| Exam Board | Edexcel |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2018 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Differentiating Transcendental Functions |
| Type | Implicit differentiation |
| Difficulty | Standard +0.3 This is a straightforward two-part question requiring standard differentiation techniques. Part (a) is a routine proof using the quotient rule to derive a standard result. Part (b) involves implicit differentiation and algebraic manipulation to express dy/dx in terms of x, which is a standard C3 technique. The question is slightly above average difficulty due to the multi-step nature and the need to manipulate exponential and trigonometric expressions, but it follows predictable patterns without requiring novel insight. |
| Spec | 1.05h Reciprocal trig functions: sec, cosec, cot definitions and graphs1.07l Derivative of ln(x): and related functions1.07s Parametric and implicit differentiation |
| Answer | Marks |
|---|---|
| \(\frac{d}{d\theta}(\sec\theta) = \frac{d}{d\theta}(\cos\theta)^{-1} = -1 \times (\cos\theta)^{-2} \times (-\sin\theta)\) | M1 |
| \(= \frac{1}{\cos\theta} \times \frac{\sin\theta}{\cos\theta}\) | A1* |
| \(= \sec\theta \tan\theta\) | A1* |
| Answer | Marks |
|---|---|
| \(x = e^{\sec y} \Rightarrow \frac{dx}{dy} = e^{\sec y} \times \sec y \tan y\) oe | M1A1 |
| \(\Rightarrow \frac{dy}{dx} = \frac{1}{e^{\sec y} \times \sec y \tan y}\) | M1 |
| Uses \(1 + \tan^2 y = \sec^2 y\) with \(\sec y = \ln x \Rightarrow \tan y = \sqrt{(\ln x)^2-1}\) | M1 |
| $\Rightarrow \frac{dy}{dx} = \frac{1}{x \ln x \sqrt{(\ln x)^2-1}} = \frac{1}{x\sqrt{(\ln x)^4-(\ln x)^2 |
**(a)**
| $\frac{d}{d\theta}(\sec\theta) = \frac{d}{d\theta}(\cos\theta)^{-1} = -1 \times (\cos\theta)^{-2} \times (-\sin\theta)$ | M1 |
| $= \frac{1}{\cos\theta} \times \frac{\sin\theta}{\cos\theta}$ | A1* |
| $= \sec\theta \tan\theta$ | A1* |
(2 marks)
**(b)**
| $x = e^{\sec y} \Rightarrow \frac{dx}{dy} = e^{\sec y} \times \sec y \tan y$ oe | M1A1 |
| $\Rightarrow \frac{dy}{dx} = \frac{1}{e^{\sec y} \times \sec y \tan y}$ | M1 |
| Uses $1 + \tan^2 y = \sec^2 y$ with $\sec y = \ln x \Rightarrow \tan y = \sqrt{(\ln x)^2-1}$ | M1 |
| $\Rightarrow \frac{dy}{dx} = \frac{1}{x \ln x \sqrt{(\ln x)^2-1}} = \frac{1}{x\sqrt{(\ln x)^4-(\ln x)^2\begin{enumerate}
\item (a) By writing $\sec \theta = \frac { 1 } { \cos \theta }$, show that $\frac { \mathrm { d } } { \mathrm { d } \theta } ( \sec \theta ) = \sec \theta \tan \theta$\\
(b) Given that
\end{enumerate}
$$x = \mathrm { e } ^ { \sec y } \quad x > \mathrm { e } , \quad 0 < y < \frac { \pi } { 2 }$$
show that
$$\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 1 } { x \sqrt { \mathrm {~g} ( x ) } } , \quad x > \mathrm { e }$$
where $\mathrm { g } ( x )$ is a function of $\ln x$.\\
\hfill \mbox{\textit{Edexcel C3 2018 Q8 [7]}}