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

Let y = sec - 1 x + 1 x - 1 + sin - 1 x - 1 x + 1 As we know, sec - 1 x = cos - 1 1 x ⇒ y = cos - 1 x - 1 x + 1 + sin - 1 x - 1 x + 1 ⇒ y = π 2 ∵ sin - 1 x + cos - 1 x = π 2 Differentiating above equation with respect to x , d y d x = d d x π 2 = 0 .

Differentiate the following w.r.t. x : 2 x cot x x .

Let y = 2 x cot x x Differentiating above equation with respect to x , d y d x = d d x 2 x cot x x Applying quotient rule of differentiation, d y d x = x d d x 2 x cot x - 2 x cot x d d x x x 2 ∵ d d x f x g x = g x f ' x - f x g ' x g 2 x Applying product rule of differentiation, d y d x = x 2 x d d x cot x + cot x d d x 2 x - 2 x cot x d d x x x 2 ∵ d d x f x g x = f ' x g x + g ' x f x ⇒ d y d x = x 2 x - c o s e c 2 x + cot x 2 x log e 2 - 2 x cot x 1 2 x x = 2 x - c o s e c 2 x x + cot x 2 x log e 2 x - 2 x cot x 1 2 x x ∵ d d x cot x = - c o s e c 2 x , d d x a x = a x ln a , d d x x n = n x n - 1 = 2 x x - cosec 2 x + cot x log e 2 - cot x 2 x .

Differentiate following w.r.t. x : sin x - x cos x x sin x + cos x

Let y = sin x - x cos x x sin x + cos x Differentiating above equation with respect to x , d y d x = d d x sin x - x cos x x sin x + cos x Applying quotient rule of differentiation, d y d x = x sin x + cos x d d x sin x - x cos x - sin x - x cos x d d x x sin x + cos x x sin x + cos x 2 ∵ d d x f x g x = g x f ' x - f x g ' x g 2 x d y d x = x sin x + cos x cos x - d d x x cos x - sin x - x cos x d d x x sin x - sin x x sin x + cos x 2 ∵ d d x sin x = cos x , d d x cos x = - sin x Applying product rule of differentiation we get, d y d x = x sin x + cos x cos x - cos x - x sin x - sin x - x cos x x cos x + sin x - sin x x sin x + cos x 2 ∵ d d x f x g x = f ' x g x + g ' x f x d y d x = x sin x + cos x x sin x - sin x - x cos x x cos x x sin x + cos x 2 d y d x = x 2 sin 2 x + cos x x sin x - x cos x sin x + x 2 cos 2 x x sin x + cos x 2 d y d x = x 2 sin 2 x + x 2 cos 2 x x sin x + cos x 2 d y d x = x 2 x sin x + cos x 2 ∵ sin 2 x + cos 2 x = 1

If y = 1 + x 1 ! + x 2 2 ! + x 3 3 ! + x 4 4 ! + . . . . . . ∞ , find the value of d y d x .

Given y = 1 + x 1 ! + x 2 2 ! + x 3 3 ! + x 4 4 ! + . . . ∞ . . . (i) Differentiating above equation with respect to x , d y d x = d d x 1 + x 1 ! + x 2 2 ! + x 3 3 ! + x 4 4 ! + . . . ∞ d y d x = d d x 1 + d d x x 1 ! + d d x x 2 2 ! + d d x x 3 3 ! + d d x x 4 4 ! + . . . ∞ d y d x = 0 + 1 1 ! d d x x + 1 2 ! d d x x 2 + 1 3 ! d d x x 3 + 1 4 ! d d x x 4 + . . . ∞ d y d x = 1 1 ! + 1 2 ! 2 x + 1 3 ! 3 x 2 + 1 4 ! 4 x 3 + . . . ∞ d y d x = 1 + 2 x 2 × 1 ! + 3 x 2 3 × 2 ! + 4 x 3 4 × 3 ! + . . . ∞ d y d x = 1 + x 1 ! + x 2 2 ! + x 3 3 ! + . . . ∞ By equation (i) , ⇒ d y d x = y .

Find the values of x for which the rate of change of x 4 4 + x 3 3 - x is more than x 4 4 .

Let y = x 4 4 + x 3 3 - x . . . i And z = x 4 4 . . . ii Let's differentiate equation (i) with respect to x , d y d x = d d x x 4 4 + x 3 3 - x d y d x = d d x x 4 4 + d d x x 3 3 - d d x x d y d x = 4 x 3 4 + 3 x 2 3 - 1 d y d x = x 3 + x 2 - 1 Let's differentiate equation (ii) with respect to x , d z d x = d d x x 4 4 d z d x = 4 x 3 4 d z d x = x 3 As we know that the rate of change is given by its derivative so, The rate of change of x 4 4 + x 3 3 - x = d y d x = x 3 + x 2 - 1 And the rate of change of x 4 4 = d z d x = x 3 Now we have to find the values of x for which d y d x > d z d x ⇒ x 3 + x 2 - 1 > x 3 ⇒ x 2 - 1 > 0 ⇒ x 2 > 1 ∴ x ∈ ( - ∞ , - 1 ) ∪ 1 , ∞ .

E M B I B E

Skills in Mathematics Differential Calculus for JEE Main & Advanced>Chapter 2 - Differentiation>Target Exercise 2.1

Amit M Agarwal Solutions for Chapter: Differentiation, Exercise 1: Target Exercise 2.1

Author:Amit M Agarwal

Amit M Agarwal Mathematics Solutions for Exercise - Amit M Agarwal Solutions for Chapter: Differentiation, Exercise 1: Target Exercise 2.1

Attempt the practice questions on Chapter 2: Differentiation, Exercise 1: Target Exercise 2.1 with hints and solutions to strengthen your understanding. Skills in Mathematics Differential Calculus for JEE Main & Advanced solutions are prepared by Experienced Embibe Experts.