Derivate 1 + sin x 1 - sin x

Let y = 1 + sin x 1 - sin x On differentiating y with respect to x we get ⇒ d y d x = d d x 1 + sin x 1 - sin x 1 2 ⇒ d y d x = 1 2 1 + sin x 1 - sin x - 1 2 d d x 1 + sin x 1 - sin x ⇒ d y d x = 1 2 1 + sin x 1 - sin x - 1 2 ( 1 - sin x ) d d x ( 1 + sin x ) - ( 1 + sin x ) d d x ( 1 - sin x ) ( 1 - sin x ) 2 ⇒ d y d x = 1 2 1 + sin x 1 - sin x - 1 2 ( 1 - sin x ) cos x + ( 1 + sin x ) cos x ( 1 - sin x ) 2 ⇒ d y d x = 1 2 1 + sin x 1 - sin x - 1 2 2 cos x ( 1 - sin x ) 2 ⇒ d y d x = 1 + sin x 1 - sin x - 1 2 cos x ( 1 - sin x ) 2 ⇒ d y d x = ( 1 + sin x ) - 1 2 cos x ( 1 - sin x ) 3 2 ⇒ d y d x = cos x ( 1 - sin x ) ( 1 - sin x ) ( 1 + sin x ) ⇒ d y d x = cos x ( 1 - sin x ) 1 - sin 2 x ⇒ d y d x = cos x ( 1 - sin x ) cos 2 x ⇒ d y d x = cos x ( 1 - sin x ) cos x ⇒ d y d x = 1 1 - sin x ⇒ d y d x = 1 1 - sin x × 1 + sin x 1 + sin x ⇒ d y d x = 1 + sin x 1 - sin 2 x ⇒ d y d x = 1 + sin x cos 2 x ⇒ d y d x = sec 2 x + sec x tan x ∴ d y d x = sec x ( sec x + tan x ) Thus, d d x ( 1 + sin x 1 - sin x ) = sec x ( sec x + tan x )

Derivate 1 - x 2 1 + x 2

Let y = 1 - x 2 1 + x 2 ⇒ d y d x = d d x 1 - x 2 1 + x 2 1 2 ⇒ d y d x = 1 2 1 - x 2 1 + x 2 - 1 2 d d x 1 - x 2 1 + x 2 ⇒ d y d x = 1 2 1 - x 2 1 + x 2 - 1 2 1 + x 2 d d x 1 - x 2 - 1 - x 2 d d x 1 + x 2 1 + x 2 2 ⇒ d y d x = 1 2 1 - x 2 1 + x 2 - 1 2 - 2 x 1 + x 2 - 2 x 1 - x 2 1 + x 2 2 ⇒ d y d x = 1 2 1 - x 2 1 + x 2 - 1 2 - 2 x 1 + x 2 + 1 - x 2 1 + x 2 2 ⇒ d y d x = 1 - x 2 1 + x 2 - 1 2 - 2 x 1 + x 2 2 ⇒ d y d x = - 2 x 1 - x 2 - 1 2 1 + x 2 3 2 ⇒ d y d x = - 2 x 1 + x 2 3 2 1 - x 2 1 2 Thus, d d x 1 - x 2 1 + x 2 = - 2 x 1 + x 2 3 2 1 - x 2

