Prove that: d d x cos - 1 x = - 1 1 - x 2 , - 1 < x < 1 .

To prove: d d x cos - 1 x = - 1 1 - x 2 , - 1 < x < 1 Let y = cos - 1 x , x ∈ ( - 1 , 1 ) ⇔ x = cos y Now, differentiate with respect to y , then we get: d x d y = - sin y ⇒ d y d x = - 1 sin y ⇒ d y d x = - 1 1 - cos 2 y = - 1 1 - x 2 ⇒ d y d x = - 1 1 - x 2 Hence, proved.

Prove that: d d x tan - 1 x = 1 1 + x 2 , x ∈ R .

To prove: d d x tan - 1 x = 1 1 + x 2 , x ∈ R Let y = tan - 1 x , x ∈ R ⇔ x = tan y , y ∈ - π 2 , π 2 Now, differentiate with respect to y , then we get: d x d y = sec 2 y ⇒ d y d x = 1 sec 2 y ⇒ d y d x = 1 1 + tan 2 y = 1 1 + x 2 Hence, proved.

Prove that: d d x cot - 1 x = - 1 1 + x 2 , x ∈ R .

To prove: d d x cot - 1 x = - 1 1 + x 2 , x ∈ R Let y = cot - 1 x , x ∈ R ⇔ x = cot y , y ∈ 0 , π Now, differentiate with respect to y , then we get: d x d y = - cosec 2 y ⇒ d y d x = 1 - cosec 2 y ⇒ d y d x = 1 - 1 + cot 2 y = - 1 1 + x 2 Hence, proved.

Prove that: d d x cosec - 1 x = - 1 | x | x 2 - 1 , | x | > 1 .

To prove: d d x cosec - 1 x = - 1 | x | x 2 - 1 , | x | > 1 Let y = cosec - 1 x , | x | > 1 , ie. x ∈ ( - ∞ , - 1 ) ∪ ( 1 , ∞ ) ⇔ x = cosec y , y ∈ 0 , π 2 ∪ π 2 , π Now, differentiate with respect to y , then we get: d x d y = - cosec y · cot y ⇒ d y d x = 1 - cosec y · cot y ⇒ d y d x = 1 - cosec y ± cosec 2 y - 1 ⇒ d y d x = - 1 ± x x 2 - 1 Where the sign before the radical depends on the sign of cot y Now, if x ∈ ( - ∞ , - 1 ) , then cosec y = x < - 1 , which implies that y ∈ π 2 , π .But then cot y < 0 Similarly, if x ∈ ( 1 , ∞ ) , then y ∈ 0 , π 2 ,where cot y > 0 Thus, d y d x = - 1 - x x 2 - 1 , if x < - 1 - 1 x x 2 - 1 , if x > 1 Combining the two cases we may write d d x cosec - 1 x = - 1 | x | x 2 - 1 Hence, proved.

Find derivative of the following function: cot - 1 1 - x 2 x .

Let y = f x = cot - 1 1 - x 2 x Now, substitute x = sin θ ⇒ θ = sin - 1 x , then y = cot - 1 1 - sin 2 θ sin θ ⇒ y = cot - 1 cos θ sin θ ⇒ y = cot - 1 cos θ sin θ [Consider the angle is in first quadrant] ⇒ y = cot - 1 cot θ ⇒ y = θ ⇒ y = sin - 1 x Now, differentiate y with respect to x , then we get: d y d x = d d x sin - 1 x ⇒ d y d x = 1 1 - x 2 Hence, the derivative of cot - 1 1 - x 2 x is 1 1 - x 2 .

Find derivative of the following function: sin - 1 1 - x 1 + x .

Let y = f x = sin - 1 1 - x 1 + x Now, differentiate y with respect to x , then we get: f ' x = d y d x = d d x sin - 1 1 - x 1 + x ⇒ d y d x = 1 1 - 1 - x 1 + x 2 d d x 1 - x 1 + x ⇒ d y d x = 1 1 - 1 - x 1 + x 2 × 1 2 1 - x 1 + x × d d x 1 - x 1 + x ⇒ d y d x = 1 1 - 1 - x 1 + x 2 × 1 2 1 - x 1 + x × 1 + x d d x 1 - x - 1 - x d d x 1 + x 1 + x 2 [Using quotient rule of differentiation] ⇒ d y d x = 1 + x 2 1 + x - 1 + x × 1 + x 1 - x - 1 - x - 1 + x 1 + x 2 ⇒ d y d x = 1 2 2 x 1 - x × - 2 1 + x ⇒ d y d x = - 1 2 x ( 1 - x ) ( 1 + x ) Hence, the derivative of sin - 1 1 - x 1 + x is - 1 2 x ( 1 - x ) ( 1 + x ) .

E M B I B E

Elements of Mathematics Class 12>Chapter 7 - Continuity and Differentiability>EXERCISE 7(d)

Odisha Board Solutions for Chapter: Continuity and Differentiability, Exercise 4: EXERCISE 7(d)

Author:Odisha Board

Odisha Board Mathematics Solutions for Exercise - Odisha Board Solutions for Chapter: Continuity and Differentiability, Exercise 4: EXERCISE 7(d)

Attempt the practice questions on Chapter 7: Continuity and Differentiability, Exercise 4: EXERCISE 7(d) with hints and solutions to strengthen your understanding. Elements of Mathematics Class 12 solutions are prepared by Experienced Embibe Experts.