dt}\cdot\frac{dt}{dx} = \frac{\text{sech}^2 x}{\sqrt{1+t^2}}$ | M1 | Uses correct form of chain rule for their variables to express $\frac{dy}{dx}$ in terms of $x$ only |
| $= \frac{\text{sech}^2 x}{\sqrt{1+\tanh^2 x}}$ * | A1* | cso. No errors, no missing/inconsistent variables, no missing h's |
### Way 3
| Answer/Working | Mark | Guidance |
|---|---|---|
| $u = \tanh x \Rightarrow \frac{du}{dx} = \text{sech}^2 x$ | B1 | Correct derivative |
| $\int \frac{\text{sech}^2 x}{\sqrt{1+\tanh^2 x}}\,dx = \int \frac{\text{sech}^2 x}{\sqrt{1+u^2}}\cdot\frac{1}{\text{sech}^2 x}\,du$ | M1A1 | M1: Complete substitution including the "dx"; A1: Fully correct substitution |
| $= \int \frac{1}{\sqrt{1+u^2}}\,du = \text{arsinh}\,u (+c)$ | M1 | Reaches $\text{arsinh}\,u$ |
| $y = \text{arsinh}(\tanh x)(+c)$ | A1* | Reaches $y = \text{arsinh}(\tanh x)$ with/without $+c$, no errors |
### Special Case
| Answer/Working | Mark | Guidance |
|---|---|---|
| $y = \text{arsinh}(\tanh x) \Rightarrow \frac{dy}{dx} = \frac{1}{\sqrt{1+\tanh^2 x}} \times \text{sech}^2 x = \frac{\text{sech}^2 x}{\sqrt{1+\tanh^2 x}}$ | M1A1 | $\text{sech}^2 x$ must appear separate from the fraction. To score more than 2 marks using a chain rule method, a third variable must be introduced |
---\begin{enumerate}
\item Given that $y = \operatorname { arsinh } ( \tanh x )$, show that
\end{enumerate}
$$\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { \operatorname { sech } ^ { 2 } x } { \sqrt { 1 + \tanh ^ { 2 } x } }$$
\section*{-}
\includegraphics[max width=\textwidth, alt={}, center]{64dc962a-1788-49ac-a4db-af1241b552a0-03_51_51_276_2012}\\
N\\
\hfill \mbox{\textit{Edexcel FP3 2017 Q1 [5]}}