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

Let y = 1 + x 1 - x On differentiating y with respect to x , we get d y d x = d d x ( 1 + x 1 - x ) ⇒ d y d x = d d x 1 + x 1 - x 1 2 We know d d x x n = n x n - 1 Now by using chain rule, ⇒ d y d x = 1 2 1 + x 1 - x 1 2 - 1 d d x 1 + x 1 - x ⇒ d y d x = 1 2 1 + x 1 - x - 1 2 d d x 1 + x 1 - x We know, u v ' = v u ' - u v ' v 2 (quotient rule) ⇒ d y d x = 1 2 1 + x 1 - x - 1 2 ( 1 - x ) d d x ( 1 + x ) - ( 1 + x ) d d x ( 1 - x ) ( 1 - x ) 2 ⇒ d y d x = 1 2 1 + x 1 - x - 1 2 ( 1 - x ) d d x ( 1 ) + d d x ( x ) - ( 1 + x ) d d x ( 1 ) - d d x ( x ) ( 1 - x ) 2 How ever, d d x ( x ) = 1 and derivative of a constant is 0 ⇒ d y d x = 1 2 1 + x 1 - x 1 2 ( 1 - x ) ( 0 + 1 ) - ( 1 + x ) ( 0 - 1 ) ( 1 - x ) 2 ⇒ d y d x = 1 2 1 + x 1 - x - 1 2 ( 1 - x ) + ( 1 + x ) ( 1 - x ) 2 ⇒ d y d x = 1 2 1 + x 1 - x - 1 2 2 ( 1 - x ) 2 ⇒ d y d x = 1 + x 1 - x - 1 2 1 ( 1 - x ) 2 ⇒ d y d x = ( 1 + x ) 1 2 ( 1 - x ) 1 2 1 ( 1 - x ) 2 ⇒ d y d x = ( 1 + x ) 1 2 ( 1 - x ) 1 2 + 2 ⇒ d y d x = ( 1 + x ) - 1 2 ( 1 - x ) 3 2 ⇒ d y d x = 1 ( 1 - x ) 3 2 ( 1 + x ) 1 2 ∴ d y d x = 1 ( 1 - x ) 3 2 1 + x Thus, d d x ( 1 + x 1 - x ) = 1 ( 1 - x ) 3 2 1 + x

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

Let y = sin 1 + x 2 1 - x 2 On differentiating y with respect to x , we get d y d x = d d x sin 1 + x 2 1 - x 2 We kniow d d x ( sin x ) = cos x Now by using chain rule ⇒ d y d x = cos 1 + x 2 1 - x 2 d d x 1 + x 2 1 - x 2 We know that u v ' = v u ' - u v ' v 2 (quotient rule) ⇒ d y d x = cos 1 + x 2 1 - x 2 1 - x 2 d d x 1 + x 2 - 1 + x 2 d d x 1 - x 2 1 - x 2 2 ⇒ d y d x = cos 1 + x 2 1 - x 2 1 - x 2 d d x ( 1 ) + d d x x 2 - 1 + x 2 d d x ( 1 ) - d d x x 2 1 - x 2 2 How ever, d d x x 2 = 2 x and derivative of a constant is 0 ⇒ d y d x = cos 1 + x 2 1 - x 2 1 - x 2 ( 0 + 2 x ) - 1 + x 2 ( 0 - 2 x ) 1 - x 2 2 ⇒ d y d x = cos 1 + x 2 1 - x 2 2 x 1 - x 2 + 2 x 1 + x 2 1 - x 2 2 ⇒ d y d x = cos 1 + x 2 1 - x 2 2 x 1 - x 2 + 1 + x 2 1 - x 2 2 ⇒ d y d x = cos 1 + x 2 1 - x 2 2 x ( 2 ) 1 - x 2 2 ⇒ d y d x = cos 1 + x 2 1 - x 2 4 x 1 - x 2 2 ∴ d y d x = 4 x 1 - x 2 2 cos 1 + x 2 1 - x 2 Thus, d d x sin 1 + x 2 1 - x 2 = 4 x 1 - x 2 2 cos 1 + x 2 1 - x 2

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MATHEMATICS CLASS XII VOLUME-1>Chapter 11 - Differentiation>EXERCISE

R D Sharma Solutions for Chapter: Differentiation, Exercise 1: EXERCISE

Author:R D Sharma

R D Sharma Mathematics Solutions for Exercise - R D Sharma Solutions for Chapter: Differentiation, Exercise 1: EXERCISE

Attempt the free practice questions on Chapter 11: Differentiation, Exercise 1: EXERCISE with hints and solutions to strengthen your understanding. MATHEMATICS CLASS XII VOLUME-1 solutions are prepared by Experienced Embibe Experts.