By writing cosec x as 1 sin x , find d d x cosec x .

To find: d d x cosec x ⇒ d d x cosec x = d d x 1 sin x ⇒ d d x cosec x = sin x d d x 1 - 1 d d x sin x sin 2 x Using quotient rule of differentiation ⇒ d d x cosec x = sin x 0 - cos x sin 2 x ∵ d d x sin x = cos x and d d x constant = 0 ⇒ d d x cosec x = - cos x sin 2 x ⇒ d d x cosec x = - cos e c x c o t x

By writing cot x as cos x sin x , find d d x cot x .

To find: d d x cot x ⇒ d d x cot x = d d x cos x sin x ⇒ d d x cot x = sin x d d x cos x - cos x d d x sin x sin 2 x Using quotient rule of differentiation ⇒ d d x cot x = sin x - sin x - cos x cos x sin 2 x ∵ d d x sin x = cos x and d d x cos x = - sin x ⇒ d d x co t x = - sin 2 x + cos 2 x sin 2 x ⇒ d d x co t x = - 1 sin 2 x ⇒ d d x co t x = - cos e c 2 x

The equation of a curve is y = 5 sin 3 x - 2 cos x Find the equation of the tangent to the curve at the point π 3 , - 1 . Give the answer in the form y = m x + c where the values of m and c are correct to 3 significant figures.

Given equation of the curve is y = 5 sin 3 x - 2 cos x We need to find the equation of tangent to the curve at the point π 3 , - 1 . We know that the gradient of tangent to the curve y = f x at point x 1 , y 1 is given by d y d x ] x 1 , y 1 . Thus, for the given curve d y d x = d d x 5 sin 3 x - 2 cos x = 5 d d x sin 3 x - 2 d d x cos x = 5 cos 3 x × 3 - 2 × - sin x = 15 cos 3 x + 2 sin x Now, d y d x ] π 3 , - 1 = 15 cos 3 π 3 + 2 sin π 3 = 15 cosπ + 2 × 3 2 = - 15 + 3 Thus, gradient of the tangent to the given curve is - 15 + 3 . Now, the equation of the tangent to the given curve with gradient - 15 + 3 at the point π 3 , - 1 is given by y - - 1 = - 15 + 3 x - π 3 ⇒ y + 1 = - 15 x - 1 . 0472 + 3 x - 1 . 0472 ⇒ y + 1 = - 15 x + 15 . 70796 + 3 x - 1 . 8138 ⇒ y = - 15 + 3 x + 12 . 9 ⇒ y = - 13 . 3 x + 12 . 9 which is of the form y = m x + c .

A curve has equation y = sin 2 x e 2 x for 0 ⩽ x ⩽ π 2 . Find the exact value for the x -coordinate of the stationary point of this curve.

Given equation of the curve is y = sin 2 x e 2 x for 0 ≤ x ≤ π 2 . We need to find the x -coordinate of the stationary point of the curve. We know that the stationary points are the points where the gradient of the curve is zero. So, d y d x = d d x sin 2 x e 2 x = e 2 x d d x sin 2 x - sin 2 x d d x e 2 x e 2 x 2 [Applying the quotient rule] = e 2 x × cos 2 x × d d x 2 x - sin 2 x × e 2 x × d d x 2 x e 4 x [Applying the chain rule for differentiation]. = 2 e 2 x cos 2 x - 2 e 2 x sin 2 x e 4 x = 2 e 2 x cos 2 x - sin 2 x e 4 x . . . . i Now, for the stationary points we must have d y d x = 0 ⇒ 2 e 2 x cos 2 x - sin 2 x e 4 x = 0 ⇒ 2 e - 2 x cos 2 x - sin 2 x = 0 ⇒ cos 2 x - sin 2 x = 0 ∵ e - 2 x ≠ 0 ⇒ tan 2 x = 1 ⇒ tan 2 x = tan π 4 ⇒ 2 x = n π + π 4 , n ∈ Z ⇒ x = n π 2 + π 8 For n = 0 , we have x = π 8 For n = 1 , we have x = π 2 + π 8 which is not the required value since 0 ≤ x ≤ π 2 . Hence, the exact value of the x -coordinate of the stationary point of the curve y = sin 2 x e 2 x for 0 ≤ x ≤ π 2 is π 8 .

A curve has equation y = e 3 x sin 3 x , for 0 < x < π 2 . Find the exact value for the x -coordinates of the stationary points of this curve.

Given equation of the curve is y = e 3 x sin 3 x for 0 < x < π 2 . We need to find the x -coordinate of the stationary point of the curve. We know that the stationary points are the points where the gradient of the curve is zero. So, d y d x = d d x e 3 x sin 3 x = sin 3 x d d x e 3 x - e 3 x d d x sin 3 x sin 3 x 2 = sin 3 x × e 3 x × d d x 3 x - e 3 x × cos 3 x × d d x 3 x sin 3 x 2 = sin 3 x × e 3 x × 3 - e 3 x × cos 3 x × 3 sin 3 x 2 = 3 e 3 x sin 3 x - cos 3 x sin 3 x 2 Now, for the stationary points we must have d y d x = 0 ⇒ 3 e 3 x sin 3 x - cos 3 x sin 3 x 2 = 0 ⇒ sin 3 x - cos 3 x = 0 ⇒ tan 3 x = 1 ⇒ tan 3 x = tan π 4 ⇒ 3 x = n π + π 4 , n ∈ Z ⇒ x = n π 3 + π 12 Since 0 < x < π 2 ∴ n = 0 ⇒ x = π 12 n = 1 ⇒ x = π 3 + π 12 = 5 π 12 Hence, the x -coordinates of the stationary points of this curve are x = π 12 , and 5 π 12 .

