Question 8 - A-Level Maths

Edexcel C3 2011 January — Question 8 9 marks

Exam Board	Edexcel
Module	C3 (Core Mathematics 3)
Year	2011
Session	January
Marks	9
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Chain Rule
Type	Inverse function differentiation
Difficulty	Moderate -0.3 This is a structured, multi-part question testing standard differentiation techniques. Part (a) is routine proof using quotient rule or chain rule on sec x = 1/cos x. Parts (b) and (c) apply the chain rule and inverse function relationship (dy/dx = 1/(dx/dy)) with straightforward algebraic manipulation. All steps are textbook exercises with clear signposting, making it slightly easier than average.
Spec	1.05h Reciprocal trig functions: sec, cosec, cot definitions and graphs 1.07k Differentiate trig: sin(kx), cos(kx), tan(kx)1.07s Parametric and implicit differentiation

8. (a) Given that $$\frac { \mathrm { d } } { \mathrm {~d} x } ( \cos x ) = - \sin x$$ show that $\frac { \mathrm { d } } { \mathrm { d } x } ( \sec x ) = \sec x \tan x$. Given that $$x = \sec 2 y$$ (b) find $\frac { \mathrm { d } x } { \mathrm {~d} y }$ in terms of $y$.
(c) Hence find $\frac { \mathrm { d } y } { \mathrm {~d} x }$ in terms of $x$. \includegraphics[max width=\textwidth, alt={}, center]{3ff6824f-9fbf-4b5b-8bab-91332c549b36-14_102_93_2473_1804}

Show mark scheme Show mark scheme source

Question 8:

Part (a)

$y = \sec x = \frac{1}{\cos x} = (\cos x)^{-1}$

Answer	Marks	Guidance
Working	Mark	Guidance
$\frac{dy}{dx} = -1(\cos x)^{-2}(-\sin x)$	M1	Writes $\sec x$ as $(\cos x)^{-1}$ and gives $\frac{dy}{dx} = \pm\left((\cos x)^{-2}(\sin x)\right)$
$-1(\cos x)^{-2}(-\sin x)$ or $(\cos x)^{-2}(\sin x)$	A1
$\frac{dy}{dx} = \left\{\frac{\sin x}{\cos^2 x}\right\} = \left(\frac{1}{\cos x}\right)\left(\frac{\sin x}{\cos x}\right) = \sec x \tan x$	A1 AG	Convincing proof. Must see both underlined steps.

Total: 3 marks

Part (b)

$x = \sec 2y$, $y \neq (2n+1)\frac{\pi}{4}$, $n \in \mathbb{Z}$

Answer	Marks	Guidance
Working	Mark	Guidance
$\frac{dx}{dy} = 2\sec 2y \tan 2y$	M1	$K\sec 2y \tan 2y$
$\frac{dx}{dy} = 2\sec 2y \tan 2y$	A1

Total: 2 marks

Part (c)

Answer	Marks	Guidance
Working	Mark	Guidance
$\frac{dy}{dx} = \frac{1}{2\sec 2y \tan 2y}$	M1	Applies $\frac{dy}{dx} = \frac{1}{\left(\frac{dx}{dy}\right)}$
$\frac{dy}{dx} = \frac{1}{2x\tan 2y}$	M1	Substitutes $x$ for $\sec 2y$
$1 + \tan^2 A = \sec^2 A \Rightarrow \tan^2 2y = \sec^2 2y - 1$, so $\tan^2 2y = x^2 - 1$	M1	Attempts to use identity $1 + \tan^2 A = \sec^2 A$
$\frac{dy}{dx} = \frac{1}{2x\sqrt{x^2-1}}$	A1	$\frac{dy}{dx} = \frac{1}{2x\sqrt{x^2-1}}$

Total: 4 marks

Question total: [9]

## Question 8:

### Part (a)

$y = \sec x = \frac{1}{\cos x} = (\cos x)^{-1}$

| Working | Mark | Guidance |
|---------|------|----------|
| $\frac{dy}{dx} = -1(\cos x)^{-2}(-\sin x)$ | M1 | Writes $\sec x$ as $(\cos x)^{-1}$ and gives $\frac{dy}{dx} = \pm\left((\cos x)^{-2}(\sin x)\right)$ |
| $-1(\cos x)^{-2}(-\sin x)$ or $(\cos x)^{-2}(\sin x)$ | A1 | |
| $\frac{dy}{dx} = \left\{\frac{\sin x}{\cos^2 x}\right\} = \left(\frac{1}{\cos x}\right)\left(\frac{\sin x}{\cos x}\right) = \sec x \tan x$ | A1 AG | Convincing proof. Must see both underlined steps. |

**Total: 3 marks**

---

### Part (b)

$x = \sec 2y$, $y \neq (2n+1)\frac{\pi}{4}$, $n \in \mathbb{Z}$

| Working | Mark | Guidance |
|---------|------|----------|
| $\frac{dx}{dy} = 2\sec 2y \tan 2y$ | M1 | $K\sec 2y \tan 2y$ |
| $\frac{dx}{dy} = 2\sec 2y \tan 2y$ | A1 | |

**Total: 2 marks**

---

### Part (c)

| Working | Mark | Guidance |
|---------|------|----------|
| $\frac{dy}{dx} = \frac{1}{2\sec 2y \tan 2y}$ | M1 | Applies $\frac{dy}{dx} = \frac{1}{\left(\frac{dx}{dy}\right)}$ |
| $\frac{dy}{dx} = \frac{1}{2x\tan 2y}$ | M1 | Substitutes $x$ for $\sec 2y$ |
| $1 + \tan^2 A = \sec^2 A \Rightarrow \tan^2 2y = \sec^2 2y - 1$, so $\tan^2 2y = x^2 - 1$ | M1 | Attempts to use identity $1 + \tan^2 A = \sec^2 A$ |
| $\frac{dy}{dx} = \frac{1}{2x\sqrt{x^2-1}}$ | A1 | $\frac{dy}{dx} = \frac{1}{2x\sqrt{x^2-1}}$ |

**Total: 4 marks**

**Question total: [9]**

Show LaTeX source

8. (a) Given that

$$\frac { \mathrm { d } } { \mathrm {~d} x } ( \cos x ) = - \sin x$$

show that $\frac { \mathrm { d } } { \mathrm { d } x } ( \sec x ) = \sec x \tan x$.

Given that

$$x = \sec 2 y$$

(b) find $\frac { \mathrm { d } x } { \mathrm {~d} y }$ in terms of $y$.\\
(c) Hence find $\frac { \mathrm { d } y } { \mathrm {~d} x }$ in terms of $x$.

\includegraphics[max width=\textwidth, alt={}, center]{3ff6824f-9fbf-4b5b-8bab-91332c549b36-14_102_93_2473_1804}

\hfill \mbox{\textit{Edexcel C3 2011 Q8 [9]}}

This paper (8 questions)

View full paper

Q1 9 Q2 9 Q3 6 Q4 8 Q5 13 Q6 13 Q7 8 Q8 9