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

Let y = f x = cos - 1 1 + x 2 Now, differentiate y with respect to x , then we get: f ' x = d y d x = d d x cos - 1 1 + x 2 ⇒ d y d x = - 1 1 - 1 + x 2 2 d d x 1 + x 2 [Using chain rule of differentiation] ⇒ d y d x = - 1 1 - 1 + x 2 2 × 1 2 1 + x 2 d d x 1 2 + x 2 [Again Using chain rule of differentiation] ⇒ d y d x = - 1 1 - 1 + x 2 2 × 1 2 1 + x 2 × 1 2 ⇒ d y d x = - 1 1 - 1 + x 2 × 1 4 1 + x 2 ⇒ d y d x = - 2 1 - x × 2 4 1 + x ⇒ d y d x = - 1 2 1 - x 2 Hence, the derivative of cos - 1 1 + x 2 is - 1 2 1 - x 2 .

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

Let y = f x = cos - 1 x - 1 x x + 1 x ⇒ y = f x = cos - 1 x 2 - 1 x 2 + 1 ⇒ y = π - cos - 1 1 - x 2 1 + x 2 [Using cos - 1 - x = π - cos - 1 x ; x ∈ - 1 , 1 ] Now, substitute x = tan θ , so that θ = tan - 1 x Therefore, y = π - cos - 1 1 - tan 2 θ 1 + tan 2 θ ⇒ y = π - cos - 1 cos 2 θ ⇒ y = π - 2 θ ⇒ y = π - 2 tan - 1 x Now, differentiate y with respect to x , then we get: f ' x = d y d x = d d x π - 2 tan - 1 x ⇒ d y d x = 0 - 2 1 + x 2 = - 2 1 + x 2 Hence, the derivative of cos - 1 x - 1 x x + 1 x is - 2 1 + x 2 .

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 ) .