A curve has equation y = tan x cos 2 x for 0 ⩽ x < π 2 . Find the gradient of this curve.

Given equation of the curve is y = tan x cos 2 x for 0 ≤ x ≤ π 2 . We need to find the gradient of this curve. So, d y d x = d d x tan x cos 2 x = tan x d d x cos 2 x + cos 2 x d d x tan x ∵ d d x u v = u d v d x + v d u d x = tan x × - sin 2 x × d d x 2 x + cos 2 x × sec 2 x = - 2 tan x sin 2 x + cos 2 x sec 2 x = - 2 sin x × 2 sin x cos x cos x + 2 cos 2 x - 1 cos 2 x ∵ tan x = sin x cos x , sin 2 x = 2 sin x cos x , cos 2 x = 2 cos 2 x - 1 = 4 sin 2 x + 2 - sec 2 x Hence, d y d x = 4 sin 2 x + 2 - sec 2 x

E M B I B E

Cambridge International AS & A Level Mathematics : Pure Mathematics 2 & 3 Course Book>Chapter 4 - Differentiation>EXERCISE 4E

Sue Pemberton, Julianne Hughes and, Julian Gilbey Solutions for Chapter: Differentiation, Exercise 6: EXERCISE 4E

Author:Sue Pemberton, Julianne Hughes & Julian Gilbey

Sue Pemberton Mathematics Solutions for Exercise - Sue Pemberton, Julianne Hughes and, Julian Gilbey Solutions for Chapter: Differentiation, Exercise 6: EXERCISE 4E

Attempt the free practice questions on Chapter 4: Differentiation, Exercise 6: EXERCISE 4E with hints and solutions to strengthen your understanding. Cambridge International AS & A Level Mathematics : Pure Mathematics 2 & 3 Course Book solutions are prepared by Experienced Embibe Experts.

Questions from Sue Pemberton, Julianne Hughes and, Julian Gilbey Solutions for Chapter: Differentiation, Exercise 6: EXERCISE 4E with Hints & Solutions

EASY

AS and A Level

IMPORTANT

Cambridge International AS & A Level Mathematics : Pure Mathematics 2 & 3 Course Book>Chapter 4 - Differentiation>EXERCISE 4E>Q 9

By writing $cosec x$ as $\frac{1}{\sin x},$ find $\frac{d}{d x} (cosec x) .$

EASY

AS and A Level

IMPORTANT

Cambridge International AS & A Level Mathematics : Pure Mathematics 2 & 3 Course Book>Chapter 4 - Differentiation>EXERCISE 4E>Q 9

By writing $\cot x$ as $\frac{\cos x}{\sin x},$ find $\frac{d}{d x} (\cot x) .$

HARD

AS and A Level

IMPORTANT

Cambridge International AS & A Level Mathematics : Pure Mathematics 2 & 3 Course Book>Chapter 4 - Differentiation>EXERCISE 4E>Q 11

The equation of a curve is $y = 5 \sin 3 x - 2 \cos x$ Find the equation of the tangent to the curve at the point $(\frac{π}{3}, - 1) .$ Give the answer in the form $y = m x + c$ where the values of $m$ and $c$ are correct to $3$ significant figures.

MEDIUM

AS and A Level

IMPORTANT

Cambridge International AS & A Level Mathematics : Pure Mathematics 2 & 3 Course Book>Chapter 4 - Differentiation>EXERCISE 4E>Q 13

A curve has equation $y = 3 \cos 2 x + 4 \sin 2 x + 1$ for $0 ⩽ x ⩽ π,$ Find the exact value of the $x$ -coordinate of the stationary point of the curve, giving your answer correct to $3$ significant figures.

MEDIUM

AS and A Level

IMPORTANT

Cambridge International AS & A Level Mathematics : Pure Mathematics 2 & 3 Course Book>Chapter 4 - Differentiation>EXERCISE 4E>Q 14

A curve has equation $y = \frac{\sin 2 x}{e^{2 x}}$ for $0 ⩽ x ⩽ \frac{π}{2} .$ Find the exact value for the $x$ -coordinate of the stationary point of this curve.

MEDIUM

AS and A Level

IMPORTANT

Cambridge International AS & A Level Mathematics : Pure Mathematics 2 & 3 Course Book>Chapter 4 - Differentiation>EXERCISE 4E>Q 15

A curve has equation $y = \frac{e^{3 x}}{\sin 3 x},$ for $0 < x < \frac{π}{2} .$ Find the exact value for the $x$ -coordinates of the stationary points of this curve.

MEDIUM

AS and A Level

IMPORTANT

Cambridge International AS & A Level Mathematics : Pure Mathematics 2 & 3 Course Book>Chapter 4 - Differentiation>EXERCISE 4E>Q 16

A curve has equation $y = \sin 2 x - x$ for $0 ⩽ x ⩽ 2 π .$ Find the $x$ -coordinates of the stationary points of the curve, and determine the nature of at least one of these stationary points.