C3 January 2011 Q8

(a) Given that\[\frac{\mathrm{d}}{\mathrm{d}x}(\cos x) = -\sin x\]show that \(\dfrac{\mathrm{d}}{\mathrm{d}x}(\sec x) = \sec x\tan x\). (3)

Given that

\[x = \sec 2y\]

(b) find \(\dfrac{\mathrm{d}x}{\mathrm{d}y}\) in terms of \(y\). (2)

Mark scheme (a)

Scheme	Marks
\(y = \sec x = \dfrac{1}{\cos x} = (\cos x)^{-1}\)
\(\dfrac{\mathrm{d}y}{\mathrm{d}x} = -1(\cos x)^{-2}(-\sin x)\)	M1 A1
\(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \left\{\dfrac{\sin x}{\cos^2 x}\right\} = \underline{\underline{\left(\dfrac{1}{\cos x}\right)\left(\dfrac{\sin x}{\cos x}\right)}} = \underline{\underline{\sec x\tan x}}\)	A1 AG
	(3)

M1: Writes \(\sec x\) as \((\cos x)^{-1}\) and gives \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \pm\left((\cos x)^{-2}(\sin x)\right)\)

A1: \(-1(\cos x)^{-2}(-\sin x)\) or \((\cos x)^{-2}(\sin x)\)

A1 AG: Convincing proof. Must see both underlined steps.

Mark scheme (b)

Scheme	Marks
\(x = \sec 2y, \quad y \ne (2n + 1)\frac{\pi}{4},\ n \in \mathbb{Z}.\)
\(\dfrac{\mathrm{d}x}{\mathrm{d}y} = 2\sec 2y\tan 2y\)	M1 A1
	(2)

M1: \(K\sec 2y\tan 2y\)

A1: \(2\sec 2y\tan 2y\)

Mark scheme (c)

Scheme	Marks
\(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{1}{2\sec 2y\tan 2y}\)	M1
\(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{1}{2x\tan 2y}\)	M1
\(1 + \tan^2 A = \sec^2 A \Rightarrow \tan^2 2y = \sec^2 2y - 1\)	M1
So \(\tan^2 2y = x^2 - 1\)
\(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{1}{2x\sqrt{(x^2 - 1)}}\)	A1
	(4)
	(9 marks)

M1: Applies \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{1}{\left(\frac{\mathrm{d}x}{\mathrm{d}y}\right)}\)

M1: Substitutes \(x\) for \(\sec 2y\).

M1: Attempts to use the identity \(1 + \tan^2 A = \sec^2 A\)

A1: \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{1}{2x\sqrt{(x^2 - 1)}}\)