Differentiate the following function w.rt. x sin - 1 2 sec x 1 + sec 2 x

The given function is, sin - 1 2 sec x 1 + sec 2 x Let y = sin - 1 2 sec x 1 + sec 2 x Let sec x = tan t ⇒ t = tan - 1 sec x y = sin - 1 2 tan t 1 + tan 2 t y = sin - 1 sin 2 t ∵ sin 2 x = 2 tan x 1 + tan 2 x y = 2 t ∵ sin - 1 sin x = x y = 2 tan - 1 sec x [ Substituting value from t ] On differentiating both sides with respect to x , we get d d x y = d d x 2 tan - 1 sec x [using chain rule] d y d x = 2 1 + sec 2 x d d x sec x ∵ d d x tan - 1 x = 1 1 + x 2 d y d x = 2 sec x tan x 1 + sec 2 x

Differentiate the following function w.r.t. x sin - 1 x + 1 - x 2 2

The given function is, sin - 1 x + 1 - x 2 2 Let y = sin - 1 x + 1 - x 2 2 Let x = sin t ⇒ t = sin - 1 x y = sin - 1 sin t + 1 - sin 2 t 2 ∵ 1 - sin 2 x = cos 2 x y = sin - 1 sin t + cos 2 t 2 y = sin - 1 sin t + cos t 2 y = sin - 1 1 2 cos t + 1 2 sin t y = sin - 1 cos π 4 sin t + sin π 4 cos t ∵ cos π 4 = 1 2 ; sin π 4 = 1 2 y = sin - 1 sin π 4 + t ∵ sin a + b = cos a sin b + sin a cos b y = π 4 + t ∵ sin - 1 sin x = x y = π 4 + sin - 1 x [substituting value of t ] On differentiating both sides with respect to x , we get d d x y = d d x π 4 + sin - 1 x [using subtraction rule] d y d x = d d x π 4 + d d x sin - 1 x d y d x = 0 + 1 1 - x 2 d y d x = 1 1 - x 2

Differentiate the following function w.rt. x sin - 1 6 x - 4 1 - 4 x 2 5

The given function is, sin - 1 6 x - 4 1 - 4 x 2 5 Let y = sin - 1 6 x - 4 1 - 4 x 2 5 Let 2 x = sin t ⇒ t = sin - 1 2 x y = sin - 1 3 2 x - 4 1 - 2 x 2 5 y = sin - 1 3 sin t - 4 1 - sin 2 t 5 ∵ 1 - sin 2 x = cos 2 x y = sin - 1 3 sin t - 4 cos 2 t 5 y = sin - 1 3 sin t - 4 cos t 5 Let 3 = r cos α , 4 = r sin α ⇒ r 2 cos 2 α + sin 2 α = 3 2 + 4 2 ∵ cos 2 x + sin 2 x = 1 ⇒ r 2 = 9 + 16 ⇒ r 2 = 25 ⇒ r = 5 and tan α = 4 3 ⇒ α = tan - 1 4 3 y = sin - 1 r sin t cos α - r sin α cos t 5 ∵ sin a - b = sin a cos b - cos a sin b y = sin - 1 r sin t - α 5 [putting value of r] y = sin - 1 5 sin t - α 5 y = sin - 1 sin t - α ∵ sin - 1 sin x = x y = t - α y = sin - 1 2 x - tan - 1 4 3 [ Substituting value of t , α ] On differentiating both sides with respect to x , we get d d x y = d d x sin - 1 2 x - tan - 1 4 3 [using chain rule and addition rule] d y d x = d d x sin - 1 2 x - d d x tan - 1 4 3 ∵ d d x c = 0 ; d d x sin - 1 x = 1 1 - x 2 d y d x = 1 1 - 4 x 2 d d x 2 x - 0 d y d x = 2 1 - 4 x 2

Differentiate the following function w.rt. x tan - 1 x 1 3 + a 1 3 1 - x 1 3 a 1 3

The given function is, tan - 1 x 1 3 + a 1 3 1 - x 1 3 a 1 3 Let y = tan - 1 x 1 3 + a 1 3 1 - x 1 3 a 1 3 ∵ tan - 1 x + tan - 1 y = tan - 1 x + y 1 - x y y = tan - 1 x 1 3 + tan - 1 a 1 3 On differentiating both sides with respect to x , we get d d x y = d d x tan - 1 x 1 3 + tan - 1 a 1 3 d y d x = d d x tan - 1 x 1 3 + d d x tan - 1 a 1 3 ∵ d d x tan - 1 x = 1 1 + x 2 d y d x = 1 1 + x 1 3 2 d d x x 1 3 + 0 d y d x = 1 1 + x 1 3 2 × 1 3 x - 2 3 d d x = 1 3 x 2 3 1 + x 2 3

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Understanding ISC Mathematics Class 12 Volume 1>Chapter 5 - Continuity and Differentiability>EXERCISE 5.7

M L Aggarwal Solutions for Chapter: Continuity and Differentiability, Exercise 7: EXERCISE 5.7

Author:M L Aggarwal

M L Aggarwal Mathematics Solutions for Exercise - M L Aggarwal Solutions for Chapter: Continuity and Differentiability, Exercise 7: EXERCISE 5.7

Attempt the practice questions on Chapter 5: Continuity and Differentiability, Exercise 7: EXERCISE 5.7 with hints and solutions to strengthen your understanding. Understanding ISC Mathematics Class 12 Volume 1 solutions are prepared by Experienced Embibe Experts.