| Exam Board | WJEC |
|---|---|
| Module | Further Unit 4 (Further Unit 4) |
| Year | 2022 |
| Session | June |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hyperbolic functions |
| Type | Differentiate inverse hyperbolic functions |
| Difficulty | Standard +0.8 This Further Maths question requires multiple advanced differentiation techniques: product rule with inverse trig (part a(i)), chain rule with hyperbolic functions and logarithms (part a(ii)), and implicit differentiation with inverse hyperbolic functions for tangent equations (part b). While the techniques are standard for Further Maths, the combination of hyperbolic functions, inverse functions, and implicit differentiation in part (b) elevates this above routine practice, requiring careful algebraic manipulation and multiple steps to find tangent equations. |
| Spec | 1.07q Product and quotient rules: differentiation1.07s Parametric and implicit differentiation4.07d Differentiate/integrate: hyperbolic functions4.08g Derivatives: inverse trig and hyperbolic functions |
| Answer | Marks | Guidance |
|---|---|---|
| \(y = e^{3x}\sin^{-1}x\) | M1 | Use of product rule while differentiating |
| \(\frac{dy}{dx} = e^{3x} \cdot \frac{1}{\sqrt{1-x^2}} + 3e^{3x}\sin^{-1}x\) | A2 | A1 each term ISW |
| Answer | Marks | Guidance |
|---|---|---|
| \(y = \ln(\cosh(2x^2 + 7x))^2 = 2\ln(\cosh(2x^2 + 7x))\) | M1 | Log rule AND chain rule |
| \(\frac{dy}{dx} = \frac{2 \times \sinh(2x^2 + 7x) \times (4x + 7)}{\cosh(2x^2 + 7x)}\) | A1 | A1 oe Fully correct ISW |
| A1 | \(\sinh(2x^2 + 7x)\); \(4x + 7\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(y = \ln(\cosh(2x^2 + 7x))^2\) | M1 | Chain rule |
| \(\frac{dy}{dx} = \frac{2\cosh(2x^2 + 7x) \times \sinh(2x^2 + 7x) \times (4x + 7)}{(\cosh(2x^2 + 7x))^2}\) | (M1)(A1) | Chain rule; \(\sinh(2x^2 + 7x)\); \(4x + 7\) |
| (A1) | ||
| \(\frac{dy}{dx} = \frac{2 \times \sinh(2x^2 + 7x) \times (4x + 7)}{\cosh(2x^2 + 7x)}\) | (A1) | oe Fully correct ISW |
| Answer | Marks | Guidance |
|---|---|---|
| \(1 = \frac{1}{\sqrt{1+(y')^2}} \times \left(2y\frac{dy}{dx}\right)\) | M1 | Must see chain rule; Differentiate \(\sinh^{-1}\) |
| A1 | \(2y\frac{dy}{dx}\) | |
| A1 | ||
| \(\sqrt{1 + y^4} = 2y\frac{dy}{dx}\) | A1 | |
| \(\frac{dy}{dx} = \frac{\sqrt{1+y^4}}{2y}\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(y^2 = \sinh x\) | (M1)(A1) | \(2y\frac{dy}{dx} = \cosh x\); Cosh |
| (A1) | ||
| \(\frac{dy}{dx} = \frac{\cosh x}{2y}\) | (A1) |
| Answer | Marks | Guidance |
|---|---|---|
| \(y = \pm\sqrt{\sinh x}\) | (M1) | |
| (A1) | \(\frac{1}{2}\sinh^{-1}x\); Cosh | |
| \(\frac{dy}{dx} = \pm\frac{1}{2}\sinh^{-\frac{1}{2}}x \cosh x\) | (A1) | \(\pm\) |
| (A1) | ||
| THEN: | ||
| When \(x = 1\), \(y = \pm 1.084\), | B1 | Both |
| \(\frac{dy}{dx} = \pm 0.7117\) | A1 | cao Both |
| \(y - 1.084 = 0.7117(x - 1)\) | B1 | FT their \(y\) and \(\frac{dy}{dx}\) |
| \(y + 1.084 = -0.7117(x - 1)\) | B1 | FT their \(y\) and \(\frac{dy}{dx}\) |
**Part a) i)**
$y = e^{3x}\sin^{-1}x$ | M1 | Use of product rule while differentiating
$\frac{dy}{dx} = e^{3x} \cdot \frac{1}{\sqrt{1-x^2}} + 3e^{3x}\sin^{-1}x$ | A2 | A1 each term ISW
**Part a) ii)**
METHOD 1:
$y = \ln(\cosh(2x^2 + 7x))^2 = 2\ln(\cosh(2x^2 + 7x))$ | M1 | Log rule AND chain rule
$\frac{dy}{dx} = \frac{2 \times \sinh(2x^2 + 7x) \times (4x + 7)}{\cosh(2x^2 + 7x)}$ | A1 | A1 oe Fully correct ISW
| A1 | $\sinh(2x^2 + 7x)$; $4x + 7$
METHOD 2:
$y = \ln(\cosh(2x^2 + 7x))^2$ | M1 | Chain rule
$\frac{dy}{dx} = \frac{2\cosh(2x^2 + 7x) \times \sinh(2x^2 + 7x) \times (4x + 7)}{(\cosh(2x^2 + 7x))^2}$ | (M1)(A1) | Chain rule; $\sinh(2x^2 + 7x)$; $4x + 7$
| (A1) |
$\frac{dy}{dx} = \frac{2 \times \sinh(2x^2 + 7x) \times (4x + 7)}{\cosh(2x^2 + 7x)}$ | (A1) | oe Fully correct ISW
**Part b)**
METHOD 1:
$1 = \frac{1}{\sqrt{1+(y')^2}} \times \left(2y\frac{dy}{dx}\right)$ | M1 | Must see chain rule; Differentiate $\sinh^{-1}$
| A1 | $2y\frac{dy}{dx}$
| A1 |
$\sqrt{1 + y^4} = 2y\frac{dy}{dx}$ | A1 |
$\frac{dy}{dx} = \frac{\sqrt{1+y^4}}{2y}$ | |
METHOD 2:
$y^2 = \sinh x$ | (M1)(A1) | $2y\frac{dy}{dx} = \cosh x$; Cosh
| (A1) |
$\frac{dy}{dx} = \frac{\cosh x}{2y}$ | (A1) |
METHOD 3:
$y = \pm\sqrt{\sinh x}$ | (M1) |
| (A1) | $\frac{1}{2}\sinh^{-1}x$; Cosh
$\frac{dy}{dx} = \pm\frac{1}{2}\sinh^{-\frac{1}{2}}x \cosh x$ | (A1) | $\pm$
| (A1) |
THEN: | |
When $x = 1$, $y = \pm 1.084$, | B1 | Both
$\frac{dy}{dx} = \pm 0.7117$ | A1 | cao Both
$y - 1.084 = 0.7117(x - 1)$ | B1 | FT their $y$ and $\frac{dy}{dx}$
$y + 1.084 = -0.7117(x - 1)$ | B1 | FT their $y$ and $\frac{dy}{dx}$
---\begin{enumerate}[label=(\alph*)]
\item Differentiate each of the following with respect to $x$.
\begin{enumerate}[label=(\roman*)]
\item $y = e^{3x}\sin^{-1}x$
\item $y = \ln\left(\cosh^2(2x^2 + 7x)\right)$ [7]
\end{enumerate}
\item Find the equations of the tangents to the curve $x = \sinh^{-1}(y^2)$ at the points where $x = 1$. [8]
\end{enumerate}
\hfill \mbox{\textit{WJEC Further Unit 4 2022 Q11 [15]}}