Differentiate the following w.r.t. x : cot x

Let, y = cot x and u = x Therefore, y = cot u Let, v = cot u , therefore, y = v Differentiating above equation w.r.t. x , d y d x = d y d v . d v d u . d u d x .................By chain rule d y d x = d d v ( v ) . d d u ( cot u ) . d d x ( x ) = 1 2 v . ( - cosec 2 u ) . 1 2 x = 1 2 cot u . ( - cosec 2 u ) . 1 2 x = 1 2 cot x . ( - cosec 2 x ) . 1 2 x = - cosec 2 x 4 x cot x .

Differentiate the following w.r.t. x : cos sin a x + b

Let, y = cos sin a x + b and u = a x + b Therefore, y = cos sin u Let, v = u , therefore, y = cos sin v Let, w = sin v , therefore, y = cos w Differentiating above equation w.r.t. x , By using chain rule d y d x = d y d w . d w d v . d v d u . d u d x d y d x = d ( cos w ) d w . d ( sin v ) d v . d ( u ) d u . d ( a x + b ) d x = ( - sin w ) . ( cos v ) . 1 2 u . d ( a x ) d x + d ( b ) d x = ( - sin ( sin v ) ) . cos u . a 2 a x + b = - sin sin u . cos ( a x + b ) . a 2 a x + b = - a . cos a x + b 2 a x + b . sin sin a x + b .

Differentiate the following w.r.t. x : cosec x 3 + 1

Let, y = cosec ( x 3 + 1 ) and u = x 3 + 1 Therefore, y = cosec u Let, v = cosec u , therefore, y = v Differentiating above equation w.r.t. x , By using chain rule d y d x = d y d v . d v d u d u d x d y d x = d d v ( v ) . d d u ( cosec u ) . d d x ( x 3 + 1 ) = 1 2 v . - cosec u · cot u d ( x 3 ) d x + d ( 1 ) d x = 1 2 cosec x 3 + 1 . - cosec x 3 + 1 . cot ( x 3 + 1 ) . ( 3 x 2 + 0 ) ) = - 3 x 2 2 . cosec ( x 3 + 1 ) . cot ( x 3 + 1 ) .

Find d y d x , when: y = sin 1 + x 2 1 - x 2

Given y = sin 1 + x 2 1 - x 2 d y d x = cos 1 + x 2 1 - x 2 · d d x 1 + x 2 1 - x 2 dy d x = cos 1 + x 2 1 - x 2 · 1 - x 2 2 x - 1 + x 2 - 2 x 1 - x 2 2 (Using quotient rule d d x u v = v u ' - u v ' v 2 ) d y d x = cos 1 + x 2 1 - x 2 · 2 x - 2 x 3 + 2 x + 2 x 3 1 - x 2 2 d y d x = cos 1 + x 2 1 - x 2 · 4 x 1 - x 2 2 .

Find d y d x , when: y = sin x + x 2 cot 2 x

Given, y = sin x + x 2 cot 2 x Differentiate with respect to x , d y d x = d d x sin x + x 2 cot 2 x = cot 2 x d ( sin x + x 2 ) d x - sin x + x 2 d cot 2 x d x cot 2 x 2 = cot 2 x ( cos x + 2 x ) - ( sin x + x 2 ) ( - 2 cosec 2 2 x ) ( cot 2 x ) 2 = cos x + 2 x cot 2 x - ( sin x + x 2 ) ( - 2 cosec 2 2 x ) ( cot 2 x ) 2 = tan 2 x cos x + 2 x + ( sin x + x 2 ) 2 sin 2 2 x cos 2 2 x sin 2 2 x = tan 2 x cos x + 2 x + 2 ( sin x + x 2 ) cos 2 2 x d y d x = tan 2 x ( cos x + 2 x ) + 2 sec 2 2 x ( sin x + x 2 )

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SENIOR SECONDARY SCHOOL MATHEMATICS FOR CLASS 12>Differential Calculus>Chapter 10 - Differentiation>EXERCISE

R S Aggarwal Solutions for Exercise 1: EXERCISE

Author:R S Aggarwal

R S Aggarwal Mathematics Solutions for Exercise - R S Aggarwal Solutions for Exercise 1: EXERCISE

Attempt the free practice questions from Exercise 1: EXERCISE with hints and solutions to strengthen your understanding. SENIOR SECONDARY SCHOOL MATHEMATICS FOR CLASS 12 solutions are prepared by Experienced Embibe Experts.