Find the derivative of the function sin ( a x + b ) cos ( c x + d ) .

Let, y = sin ( a x + b ) cos ( c x + d ) Differentiating w.r.t x , we get d y d x = cos c x + d d d x sin ( a x + b ) - sin a x + b d d x cos ( c x + d ) cos 2 ( c x + d ) = cos ( c x + d ) cos ( a x + b ) · a - sin ( a x + b ) - sin ( c x + d ) · c cos 2 ( c x + d ) ∴ d y d x = a · cos ( a x + b ) · cos ( c x + d ) + c · sin ( a x + b ) sin ( c x + d ) cos 2 ( c x + d )

Find the derivative of the function tan - 1 tan h x 2 .

Let, y = tan - 1 tan h x 2 Differentiating w.r.t x , we get d y d x = d d x tan - 1 tan h x 2 = 1 1 + tan h 2 x 2 · d d x tan h x 2 = sec h 2 x 2 · 1 2 1 + tan h 2 x 2 ∴ d y d x = sec h 2 x 2 2 1 + tan h 2 x 2

Find the derivative of the function sin x · tan - 1 x 2 .

Let, y = sin x · tan - 1 x 2 Differentiating w.r.t x , we get d y d x = d d x sin x · tan - 1 x 2 = sin x · d d x tan - 1 x 2 + tan - 1 x 2 · d d x sin x = sin x · 2 tan - 1 x · 1 1 + x 2 + tan - 1 x 2 · cos x ∴ d y d x = 2 sin x tan - 1 x 1 + x 2 + cos x · tan - 1 x 2

Find the derivative of the function sin - 1 b + a sin x a + b sin x a > 0 , b > 0 .

Let, y = sin - 1 b + a sin x a + b sin x a > 0 , b > 0 Let, u = b + a sin x a + b sin x Differentiating w.r.t x , we get d u d x = d d x b + a sin x a + b sin x = ( a + b sin x ) ( a cos x ) - ( b + a sin x ) ( b cos x ) ( a + b sin x ) 2 ∵ d d x f x g x = g x f ' x - f x g ' x g x 2 = a 2 cos x + a b sin x cos x - b 2 cos x - a b sin x cos x ( a + b sin x ) 2 = a 2 - b 2 cos x ( a + b sin x ) 2 Now, y = sin - 1 u ⇒ d y d x = d d x sin - 1 u = 1 1 - u 2 · d u d x ⇒ d y d x = 1 1 - b + a sin x a + b sin x 2 · a 2 - b 2 cos x ( a + b sin x ) 2 = a + b sin x ( a + b sin x ) 2 - ( b + a sin x ) 2 · a 2 - b 2 cos x ( a + b sin x ) 2 = a 2 - b 2 cos x a 2 + b 2 sin 2 x - b 2 - a 2 sin 2 x · 1 ( a + b sin x ) = a 2 - b 2 cos x a 2 - b 2 - a 2 - b 2 sin 2 x · 1 ( a + b sin x ) ∴ d y d x = a 2 - b 2 ( a + b sin x )

Find the derivative of the function cos - 1 b + a cos x a + b cos x a > 0 , b > 0 .

Let, y = cos - 1 b + a cos x a + b cos x a > 0 , b > 0 Let, u = b + a cos x a + b cos x Differentiating w.r.t x , we get d u d x = d d x b + a cos x a + b cos x = ( a + b cos x ) ( - a sin x ) + ( b + a cos x ) ( - b sin x ) ( a + b cos x ) 2 = sin x a 2 + a b cos x - b 2 - a b cos x ( a + b cos x ) 2 = sin x a 2 - b 2 ( a + b cos x ) 2 Now, y = cos - 1 u ⇒ d y d x = d d x cos - 1 u = - 1 1 - u 2 · d d x u = - 1 1 - b + a cos x a + b cos x 2 · sin x · a 2 - b 2 ( a + b cos x ) 2 = - sin x · a 2 - b 2 ( a + b cos x ) 2 - ( b + a cos x ) 2 · 1 ( a + b cos x ) = - a 2 - b 2 sin x a 2 + b 2 cos 2 x - b 2 - a 2 cos 2 x · 1 ( a + b cos x ) ∴ d y d x = - a 2 - b 2 a + b cos x

Find the derivative of the function tan - 1 cos x 1 + cos x .

Let, y = tan - 1 cos x 1 + cos x Let, u = cos x 1 + cos x Differentiating w.r.t x , we get d u d x = d d x cos x 1 + cos x = ( 1 + cos x ) ( - sin x ) - cos x ( - sin x ) ( 1 + cos x ) 2 = - sin x - sin x cos x + sin x · cos x ( 1 + cos x ) 2 = - sin x ( 1 + cos x ) 2 Now, y = tan - 1 u ⇒ d y d u = 1 1 + u 2 d u d x = 1 1 + cos 2 x ( 1 + cos x ) 2 · - sin x ( 1 + cos x ) 2 = 1 + cos x 2 ( 1 + cos x ) 2 + cos 2 x · - sin x ( 1 + cos x ) 2 = ( 1 + cos x ) 2 2 cos 2 x + 2 cos x + 1 · - sin x ( 1 + cos x ) 2 = ( 1 + cos x ) 2 2 cos 2 x + 2 cos x + 1 × - sin x ( 1 + cos x ) 2 ∴ d y d x = - sin x 2 cos 2 x + 2 cos x + 1

E M B I B E

Intermediate First Year Mathematics Paper 1B>Chapter 9 - Differentiation>Exercise 9(b)

Telangana Board Solutions for Chapter: Differentiation, Exercise 2: Exercise 9(b)

Author:Telangana Board

Telangana Board Mathematics Solutions for Exercise - Telangana Board Solutions for Chapter: Differentiation, Exercise 2: Exercise 9(b)

Attempt the free practice questions on Chapter 9: Differentiation, Exercise 2: Exercise 9(b) with hints and solutions to strengthen your understanding. Intermediate First Year Mathematics Paper 1B solutions are prepared by Experienced Embibe Experts.