Differentiate with respect to x . tan - 1 2 x x + 1

Let y = tan - 1 2 x x + 1 Differentiate both sides, with respect to x , we get d y d x = d d x tan - 1 2 x x + 1 ⇒ d y d x = 1 1 + 2 x x + 1 2 × d d x 2 x x + 1 ⇒ d y d x = ( x + 1 ) 2 ( x + 1 ) 2 + 4 x 2 × ( x + 1 ) · d d x 2 x - ( 2 x ) · d d x ( x + 1 ) x + 1 2 [Applying quotient rule] ⇒ d y d x = ( x + 1 ) 2 x 2 + 2 x + 1 + 4 x 2 × 2 x + 1 - 2 x ( x + 1 ) 2 ⇒ d y d x = 2 5 x 2 + 2 x + 1

Differentiate with respect to x . tan - 1 2 x x

Let y = tan - 1 2 x x Differentiate both sides, with respect to x , we get d y d x = d d x tan - 1 2 x x ⇒ d y d x = x · d d x tan - 1 2 x - tan - 1 2 x · d d x ( x ) x 2 [Applying quotient rule] ⇒ d y d x = x 1 + 2 x 2 × d d x ( 2 x ) - tan - 1 2 x x 2 ⇒ d y d x = 2 x 1 + 4 x 2 - tan - 1 2 x x 2 ⇒ d y d x = 2 x - ( 1 + 4 x 2 ) tan - 1 2 x x 2 ( 1 + 4 x 2 )

Find the equation of the tangent to the curve y = tan - 1 x 2 at the point where x = 2 .

The equation of the given curve is y = tan - 1 x 2 ⋯ 1 Putting x = 2 in 1 , we get y = tan - 1 2 2 = tan - 1 1 = π 4 So, the point of contact is 2 , π 4 Now, y = tan - 1 x 2 Differentiate both sides with respect to x , we get ⇒ d y d x = 1 1 + x 2 2 × d d x x 2 ⇒ d y d x = 4 4 + x 2 × 1 2 ⇒ d y d x = 2 4 + x 2 ⇒ d y d x at x = 2 = 2 4 + 2 2 = 1 4 Thus, the required equation of tangent at 2 , π 4 is y - π 4 = d y d x at x = 2 x - 2 ⇒ y - π 4 = 1 4 x - 2 ⇒ 4 y - π = x - 2 ⇒ x - 4 y = 2 - π

E M B I B E

Cambridge International AS & A Level Mathematics : Pure Mathematics 2 & 3 Course Book>Chapter 8 - Further Calculus>EXERCISE 8A

Sue Pemberton, Julianne Hughes and, Julian Gilbey Solutions for Chapter: Further Calculus, Exercise 2: EXERCISE 8A

Author:Sue Pemberton, Julianne Hughes & Julian Gilbey

Sue Pemberton Mathematics Solutions for Exercise - Sue Pemberton, Julianne Hughes and, Julian Gilbey Solutions for Chapter: Further Calculus, Exercise 2: EXERCISE 8A

Attempt the practice questions on Chapter 8: Further Calculus, Exercise 2: EXERCISE 8A 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: Further Calculus, Exercise 2: EXERCISE 8A with Hints & Solutions

EASY

AS and A Level

IMPORTANT

Cambridge International AS & A Level Mathematics : Pure Mathematics 2 & 3 Course Book>Chapter 8 - Further Calculus>EXERCISE 8A>Q 1

Differentiate with respect to $x .$

$\tan^{- 1} (x - 1)$

EASY

AS and A Level

IMPORTANT

Cambridge International AS & A Level Mathematics : Pure Mathematics 2 & 3 Course Book>Chapter 8 - Further Calculus>EXERCISE 8A>Q 1

Differentiate with respect to $x .$

$\tan^{- 1} x^{2}$

MEDIUM

AS and A Level

IMPORTANT

Cambridge International AS & A Level Mathematics : Pure Mathematics 2 & 3 Course Book>Chapter 8 - Further Calculus>EXERCISE 8A>Q 1

Differentiate with respect to $x .$

$\tan^{- 1} (\frac{2 x}{x + 1})$

EASY

AS and A Level

IMPORTANT

Cambridge International AS & A Level Mathematics : Pure Mathematics 2 & 3 Course Book>Chapter 8 - Further Calculus>EXERCISE 8A>Q 2

Differentiate with respect to $x .$

$x \tan^{- 1} x$

MEDIUM

AS and A Level

IMPORTANT

Cambridge International AS & A Level Mathematics : Pure Mathematics 2 & 3 Course Book>Chapter 8 - Further Calculus>EXERCISE 8A>Q 2

Differentiate with respect to $x .$

$\frac{\tan^{- 1} 2 x}{x}$

MEDIUM

AS and A Level

IMPORTANT

Cambridge International AS & A Level Mathematics : Pure Mathematics 2 & 3 Course Book>Chapter 8 - Further Calculus>EXERCISE 8A>Q 2

Differentiate with respect to $x .$

$e^{x} \tan^{- 1} x$

HARD

AS and A Level

IMPORTANT

Cambridge International AS & A Level Mathematics : Pure Mathematics 2 & 3 Course Book>Chapter 8 - Further Calculus>EXERCISE 8A>Q 3

Find the equation of the tangent to the curve $y = \tan^{- 1} (\frac{x}{2})$ at the point where $x = 2 .$

HARD

AS and A Level

IMPORTANT

Cambridge International AS & A Level Mathematics : Pure Mathematics 2 & 3 Course Book>Chapter 8 - Further Calculus>EXERCISE 8A>Q 4

Show that the tangent to the curve $y = \tan^{- 1} x$ at the point where $x = - 1$ is perpendicular to the normal to the curve at the point where $x = 1 .$ Find the $x$ -coordinate of the point where this tangent and normal intersect.

Learn and Prepare for any exam you want !

|

|

|

|

|

|

|

|