Differentiate the following w.r.t x : sin - 1 { 1 - x 2 }

We have, sin - 1 { 1 - x 2 } Putting x = cos θ in the given, θ = cos - 1 x .........(i) sin - 1 { 1 - x 2 } = sin - 1 { 1 - cos 2 θ } = sin - 1 ( sin 2 θ ) = sin - 1 ( sin θ ) = θ = cos - 1 x [From (i)] On differentiating, d d x sin - 1 { 1 - x 2 } = d cos - 1 x d x = - 1 1 - x 2 .

Differentiate the following w.r.t x : sin - 1 1 - x 2

We have, sin - 1 1 - x 2 Put x = cos θ in the given expression, θ = cos - 1 x … ( i ) = sin - 1 1 - cos θ 2 = sin - 1 sin 2 θ 2 = sin - 1 sin θ 2 = θ 2 Now, sin - 1 ( 1 - x 2 ) = θ 2 And from (i), we have θ = cos - 1 x ⇒ sin - 1 ( 1 - x 2 ) = cos - 1 x 2 Thus, ⇒ d d x sin - 1 ( 1 - x 2 ) = 1 2 d cos - 1 x d x = - 1 2 1 - x 2 .

Differentiate the following w.r.t x : cos - 1 { 1 - x 2 }

We have, cos - 1 ( 1 - x 2 ) x = sin θ ⇒ θ = sin - 1 x … . . . . . i Put x = sin θ in the given expression, cos - 1 ( 1 - x 2 ) = cos - 1 ( 1 - sin 2 θ ) = cos - 1 ( cos θ ) = θ Now, cos - 1 ( 1 - x 2 ) = θ = sin - 1 x [From (i)] Thus, d d x cos - 1 ( 1 - x 2 ) = d sin - 1 x d x = 1 1 - x 2 .

Differentiate the following w.r.t x : sin - 1 { 2 x 1 - x 2 }

We have, sin - 1 ( 2 x 1 - x 2 ) x = sin θ ⇒ θ = sin - 1 x … . . . . ( i ) Put x = sin θ in the given expression, ⇒ sin - 1 ( 2 x 1 - x 2 ) = sin - 1 ( 2 sin θ 1 - ( sin θ ) 2 ) = sin - 1 ( 2 sin θ 1 - sin 2 θ ) = sin - 1 ( 2 sin θ cos θ ) = sin - 1 ( sin 2 θ ) = 2 θ Now, sin - 1 ( 2 x 1 - x 2 ) = 2 sin - 1 x [From (i)] Now Differentiating d d x sin - 1 ( 2 x 1 - x 2 ) = 2 d sin - 1 x d x = 2 1 - x 2